C.B.S.E Study Material 2 (Unit-1)

CURRENT ELECTRICITY

1. A wire of resistivity is stretched to double of its length. What will be its new resistivity?Ans. The resistivity remains the same as it does not depend upon the length of the wire.2. A student obtains resistances 3, 4, 12 and 16 ohm using only two metallic resistors either

separately or joined together. What is the value of resistance of each of the resistors?Ans. 4 and 12 ohm.3. If the length of a wire conductor is doubled by stretching it, keeping the potential difference

across it constant, by what factor does the drift speed of the electrons, change.

Ans. The drift velocity is given by the expression vd = Now if the length is doubled the drift

velocity becomes half.4. How does the electrical conductivity of an electrolyte change with decrease in temperature?Ans. The electrical conductivity decreases with decrease in temperature.5. A carbon resistor is marked in red, yellow and orange bands. What is the approximate

resistance of the resistor?Ans. 24 103 ohm 20% 6. What is the effect of temperature on the relaxation time of electrons in a metal?Ans. The relaxation time of electrons decreases with the rise in temperature of the metal.7. How does the resistance of an ohmic conductor depend on the applied voltage?Ans. It is independent of the applied voltage.

8. The figure shows the V – I graph for a parallel and series combination of two resistors A and B. Which line represents the parallel combination?

Ans. For the same potential the current is less in series combination than parallel combination. Therefore from the graph it is apparent that for the same potential current is less in A. Therefore B is the parallel combination.

9. What is the effect of heating a conductor on the drift velocity of free electrons?Ans. Heating decreases the relaxation time, therefore the drift velocity of electrons will decrease.10. If potential difference V applied across a conductor is increased to 2V. How will the drift

velocity of the electrons change?Ans. The drift velocity is given by the expression.

vd =

Therefore, if potential is doubled drift velocity is also doubled.11. A carbon resistor is marked in coloured bands of red, black, orange and silver. What is the

resistance and tolerance value of the resistor?Ans. 20 103 10%12. Which physical quantity does the voltage versus current graph for a metallic conductor depict?

Give its SI unit.Ans. It represents resistance. It is measured in ohm.

13. What are superconductors?Ans. These are the substances which lose their electrical resistance below a certain temperature

called critical temperature.

C.B.S.E Study Material

14. The given graph shows the variation of resistance of mercury in the temperature range 0 < T < 4 K. Name the phenomenon shows by the graph.

Ans. Superconductivity.

15. Two electric bulbs A and B are marked 220 V, 60 W and 220 V, 100 W respectively. Which one of the two has greater resistance?

Ans. The resistance of a bulb is given by the expression . For the same voltage the bulb having

a smaller power has more resistance. Therefore the 60 W, 220 V bulb has a greater resistance.16. The applied potential difference across a given resistor is altered so that the heat produced per

second increases by a factor of 16. By what factor the applied potential difference changes.

Ans. The heat produced across a resistor is given by H = It is directly proportional to the

square of potential. Therefore is the heat becomes 16 times the voltage must have been increased 4 times.

17. Two bulbs are marked 60 W, 220 V and 100 W, 220 V. These are connected in parallel to 220 V mains. Which one out of the two will glow brighter?

Ans. In parallel combination, the bulb having more power glows more. Therefore the bulb marked 100 W, 220 V glows brighter.

18. A heater is joined in parallel with a 60 W bulb is connected to the mains. If the 60 W bulb is replaced by a 100 W bulb, will the rate of heat produced by the heater be more, less or remain the same?

Ans. The rate of heat produced in the heater will be same as the two are connected in parallel.19. What is the larger voltage you can safely put across a resistor marked 98 ohm –0.5 W?Ans. Using the expression

V = 20. Two conductors, one having resistance R and another 2R are connected in turn across a dc

source. If the rate of heat produced in the two conductors is Q1 and Q2 respectively, what is the value of q1/q2?

Ans. We know that Q = therefore

21. State the condition in which terminal voltage across a secondary cell is equal to its emf.Ans. When the cell is in an open circuit i.e., when no current is being drawn from the cell.22. Under what condition can we draw maximum current from a secondary cell?Ans. When external resistance present in the circuit is zero.23. Define drift velocity of electrons.Ans. The mean velocity acquired by electrons in a conductor when an external electric field is

applied on it.24. Define conductance. What is its unit?

Ans. Reciprocal of resistance is called conductance. Thus, conductance is C = . SI unit of

conductance is siemen (S).25. Two wires A and B are of same metal, have the same area of cross-section and have their

lengths in the ratio 2 : 1.What will be the ratio of currents flowing through them respectively, when the same potential difference is applied across length of each of them?

Ans.

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26. V – I graph for a given metallic wire at two temperatures are shown, which of these is for a higher temperature?

Ans. At higher temperature resistance of metallic wire is more or its conductance is low. Hence, graph (2) is at a higher temperature i.e., T2 > T1.

27. Two wires of equal lengths, one of copper and the other of manganin have the same resistance. Which wire is thicker?

Ans. In accordance with the formula R = L/A, manganin wire will be thicker because its resistivity is more.

28. Two wires of equal cross-sectional area, one of copper and the other of manganin have the same resistance. Which one will be longer?

Ans. The copper wire will be longer because resistivity of copper is less than that of manganin.29. Of metal and alloys, which have greater value of temperature coefficient of resistance?Ans. Pure metals have a greater value of temperature coefficient of resistance than their alloys.30. Of copper and nichrome, which one has possibly larger value of temperature coefficient of

resistance.Ans. Copper31. How does the heat produced in a resistor depend on its resistance when (i) a constant current is

passed through it, (ii) a constant potential difference is applied across its ends?Ans. (i) For I = constant,

heat produced H R(ii) For constant potential difference V.

heat produced H

32. What happens to the power dissipation if the value of electric current passing through a conductor of constant resistance is doubled?

Ans. In accordance with formula P = I2R, the power dissipation becomes 4 times if the current passing through given resistance is doubled.

33. Sketch a graph showing variation of resistivity of carbon with temperature.

Ans. Carbon is a non-conductor, therefore its resistivity decreases with increase in temperature as shown in the figure below.

34. State kirchoff’s laws for electrical circuits. Derive the balance condition for a Wheatstone bridge using these laws.

Ans. Kirchoff’s first rule: The algebraic sum of currents into any junction is zero. That isI = 0

Kirchoff’s second rule: The algebraic sum of the potential differences in any loop, including those with emfs and those of resistive elements, must equal zero.Consider the diagram as shown below.Let I be the total current in the circuit, then by Kirchoff’s junction rule the following current

distribution can be made. Let current I1 flow through arm AB, such that current I – I1 flow through the arm AD. At junction B the current I1 gets divided, Ig flows through the galvanometer and current I1 – Ig flows through the arm BC. At junction D currents I – I1 and Ig add up such that current I – I1 + Ig flows through the arm DC. At junction C the currents from arms BC and DC combine such that the current in the circuit is again I. Now applying Kirchoff’s loop rule to the closed loop ABDA we have

– I1P – IgG + (I – I1) R = 0 ….(1)Again applying Kirchoff’s loop rule to closed loop BCDB we have

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– (I1 – Ig)Q + (I – I1 + Ig) X + IgG = 0 ….(2)The value of P, Q, R and X are so adjusted that the

galvanometer gives zero deflection. This means that both B and D will be at the same potential and hence no current will flow through the galvanometer. i.e., Ig = 0. In this situation the Wheatstone bridge is said to be balanced. Putting Ig = 0 in equations 1 and 2 we have

– I1P + (I – I1) R = 0 ….(3)and – I1Q + (I – I1) X = 0 ….(4)Rewriting the above two equations we have

I1P = (I – I1) R ….(5)and I1Q = (I – I1) X ….(6)dividing the above equations we have

….(7)

The above expression gives the condition for the balance of a Wheatstone bridge.35. How will you use a meter bridge to measure an unknown resistance? Draw the necessary

circuit diagram. Explain the principle of the experiment. Give the formula used.Ans. The experiment is based on Wheatstone bridge principle.

The circuit diagram is as shown below.The connections are made as shown in figure. A resistance R is introduced from the resistance box and the key K is closed. The jockey is moved on the wire to the point where there is no deflection in the galvanometer. In such a case points B and D are at the same potential. The point B is called the “null” point.Let in this position AB = L cm and BC = (100 – L) cm. Therefore resistance of AB i.e. P L and resistance of BC i.e. Q (100 – L) hence

….(1)

In the balanced state by the Wheatstone bridge Principle we have

….(2)

Substituting equation (1) in equation (2) we have

….(3)

Rewriting equation (3) we have

X = R

36. With the help of a circuit diagram explain how the internal resistance of a cell can be determined by using a potentiometer. Write the formula used.

Ans. The circuit shown below is used to measure the internal resistance of a cell.

The formula used is r = R

First, the key K1 is inserted and key K2 is not inserted. In this state no current is drawn from the cell. The jockey is moved on the wire to find the balance point. Let the balancing length be L1. This length balances the emf E of the cell. Therefore the potentiometer principle we have

E L1 ….(1)

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Now key K2 is also closed. A suitable resistance R is introduced in the circuit with the help of the resistance box (R.B). In this state current is drawn from the cell. The jockey is moved on the wire to find the balance point. Let the balancing length be L2. This length balances the terminal potential difference (V) of the cell. Therefore by potentiometer principle we have

V L2 ….(2)From equations 1 and 2 we have,

….(3)

The internal resistance of the cell is given by the expression

….(4)

Substituting equation 3 in 4 we have

This gives the value of the internal resistances of the cell 37. A potential difference V is applied across a conductor of length L and diameter D. How are the

resistance R and electric field E of the conductor affected, when in turn (i) v is halved (ii) L is halved and (iii) D is doubled. Justify your answer in each case.

Ans. The table below gives the variationV is halved L is halved D is doubled

No change R is halved R becomes on-fourth

E is halved E is doubled No change

38. Define resistivity of a substance. Give its unit. How does it vary with temperature in (i) good conductor (ii) ionic conductors and (iii) semiconductors?

Ans. It is defined as the resistance of a conductor of unit length and unit cross sectional area. In SI it is measured in ohm-m

(i) It increases with increase in temperature for conductors.(ii) It decreases with increase in temperature for ionic conductors.(iii) It decreases with increase in temperature for semiconductors.

39. Explain the principle on which the working of a potentiometer is based. Why is the use of a potentiometer preferred over that of a voltmeter for the measurement of emf of a cell?

Ans. It is based on the principle that if a wire of uniform area of cross section carries a constant current the potential drop across any portion of the wire is directly proportional to the length of that portion of the wire.Consider a wire of uniform area of cross section A. Let I be the current through the wire. Then by Ohm’s law

V = IR ….(1)

R = ….(2)

Substituting equation (2) in (1) we have

V = ….(3)

Since I, A and are constant therefore we have V LA potentiometer is preferred over a voltmeter, because it is based on the null method i.e., it does not draw any current while measuring the potential.

40. The following circuit shows the use of potentiometer to measure the internal resistance of a cell (i) when the key K is

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open, how does the balance point change, if the current from the driver cell decreases. (ii) When the key K is closed, how does the balance point change if R is increased, keeping the current from the driver cell constant.

Ans. (i) When the key K is open, a decrease in the current from the driver cell decreases the potential drop across the potentiometer wire. Therefore to balance the same emf again more length of the wire will be required. Thus the balance point will shift towards point B.(ii) When key K is closed and R is increased, it increases the terminal potential difference of the cell. Thus to balance the new terminal potential difference more length of the wire will be required. Thus the balance point will shift towards point B.

41. A wire of 10 ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?

Ans. (i) There is no change in the resistivity (ii) The new resistance will become n2 times the old resistance. Therefore new resistance becomes R = (3)2 10 = 90 ohm.

42. With the help of a circuit diagram, explain how the emfs of two primary cells can be compared using a potentiometer. State the formula used. How can the sensitivity of a potentiometer be increased?

Ans. The circuit diagram for comparing the emfs of two cells is given below.

First, the key K1 is inserted. This brings the cell of emf E1 in the circuit. The jockey is moved on the wire to obtain a balance point i.e. a point on the wire where the galvanometer gives zero deflection. Let the balancing length be L1. Therefore by the potentiometer principle we have

E1 L1 ….(1)Now, the key k2 is inserted. This brings the cell of emf E2 in the circuit. The jockey is again moved on the wire to obtain the balance point. Let the balancing length be L2. Then by potentiometer principle we have

E2 L2 ….(2)dividing equation (1) by (2) we have

….(3)

Knowing the values of L1 and L2 the emf’s can be compared.

Sensitivity can be increased by increasing the length of the potentiometer wire or decreasing the current through the wire.

43. Define the term resistivity of a conductor. Give its SI unit. Show that the resistance R of

conductor is given by where the symbols have their usual meaning.

Ans. It is defined as the resistance of the material of the material of the conductor of unit length and unit area of cross section. It is measured in ohm metre.Let vd be the drift velocity of the electrons, then its relation with electric field is

vd = ….(1)

Let V be the potential difference applied across the two ends of a conductor of length L and area of cross-section A, Then

E = ….(2)

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substituting equation (1) in (2) we have

vd = ….(3)

Also I = neAvd ….(4)substituting equation (3) in equation (4) we have

I = neA ….(5)

Comparing with Ohm’s law i.e.

I = ….(6)

We have ….(7)

44. Are the paths of electrons straight lines between successive collisions (with positive ions of the metals) in the (i) absence of electric field (ii) presence of electric field? Establish a relation between drift velocity vd of an electron in a conductor of cross-section A carrying current I and concentration ‘n’ of free electrons per unit volume of conductor. Hence obtain the relation between current density and drift velocity.

Ans. (i) In the absence of electric field the path of electrons is a straight line.(ii) In the presence of an electric field the path of the electrons is parabolic.

Suppose there are n electrons per unit volume. Here n is called the number density of electrons. Assume that all electrons move with the same drift velocity vd. In a time interval dt, each electron moves a distance vddt. Now the volume of the cylinder covered by the electrons in time dt is

V = A vddt ….(1)and the number of electrons in this volume are

N = nV = nA vddt ….(2)If e is the charge on the electron, then chare flowing through the conductor in small time dt is

dQ = e(nA vddt0 ….(3)hence the current through the conductor is

I = = nAevd

Since n, e and A are constant, therefore current flowing in a conductor is directly proportional to the drift velocity.

The current per unit cross sectional area is called current density (J). Therefore J = = nevd

45. Briefly explain how the drift velocity of electrons in a metallic conductor varies when (i) the temperature of the conductor is increased, and (ii) applied potential difference is decreased, keeping temperature constant.

Ans. The drift velocity is given by the expression vd =

(i) With an increase in the temperature of the conductor, the average relaxation time decreases. Since drift velocity is directly proportional to the average relaxation time, therefore it decreases with increase in temperature.

(ii) Drift velocity is directly proportional to the potential difference, therefore an increase in the value of potential difference increase the value of drift velocity.

46. The variation of potential difference with length in case of two potentiometers A and B is shown in figure. Which of the two is preferred to find the emf of a cell? Give reason for your answer?

Ans. A potentiometer having a small value of potential gradient is more accurate thant the potentiometer having a large value of potential gradient

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(V/L). As seen from the graph potentiometer B has a smaller value of V/L i.e., potential gradient, therefore it will be preferred gradient, therefore it will be preferred over potentiometer A.47. Explain how the resistivity of a conductor depends upon (i) number density ‘n’ of free

electrons, and (ii) relaxation time .

Ans. The resistivity of a conductor is given by the relation =

(i) It depends inversely on the number density of the electrons. An increase in number density leads to a decrease in the resistivity.

(ii) It depends inversely on the average relaxation time of the electrons. An increase in average relaxation time leads to a decrease in the resistivity.

48. Explain with the help of a graph the variation of conductivity with temperature for a metallic conductor.

Ans. The conductivity of a metallic conductor depends inversely on its absolute temperature. Therefore the graph of the above situation is shown below

49. State Joule’s law of heating by electric current. Name the materials used for making (i) standard resistors (ii) heater element.

Ans. Joule found that when current is passed through a conductor the heat produced across it is(i) Directly proportional to the square of current through the conductor i.e.

H I2 …..(1)(ii) Directly proportional to the resistance of the conductor i.e.

H R …..(2)(iii) Directly proportional to the time for which the current is passed i.e.

H t …..(3)Combining the above three equations we have

H I2Rt(a) For making standard resistors manganin, is used.(b) For heater element nichrome is used.

50. Name the two factors on which the resistivity of a given material depends. A carbon resistor has a value of 62 k with a tolerance of 5%. Give the colour code of the resistor.

Ans. Resistivity depends upon (i) The number density of electrons and (ii) The average relaxation time.The colour code of the cabon resistor is blue, red, orange and gold.

51. Draw a circuit diagram for a metre bridge to determine the unknown resistance of a resistor. Obtain the balance condition for a metre bridge. Why are the connections between the resistors of a metre bridge made of thick copper strips? Find the shift in the balance point of a metre bridge, when the two resistors in its two gaps, are interchanged. Take the values of the two resistors as R and S.Ans. A metre bridge is a practical form of Wheatstone bridge and the circuit diagram is shown in figure below.

Connections between resistors are made of thick copper strips so that the resistances of the resistors remain unchanged.

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Let initially with resistance R in left gap and S in right gap, null point is obtained at a length L then

….(i)

If R and S are interchanged then null point will be obtained at L’ such that

….(ii)

A comparison of (i) and (ii) shows that L’ = 100 – L. Therefore shift in position of balance point.

L’ – L = (100 – L) – L = (100 – 2L)cm52. A potentiometer wire has a length L and resistance R0. It is

connected to a battery and a resistance combination as shown. obtain an expression for the potential drop per unit length of the potentiometer wire. What is the maximum emf of a ‘test cell’ for which one can get a ‘balance point’ on this potentiometer wire? What precautions should one take, while connecting this ‘test cell’ in the circuit?

Ans. Here total resistance of potentiometer circuit

R = R0 +

Therefore current in the circuit

I =

Therefore, total potential difference across the ends of potentiometer wire AB

V = I R0 = I =

Therefore, potential drop per unit length of the potentiometer wire

k =

Maximum emf of a test cell, for which one can get a balance point on this potentiometer wire, should be less than V.Precaution: Positive terminal of the test cell must be connected to end A of the potentiometer i.e., +ve terminals of battery and test cell must be connected at the same end.

53. A carbon resistor has the following colour bands drawn on it. Find it value. Define the term resistivity of a material. Give its SI unit.

Ans. 64 101 5%Resistivity is defined as the resistance of a conductor per unit length per unit area of cross section. It is measured in ohm-m.

54. How does the resistivity of (i) conductor and (ii) semiconductor vary with temperature? Give reason for each case.

Ans. The resistivity of conductor is given by the expression = The resistivity of conductors

increases with the increase in the temperature. This is because increase in temperature increases the amplitude of vibration of lattice ions, which in turn increases the number of collisions. This decreases the average relaxation time. Since resistance of a conductor is inversely proportional to the average relaxation time, therefore the resistance increases.

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(ii) Semiconductors: In case of semiconductors, resistance decreases with the increases in the temperature. An increase in temperature breaks a large number of covalent bonds. This results in a large number of majority carriers, which in turn increase the current.

55. When two known resistances R and S are connected in the left and right gaps of a metre bridge, the balance point in found of a distance L1 from the zero end of the metre bridge wire. An unknown resistance X is now connected in parallel to the resistance S and the balance point is now found at a distance L2 from the zero end of the metre bridge wire. Obtain a formula for X in terms of L1, L2 and S.

Ans. The arrangement is as shown.Applying the formula for balanced metre bridge in first case, we have

….(i)In second case in right gap S and X are arranged in parallel and have a net resistance of

hence, now, we have

….(ii)

Dividing (i) by (ii) we have

On simplification we get

X =

56. Define the term temperature coefficient of resistivity. Show graphically the variation of resistivity with temperature for nichrome/copper.

Ans. The temperature coefficient of resistance of a conductor is defined as the ratio of the increase in resistance per unit original resistance per unit rise in temperature.

Variation of resistivity of nichrome with temperature.

Variation of resistivity of copper with temperature.

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57. You are given ‘n’ resistors each of resistance ‘r’. These are first connected to get minimum possible resistance. In the second case it is connected to get maximum possible resistance. Compute the ratio between the minimum and the maximum values of resistance so obtained.

Ans. For obtaining minimum possible resistance the resistors are connected in parallel. Therefore, in parallel combination the total resistance is

or RP =

For obtaining maximum possible resistance the resistors are connected in series. Therefore, in series combination the total resistance is

RS = r + r + … n times = n rTherefore ratio of minimum to maximum resistance is

58. Draw a circuit diagram using a metre bridge and write the necessary mathematical relation used to determine the value of an unknown resistance. Why cannot such an arrangement be used for measuring very low resistances?

Ans. The circuit diagram is a shown below.

The necessary formula is X =

The bridge is also unsuitable to measure very small resistances, this is because in this case all the other resistances should also be small to ensure the sensitivity of the bridge which would require a galvanometer of very low resistance which itself would be very insensitive. In addition to this, the contract and the lead resistance will also be comparable with the resistance to be measured.

59. A resistance of R draws current from a potentiometer. The potentiometer has a total resistance R0 as shown in fig below. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Ans. While the slide is in the middle of the potentiometer only half of its resistance (R/2) will be between the points A and B. Hence, the total resistance between A and B, say, R1 will be given by the following expression;

or

The total resistance between A and C will be sum of resistance between A and B and B and C. i.e., R1 + R0/2Therefore, the current flowing through the potentiometer will be

I =

The voltage V taken form the potentiometer will be the product of current I and resistance R1,

V1 = IR1 =

Substituting for R1 and solving, we have

V1 =

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60. Explain how electron mobility changes for a good conductor when (i) the temperature of the conductor is decreased at constant potential difference and (ii) applied potential difference is doubled at constant temperature.

Ans. The mobility of a conductor is given by the expression = where L is the

length of the conductor.(i) When temperature of a conductor is decreased at constant potential difference the average

relaxation time increases. Since mobility is directly proportional to the average relaxation time, therefore mobility increases.

(ii) When the potential difference across the conductor is doubled at constant temperature, the mobility becomes half.

61. A parallel combination of 4 cells of identical emf E, internal resistance r, are connected in series to a variable resistor. The following graph shows the variation of terminal voltage of the combination with the current output:(i) What is the emf of each cell used?(ii) For what current from the cells, does maximum power

dissipation occur in the circuit?Ans. (i) From the graph we find that when I = 0, V = 5.6 V, therefore emf of each cell is = 5.6 V

(ii) Now slope of the graph gives the total resistance of the combination, therefore

R +

or 4R + r = 11.2Power dissipated is maximum when internal resistance is equal to the external resistance i.e., R = r/4 or 4R = rTherefore,we have r + r = 11.2 or r = 5.6 Hence current in the circuit for maximum power dissipation is

I =

62. For the potentiometer circuit shown in the given figure, points X and Y represent the two terminals of an unknown emf E’. A student observed that when the jockey is moved from the end A to the end B of the potentiometer wire, the deflection in the galvanometer remains in the same direction. What may be the two possible faults in the circuit that could result in this observation?If the galvanometer deflection at the end B is (i) more, (ii) less, than that at the end A, which of the two faults, listed above, would be there in the circuit?Give reasons in support of your answer in each case.

Ans. (a) The two possible faults are(i) The negative terminal of E’ is connected to point X. In other words the positive

terminals of E and E’ are not connected to the same point.(ii) The emf of the auxiliary battery E is less than the emf of the cell E’ i.e., E’ > E

(b) If galvanometer shows more deflection at BCause: Negative terminal of E’ is connected to point X.Reason: Potential difference across wire and the cell E’ sends current through the galvanometer in the same sense and the galvanometer deflection increases with increase in the length of the wire.(c) If galvanometer shows less deflection at B

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Cause: The emf of the auxiliary battery E is less than the emf of the cell E’ i.e., E’ > EReason: Potential difference across the potentiometer wire opposes E’ and the current in the galvanometer decreases with an increase in the length of the wire.

63. What is drift velocity? Derive an expression for it.Ans. In an ordinary metal such as copper or aluminium, some of the electrons move randomly in all

directions with speeds of the order of 106 ms–1. The electrons nonetheless, do not escape from the conducting material, because they are attracted to the +ve ions of the material. The motion of the electrons is random so there is no net flow of charge in any direction and hence no current. Consequently the average thermal velocity of the electrons is zero. If there are n electrons in a conductor having thermal velocities u1, u2, u3, ……un then the average thermal velocity of the electrons is

…..(1)

Suppose a steady electric field E is established across the ends of the conductor. This subjects the electrons inside the conductor, to an electric force given by F = –eE …..(2)This force causes a steady acceleration of the electrons opposite to the direction of the electric field. The acceleration being given by

a = ….(3)

where m is the mass of the electron. Due to this acceleration the electron, in addition to the random motion within a conductor, acquired a very slow net motion or drift opposite to the direction of the electric field. This motion of the electrons is described in terms of the drift velocity (vd) of the electrons. This additional acquired velocity is lost in the next collision. Suppose that the electron remains accelerated for a time t called the relaxation time of the electron. Let v1, v2, v3,…… vn be the velocities acquired by the electrons due to this acceleration. Therefore,v1 = u1 + at1. v2 = u2 + at2, v3 = u3 + at3 ………… and vn = un + atn respectively ….(4)Now drift velocity (vd) is defined as the average velocity with which the free electrons get drifted inside a conductor under the effect of the electric field, opposite to the direction of the field.Therefore

vd = ….(5)

or vd = ….(6)

or vd = ….(7)

using equation (1) we have vd = a ….(8)

here = is called average relaxation time. Substituting equation (3) in

equation (8) we have

vd =

Since E = –V/L, therefore we have

Vd = ….(9)

This gives a relation between drift velocity and electric field and drift velocity and electric potential.

64. Define the term ‘resistivity’ and write its S.I. unit. Derive the expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.

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Ans. It is defined as the resistance of a conductor of unit length and unit cross sectional area. In SI it is measured in ohm-m. Let vd be the drift velocity of the electrons, then its relation with electric field is

vd = ….(1)

Let V be the potential difference applied across the two ends of a conductor of length L and area of cross-section A, then

E = ….(2)

substituting equation 1 in 2 we have

vd = ….(3)

Also I = neAvd ….(4)Substituting equation (3) in equation (4) we have

I = neA ….(5)

Comparing with Ohm’s “law” i.e.

I = ….(6)

We have R = ….(7)

Comparing with the equation R = we have =

65. Draw V – I graph for ohmic and non-ohmic materials. Give one example for each.Ans. Ohmic materials: Those materials in which Ohm’s law is followed are called ohmic materials.

For such circuits the V – I graph is a straight line passing through the origin. The reciprocal of the slope of the graph gives the resistance of the circuit e.g., metallic conductors.None-ohmic materials: Those materials in which Ohm’s “law” is not obeyed are called non-ohmic matierals. The V–I graph is a curve. e.g. electrolytes, semiconductors, thermionic values etc.

66. Write the mathematical relation between mobility and drift velocity of charge carriers in a conductor. Name the mobile charge carriers responsible for conduction of electric current in (i) an electrolyte (ii) and ionized gas.

Ans. Drift velocity vd = Mobility/Electric field or =

(i) The mobile charge carriers in an electrolyte are positive and negative ions called cations and anions.

(ii) In ionized gas the mobile charge carriers are electrons are positively charged ions.67. Define the term “mobility” for a charge carrier and state its SI unit.

Name the mobile charge carriers in (i) an electrolyte (ii) a semiconductor (iii) an ionized gas.

Ans. Mobility is defined as the ratio of drift velocity of the charge to the applied electric field.

14


(i) Anions and cations(ii) Electrons and holes(iii) Free electrons

68. State the principle of the device used for comparing the emf of two cells. Draw the relevant circuit diagram and explain how the emf’s of the two cells are compared. How can sensitivity of such a device be increased? In what way is this method different from the one using voltmeter for comparing the emf’s?

Ans. Principle: It is based on the principle that the potential across any section of the wire is directly proportional to the length of that section of the wire.The circuit diagram for comparing the emf’s of two cells is given below:

First, the key K1 is inserted. This brings the cell of emf E1 in the circuit. The jockey is moved on the wire to obtain a balance point i.e. a point on the wire where the galvanometer givens zero deflection. Let the balancing length be L1. Therefore by the potentiometer principle we have

E1 L1 ….(1)Now, the key K2 is inserted. This bridges the cell of emf E2 in the circuit. The jockey is again moved on the wire to obtain the balance point. Let the balancing length be L2. Then by potentiometer principle we have

E2 L2 ….(2)Dividing equation (1) by (2) we have

…(3)

Knowing the values of L1 and L2 the emf’s can be compared.The sensitivity of the potentiometer can be increased by increasing the length of the potentiometer wire.The potentiometer method is a null method i.e. no current is drawn form the cell when its emf is being measured. It measures emf more accurately than a voltmeter.

69. State Kirchoff’s rules for an electrical network. Explain their use by drawing a simple circuit diagram. Find the total current I, supplied to an external resistor R, connected as shown, across a parallel combination of three cells of equal emf E and same internal resistance r.

Ans. Kirchoff’s first rule: The algebraic sum of currents into any junction is zero.

15


Kirchoff’s second rule: The algebraic sum of the potential differences in any loop, including those with emfs and those of resistive elements, must equal zero.A simple circuit diagram shows the distribution of current in accordance with the first rule.In accordance with the second law the equations for two loops is as written below.

–I1P + (I – I1) R = 0 ….(3)and –I1Q + (I – I1) X = 0 ….(4)These tow equations will be solved fro the unknown.Since the cells are in parallel, therefore net emf of the circuit is E.Total internal resistance of the circuit is r/3. Hence total resistance of the circuit is

RT = R + r/3Hence current in the circuit is

I =

70. The given figure shows a network of resistances R1, R2, R3 and R4.Using Kirchoff’s laws, establish the balance condition for the network.

Ans. Consider the diagram as shown below.Now applying Kirchoff’s loop rule to the closed loop ABDA we have–I1R1 – IgG + (I – I1) R3 = 0 ….(1)Again applying Kirchoff’s loop rule to closed loop BCDA we have–(I1 – Ig) R2 + (I – I1 + Ig) R4 + IgG = 0 ….(2)The value of R1, R2, R3 and R4 are so adjusted that the galvanometer gives zero deflection. This means that both B and D will be at the same potential and hence no current will flow through the galvanometer. i.e., Ig = 0. In this situation the Wheatstone bridge is said to be balanced. Putting Ig = 0 in equations 1 and 2 we have

–I1R1 + (I – I1) R3 = 0 ….(3)and –I1R2 + (I – I1) R4 = 0 ….(4)Rewriting the above two equations we have

I1R1 = (I – I1) R3 ….(5)and I1R2 = (I – I1) R4 ….(6)dividing the above equation we have

….(7)

The above expression gives the condition for the balance of a Wheatstone bridge.71. Define the term current density of a metallic conductor. Deduce the relation connecting current

density (J) and the conductivity () of the conductor, when an electric field E, is applied to it.Ans. Current density is defined as the current flowing per unit area of the conductor.

Mathematically, current density is given by the expression J =

But I = V/R and R =

Substituting in the above relation we have

16


J =

72. Under what condition is the heat produced in an electric circuit (i) directly proportional, (ii) inversely proportional to the resistance of the circuit?

Ans. (i) Heat produced in an electric circuit is directly proportional to resistance of the circuit when current flowing through it is kept constant, as

H = I2Rt, and when I is constant H R.(ii) Heat produced in electrical circuit is inversely proportional to the resistance of the circuit, when potential difference across it is kept constant as

H = and when V is constant H

ELECTROSTATICS 1. Define the term electric dipole moment. Is it a scalar or vector quantity?Ans. It is defined as the product of either charge and the distance between them. It is a vector

quantity.2. Two field lines never cross each other. Why?Ans. It is because at the point of intersection two tangents can be drawn. Thus there will be two

direction of electric field at that point which is not possible.3. How does the Coulomb force between two point charges depend upon the dielectric constant of

the intervening medium?Ans. It decreases with the introduction of dielectric. It becomes 1/K times the original value.4. Name the physical quantity which has joule per coulomb as its unit. Is it a scalar or a vector

quantity?Ans. The quantity is electric potential. It is a scalar quantity.5. Sketch a graph to show how a charge Q acquired by a capacitor of capacitance C, varies with

the increase in potential difference between the plates.Ans.

6. Write the SI unit of (i) Electric field intensity and (ii) Electric dipole moment.Ans. (i) volt per metre (ii) coulomb metre7. Define one coulomb charge.Ans. It is that much charge which when placed at a distance of 1 m in a vacuum from an identical

charge applies a force of 9 109 N. 8. How much work is done in moving a 500 C charge between two points on an equipotential

surface?Ans. Zero9. What does q1 + q2 = 0 signify in electrostatics?Ans. It signifies an electric dipole.10. What orientation of an electric dipole in a uniform electric field corresponds to its stable

equilibrium?Ans. When the dipole is aligned, parallel to the direction of the electric field.11. What is the work done in moving a 100 nC charge between two points 5 cm apart on an

equipotential surface?

17


Ans. Zero12. Name the physical quantity, which has, as its units, Nm2C–1. Is it a scalar or vector quantity?Ans. Electric flux; It is a scalar quantity.13. Name the physical quantity which has as its units, coulomb-metre. It is a scalar or vector

quantity?Ans. Electric dipole moments. It is a vector quantity.14. Name the physical quantity, which has, as its unit newton per coulomb. Is it a scalar or a vector

quantity?Ans. Electric field; it is a vector quantity.15. How does the force between two point charges change, if the dielectric constant of the medium

in which they are kept, increases.Ans. The force decreases.16. Two point charges +10 C and +20 C are separated by a distance ‘r’ in air. If an additional

charge of –8 C is given to each, by what factor does the force between the charges, change.Ans. Initial force is given by

Final force is given by

Ff =

Change in force

17. In a medium the force of attraction between two point electric charges, distance ‘d’ apart is F. what distance apart should these be kept in the same medium so that the force between them becomes 3F?

Ans. Let the new distance be d, since F therefore or d = or r = d

18. Sketch two equipotential surfaces for a uniform electric field.Ans.

The dotted lines show the equipotential surfaces.

19. Sketch three equipotential surfaces for a point charge.Ans.

20. In a parallel plate capacitor the capacitance increases from 4F to 80F, on introducing a dielectric medium between the plates. What is the dielectric constant of the medium?

Ans. The dielectric constant is given by K = 80/4 = 2021. What is the dielectric constant of the medium?Ans. It is the ratio of the force between two charges placed a certain distance apart in vacuum to

that placed same distance apart in the medium.

18


22. Draw lines of force to represent a uniform electric field.Ans.

23. Sketch the electric lines of force due to point charges (i) q > 0 and (ii) q < 0. Ans.

24. What kind of charges are produced on each when (i) a glass rod is rubbed with silk and (ii) an ebonite rod is rubbed with wood?

Ans. (i) Glass rod: positive charge, silk: negative charge(ii) Ebonite rod: negative charge, wool: positive charge

25. What is a conservative field?Ans. A field is said to be conservative if work done in the field does not depend upon the path

followed but depends only upon the initial and final positions.26. Find the value of electric field that would completely balance the weight of an electron.Ans. mg = eE

= 5.57 10–11 Vm–1

27. Define electric field at a point.Ans. Electric field at a point is defined as the force experienced by a unit positive charge placed at

that point. Mathematically we have

28. Is the electric potential necessarily zero at a place where the electric field is zero?Ans. No, it is not necessary. The electric field inside a hollow metallic conductor is zero but the

electric potential is not zero.29. Express dielectric constant in terms of capacitance of a capacitor.

Ans. It is given by the expression K = where C is the capacitance of the capacitor with

dielectric and C0 is the capacitance without dielectric.30. Find the electric field between two metal plates placed 3 mm apart and connected to a battery of

12V.

Ans. The electric field between the plates is given by E = = 4 103 Vm–1

31. Define electric potential at a point. Is it a vector or a scalar quantity?Ans. It is defined as the work done in moving a unit charge from infinity to that point without

acceleration or without change in kinetic energy.32. How much energy will be stored by a capacitor of 470F when charged with a battery of 20V?

Ans. U = CV2 = 470 10–6 (20)2

= 9.4 10–2 J33. Two charges, one +5C and the other –5C are placed 1mm apart. Calculate the electric dipole

moment of the system?Ans. p = q 2a = 5 10–6 10–3

19


= 5 10–9 Cm34. Write down the relation between field and electric potential at a point.

Abs. Electric field and electric potential are related as E =

35. Define capacitance of a capacitor.Ans. It is defined as the ability of a capacitor to store charge.36. Name any two basic properties of electric charge.Ans. (i) Quantisation of charge and (ii) Conservation of charge.37. Can electric potential at a point in space be zero while intensity of electric field at that point is

not zero?Ans. Yes, at a point on the equatorial line of a dipole, electric potential is zero but electric field is

not zero.38. In a parallel plate capacitor the potential difference of 102 V is maintained between the plates.

What will be the electric field at points A and B as shown in the figure below?

Ans. The electric field between the plates of a capacitor is uniform; therefore the electric field at points A and B will be same.

39. Two protons A and B are placed between the two plates of a parallel plate capacitor having a potential difference V as shown. Will these protons experience equal or unequal forces?

Ans. Both protons will experience the same force as electric field between the plates of the capacitor is same.

40. Sketch the field lines around a system of two equal and opposite point charges.Ans. The sketch is as shown below.

41. In an electric field an electron is kept freely. If the electron is replaced by a proton, what will be the relationship between the forces experience by them?

Ans. Magnitude of force will be same but direction will be reversed.42. Give the SI unit of electric field intensity. Is electric field intensity a scalar or a vector quantity?Ans. NC–1, vector quantity.43. Which physical quantity has unit newton (coulomb)–1? Is it a vector or scalar quantity?Ans. Electric field intensity, vector quantity.44. Two point charges of +3C each are 100cm apart. At what point on the line joining the charges

will the electric intensity be zero?Ans. At the mid-point of the line joining the two point charges.45. What is nature of symmetry of electric field due to a point charge?

Ans. Spherical symmetry as E .

46. Consider the situation shown in the fig given below. What are the signs of q1 and q2?

20

A

B

+ –

Fig. for Q.38, Q.39


Ans. q1 is negative and q2 is positive.47. What orientation of an electric dipole in a uniform electric field corresponds to its stable

equilibrium?Ans. When dipole has its dipole moment along the direction of electric field.48. Four charges of same magnitude and same sign are placed at the corners of a square, of each

side 0.1m. What is electric field intensity at the centre of the square?Ans. Zero.49. What is the net force on a dipole in a uniform electric field?Ans. Zero50. In figure given below, at which point electric field is maximum?

Ans. Electric filed is maximum at point C.51. Force between two point electric charges kept at a distance d apart in air is F. If these charges

are kept at the same distance in water, how does the force between them change?Ans. New force will become F/K, where K is the dielectric constant of water.

52. What would be the work done if a point charge +q, is taken from a point A to the point B on the circumference of a circle drawn with another point charge +q at the centre?

Ans. Zero.

53. If a point charge +q is taken first from A to C and then from C to B of a circle drawn with another point charge +q at centre, then along which path more work will be done?

Ans. As VA = VB hence work done WAC (=VC – VA) = WBC (VC – VB).Therefore WAC = –WCB

54. A uniform electric field E exists between two charged plates as shown in Fig. What would be the work done in moving a charge q along the closed rectangular path ABCDA?

Ans. Work done is zero because electric field is a conservative field and work done for describing a closed path in the conservative field is always zero.

21


55. Draw four equipotential surfaces for a point charge q > 0.Ans. The figure is as shown.

56. An uncharged insulated conductor A is brought near a charged insulated conductor B. What happens to charge and potential of B?

Ans. On bringing in uncharged conductor a near a charged conductor B, charges are induced on A as shown in fig. below. As a result of it, the potential of conductor B is slightly lowered by charge on it remains unchanged.

57. In a parallel plate capacitor the potential difference of 100 V is maintained between the plates. If distance between the plates be 5mm, what will be the electric field at points A and B?

Ans. Electric field at both points A and B is same having a value

E = 2 104 NC–1

58. On what factors does the capacitance of a parallel plate capacitor depend?(i) Area of plates, (ii) Separation between the plates, and(iii) Nature of dielectric medium between the plates.

59. Define the term ‘dielectric constant’ of a medium in terms of capacitance.Ans. Dielectric constant of a medium may be defined as the ratio of the capacitance of a given

parallel plate capacitor in the presence of dielectric to its capacitance in the absence of dielectric medium.

60. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Ans. On account of polarisation of dielectric.61. Why does a given capacitor store more charge at a given potential difference when a dielectric

is filled in between the plates?Ans. Because capacitance of capacitor increases on filling the dielectric medium in between the

plates.62. For the same quantity of charge will the potential difference across the plates of a capacitor

increase or decrease on inserting a dielectric in between?Ans. The potential difference decreases.63. If the plates of a charged capacitor be suddenly connected to each other by a wire, what will

happen?Ans. The capacitor is discharged immediately.64. Can you place a parallel plate capacitor of one farad capacity in your house?Ans. Ordinarily, it is not possible because surface area of such a capacitor will be extra large.65. Is there any conductor which can be given almost unlimited charge?Ans. The earth.66. The dielectric constant of a media is unity, what is its permittivity?Ans. We know that K = /0

22


Here K = 1,therefore = 0 = 8.854 10–12 C2N–1 m–2

67. Force of attraction between two point charges placed at a distance ed apart in a medium is F. What would be the distance apart in the same medium so that the force of attraction between them becomes one fourth its previous value?

Ans. 2d.

68. Sketch the lines of force due to two equal positive point charges placed near each other.Ans. The field lines are as shown below

69. What is the ratio of electric field intensities at any two points between the plates of a capacitor?Ans. The ratio is one, as the electric field is same at all points between the plates of a capacitor.70. An electric dipole of dipole moment 20 10–6 Cm is enclosed by a closed surface. What is the

net flux coming out of the surface?Ans. The net charge of a dipole is zero as it consists of two equal and opposite charges. Hence, the

net flux coming out of the surface is zero as the net charge enclosed is zero.71. Define the term ‘dielectric constant’ of a medium in terms of capacitance of a capacitor.Ans. Dielectric constant (K) of a medium is defined as the ratio of the capacitance with the dielectric as the medium between the two plates (C) of a capacitor to its capacitance with free space

(or vacuum) as the medium between its plates (C0). That is, K =

72. Two point electric charges of unknown magnitude and sign are place a distance’d apart. The electric field intensity is zero at a point, not between the charges but on the line joining them. Write two essential conditions for this to happen.

Ans. The electric field intensity may be zero at a point on the axial line of charges, if(i) The two charges are of opposite signs i.e., one charge is positive and the other is

negative.(ii) The magnitude of charge nearer the null-point under consideration should be smaller

than the magnitude of the other charge.73. What do you understand by conservation of charge? Give one example to illustrate.Ans. According to the law of conservation of electric charge we can neither create nor destroy the

electric charge and total electric charge of an isolated system must remain conserved. To illustrate the principle consider a glass rod and silk cloth. Initially both are uncharged and total charge of the system is zero. On rubbing glass rod with silk some positive charge is developed on glass rod and an equal amount of negative charge is developed on silk, so that net charge of the system is still zero.

74. An electric dipole is free to move in a uniform electric field (i) parallel to the field, and (ii) perpendicular to the field. Explain what happens?

Ans. (i) When an electric dipole is placed parallel to a uniform electric field, net force as well as net torque acting on the dipole is zero and, thus, the dipole remains in equilibrium.

23


(ii) When dipole is placed perpendicular to the field, a torque acts on it. Under its influence the dipole executes SHM about the direction of electric field and finally aligns itself along the direction of electric field.

75. A point charge ‘q’ is placed at O as shown in fig. Is VA – VB positive, negative or zero, if ‘q’ is a (i) positive, (ii) negative charge?

Ans. (i) If q is positive then VA – VB = WBA for unit positive test charge is positive. This is bcause work will have to be done against the fore of repulsion.

(ii) If q is negative then electric field is directed towards O and test charge will move itself under the attractive force and hence VA – VB = –ve.

76. What is an equipotential surface? Show that the electric field at a point on the surface of a charged conductor or just outside it is perpendicular to the surface?

Ans. An equipotential surface is that at every point of which electric potential is the same. By definition potential difference between two points is equal to the work done in carrying unit positive test charge from one point to another. As for an equipotential surface potential difference is zero hence no work is to be done in moving a test charge from one point to another of an equipotential surface. Thus, dW = = E dr cos = 0. Therefore cos = 0 if = 90o.Thus, electric field intensity E on the surface of a conductor is always perpendicular to the surface.

77. Two insulated identical sized spheres A and B carrying charges +30 C and –10 C are separated by a distance of 1 metre in air, are made to touch each other. They are then again separated by the same distance. Compare the force between them in the two cases.

Ans. When the two spheres are touched, they share charge. The net charge on each after touching becomes + 30 – 10 = 20 C. This is shared equally between the two spheres, therefore charge on each after touching

Q = 10 CInitial force is given by

Fi = k = 300k,

Final force is given by

Ff = k = 100k

Change in force

78. A parallel plate capacitor with air between the plates is charged. A dielectric is inserted between the plates. What will happen to its electrostatic potential? Give reasons for your answer.

Ans. When the capacitor is charged the charge on it remains the same. Let this charge be Q. On the insertion of a dielectric of dielectric constant K, the capacitance becomes C = K C0. Now we

know that C = Q/V or V = Q/C. Since charge remains constant therefore, V As C

increases therefore there is a decrease in the potential between the plates.79. Define electric field intensity at a point. Derive an expression for the dipole field intensity at a

point on the axial line of a short dipole.Ans. Electric field intensity at a point is defined as the force experienced by a unit positive charge

placed at that point.Consider an electric dipole consisting of –q and +q charges separated by a distance 2a as shown in figure below. Let P be the point of observation on the axial line where the electric field has to be found. Let it be at a distance r from the centre O of the dipole. Let us suppose that the dipole is placed in vacuum.

24

O A B


Let EA and EB be the electric fields at point P due to the charges at A and B respectively. Therefore

EA = ….(1)

and EB = ….(2)

The two fields at P are in opposite directions. Thus, the resultant electric field at P is given byE =

=

or E = EB – EA

since = 180o, Therefore the resultant electric field is

E =

=

=

E =

=

= ….(3)

where p = q 2aFor a short dipole r >> a, therefore we have

E =

80. Define the term ‘electric field intensity’. Electric field inside a conductor is zero. ExplainAns. Electric field intensity at a point is defined as the force experienced by a unit positive charge

placed at that point.

By Gauss theorem Since there is no charge inside a conductor, therefore in

accordance with the above equation the electric field inside the conductor is zero.81. Sketch a graph to show how charge Q given to a capacitor of capacity C varies with potential

difference V. Prove that the total energy stored in a parallel plate capacitor is ½ CV2.Ans. The sketch between Q and V for a capacitor is as shown below.

25

Q

V


Suppose the capacitor is charged fully, its final charge is Q and final potential difference is V. These are related as

Q = CV ….(1)Let q and v be the charge and potential difference, respectively, at an intermediate stage during charging process then q = Cv. At this stage that small work done dW required to transfer an additional charge dq is

dW = vdq = ….(2)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by

W = ….(3)

or W = ….(4)

This work is stored in the capacitor in the form of its electric potential energy. Hence,

….(5)

Substituting Q = CV in equation 5 we have

U = CV2 ....(6)

Which is the required expression82. Derive an expression for potential at any point due to an electric dipole and hence show that the

potential on the equatorial line of an electric dipole is zero.Ans. Consider an electric dipole consisting of –q and +q charges separated by a distance 2a as

shown in figure below. Let P be the point of observation where the electric potential has to be found. Let it be at a distance r from the centre O of the dipole. Let us suppose that the dipole is placed in vacuum. Let angle BOP = . Let AP = r1 and BP = r2. Draw AC PO produced and BD PW. In AOC

cos = OC = a cos ,

similarly OD = a cos Net potential at P due to the dipole is

V =

….(1)Now r1 = AP CP = OP + OC

= r + a cos and r1 = BP DP = OP – OD

= r – a cos ,Substituting in equation 1 we have

V =

V =

=

V = ….(2)

26


This gives an expression for the potential at point P due to an electric dipole. Here p = q 2a is the dipole moment of the dipole. When the point lies on the equatorial line of the electric dipole, in this situation = 90o and cos 90o = 0, thus equation (2) becomes V = 0

83. Define capacitance of a capacitor. Give its SI unit. Prove that the total electrostatic energy stored in a parallel plate capacitor is ½ CV2.

Ans. (i) It is the ability of a conductor to store charge. In SI units it is measured in farad.(ii) See question 81 above.

84. S1 and S2 are two hollow concentric spheres enclosing charges Q and 2Qas shown in the figure (i) what is the ratio of the electric flux through S1 and S2? (ii) How will the electric flux through sphere S1 change, if a medium of dielectric constant 5 is introduced in the space inside S1 in place of air?

Ans. (i) According to Gauss theorem electric flux through S1

1 = and through S2

2 = therefore ratio of flux

(ii) The electric flux through S1 becomes 1/5 times the flux in air.85. What is an equipotential surface? Show that no work is done in moving a test charge from one

point to another over an equipotential surface.Ans. (i) It is a surface at every point of which the electric potential is same.

(ii) No work is done in moving a test charge over an equipotential surface. This cane be

understood as follows. By definition of electric potential we have VB – VA = Since the

surface is equipotential therefore potential at A and B are equal, therefore we have = 0

or WAB = 086. What is an electric dipole? Derive an expression for the torque acting on an electric dipole,

when held in a uniform electric field. Hence define the dipole moment.Ans. An electric dipole is a pair of point charges with equal magnitude and opposite sign separated

by a small distance.Consider an electric dipole consisting of charges –q and +q and dipole length d placed in a uniform electric field as shown figure below. Let the dipole moment make an angle with the direction of the electric field.The two charges experience force qE each. These forces are equal, parallel and opposite. Therefore the net force is zero. But these two forces constitute a couple. This applies a torque on the dipole given by

= Either force arm of the couple. = qE d sin

where d sin is the arm of the couple. = q

dE sin where p = qd, dipole moment = pE sin .In vector from The direction of torque is perpendicular to both and If E = 1 and = 90o, then = p, thus electric dipole moment is defined as numerically equal to the torque experienced by an electric dipole placed perpendicular to a unit electric field.

87. State Gauss theorem in electrostatics. Using this theorem, derive an expression for the electric field intensity due to an infinite plane sheet of charge of charge density Cm–2.

Ans. It states that “The net electric flux through any Gaussian surface is equal to 1/0 times net electric charge enclosed by the surface”.

27


Consider an infinite plane sheet of charge. Let be the uniform surface charge density i.e. the charge per unit surface area. From symmetry we find that the electric field must be perpendicular to the plane of the sheet, and that the direction of on one side of the plane must be opposite to its direction on the other side as shown in figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A. The charged sheet passes through the middle of the cylinder’s length, so that the cylinder’s ends are equidistant from the sheet.The electric field has a normal component along the curved surface of the cylinder. As a result the electric flux is linked with only the ends and not the curved surface.Therefore by the definition of electric flux, the flux linked with the Gaussian surface is given by

= EA + EA = 2EA ….(1)But by Gauss’s law

….(2)

From equations 1 and 2 we have

2EA = ….(3)

E = ….(4)

This gives the electric field due to an infinite plane sheet of charge.88. State Gauss theorem. Using Gauss’s theorem, show mathematically that for any point outside

the shell, the field due to a uniformly charged thin spherical shell is the same as if the entire charged thin spherical shell is the same as if the entire charge of the shell is concentrated at the centre. Why do you expect the electric field inside the shell to be zero according to the theorem?

Ans. It states that “The net electric flux through any Gaussian surface is equal to 1/0 times net electric charge enclosed by the surface”.Consider a thin spherical shell of radius R and centre at O. Let Q be the total charge on it. The charge distribution is spherically symmetric. In order to find the electric field at a point outside the sell let us consider a Gaussian surface in the form of a sphere of radius r (r >> R). By symmetry we find that the electric field acts radically outwards and has a normal component at all points on the Gaussian sphere. Therefore by definition of electric flux we have

= E A where A is the surface area of the Gaussian sphere therefore = E 4r2 ….(1)

But by Gauss’s law

= ….(2)

from equations (1) and (2) it follows that

E 4r2 =

This is the same as for a point charge.(ii) Since there is no charge inside the shell then in accordance with Gauss theorem =

Hence electric field inside is zero.

28


89. Name the dielectric whose molecules have (i) Non-zero (ii) Zero dipole moment. Define the term ‘dielectric constant’ for a medium.

Ans. (i) HCl and H2O (ii) Oxygen and hydrogenDielectric constant is defined as the ratio of the capacitance of a capacitor with dielectric between plates to the capacitance without dielectric between plates.

90. Write the ratio of the (i) Charges on the plates, when two capacitors, of capacitance C1 and C2, are put in series,(ii) Potential differences, between the plates, when two capacitors of capacitances C1 and C2,

are put in parallel. Give also the ratio of the equivalent capacitances of the series and parallel combinations of these two capacitors.

Ans. (i) In series both capacitors have the same charge. Therefore the ratio of the charges are 1 : 1.(ii) In parallel the potential across each capacitor is same. Therefore the ratio of the potentials is 1 : 1.

(iii) In series we have CS = and in parallel CP = C1 + C2. Therefore the ratio of

capacitance in series and parallel is

91. Define electric field intensity. Write its SI unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length 2a at the mid-point of the line joining the two charges.

Ans. Electric field at a point is defined as the force experienced by a unit charge placed at that point. The expression for the electric field due to an electric dipole at the midpoint of the line

joining the two charges is This is towards the negative charge.

92. State Gauss’ theorem. Apply this theorem to obtain the expression for the electric field intensity at a point due to an infinitely long, thin, uniformly charged straight wire.

Ans. It states that “The net electric flux through any Gaussian surface is equal to 1/0 times net electric charge enclosed by the surface”.Consider an infinitely long, thin wire charged +vely and having uniform linear charge density . The symmetry of the charge distribution shows that must be perpendicular to the line charge and directed outwards. As a result of this symmetry we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary length L with its ends perpendicular to the wire as shown in figure below.

For the cylindrical part of this Gaussian surface, is constant in magnitude and perpendicular to the surface at each point. Furthermore, the flux through the ends of the Gaussian cylinder is zero since is parallel to these surfaces. i.e., there is no normal component of the electric field at these faces. Therefore by the definition of electric flux we have

= E A ….(1)where A is the curved surface area of the cylinceror = E 2rL ….(2)

By Gauss’s law the electric flux is given by

= ….(3)

from equation 2 and 3 we have

29


E 2rL = ….(4)

or E = ….(5)

This is the expression for the electric field due to an infinitely long thin wire.93. A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation

between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in the second case.

Ans. Given C = 8pF. We know that C = . When the distance between the plates is reduced

to half and a dielectric is inserted then the new capacitance becomes

C’ = = 10 8 = 80 pF.

94. A point charge ‘q’ is placed at O as shown in the figure. Is VP – VQ positive or negative when (i) q > 0 and (ii) q < 0? Justify your answer.

Ans. (i) If q > 0, then the potential at point P will be greater than that at point Q, since the distance of P is lesser than that of Q from point O. Hence VP > VQ or VP – VQ is positive.

(ii) If q < 0, then the potential at point P will be less than that a point Q, since the distance of P is lesser than that of Q from point O. Hence VP < VQ or VP – VQ is negative.

95. On what factors does the capacitance of a capacitor depend? Give the formula used.Ans. The capacitance of a capacitor depends upon (i) shape (ii) size and (iii) nature of medium

between the plates. The required formula is C =

96. Two point electric charges of unknown magnitude and sign are placed a distance ‘d’ apart. The electric field intensity is zero at a point, not between the charges but on the line joining them Write two essential conditions for this to happen.

Ans. The electric field intensity may be zero at a point on the axial line of charges, if(i) The two charges are of opposite signs i.e., one charge is positive and the other is negative.(ii) The magnitude of charge nearer the null-point under consideration should be smaller than

the magnitude of the other charge.97. The graph shows the variation of voltage V across the plates of two capacitors A and B versus

increase of charge Q stored on them. Which of the two capacitors has higher capacitance? Given reason for your answer.

Ans. The capacitor A has higher capacitance. We know that capacitance C = Q/V and in the graph shown here slope of the graph gives the value of V/Q (i.e. 1/C). Thus, capacitance is obtained by reciprocal of the slope of V – Q graph. As slope of graph for capacitor A is less, it means that capacitance of A is higher.

98. An isolated air capacitor of capacitance C0 is charged to a potential V0. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how do the following change, when the battery is disconnected (i) capacitance, (ii) potential difference, (iii) field between the plates, (iv) energy stored by the capacitor.

Ans. When the battery is disconnected, the charge on the capacitor does not change.

30

+qO P Q

V

Q

A

B


(i) The capacitance of the capacitor becomes K times the original value i.e., C = K C0

(ii) Now new potential difference is

V =

(iii) The field between the plates becomes

E =

(iv)The energy stored becomes

U =

99. An isolated air capacitor of capacitance C0 is charged to a potential V0. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then now do the following change, when the battery remains connected (i) capacitance, (ii) charge, (iii) field between the plates, (iv) energy stored by the capacitor.

Ans. When the battery remains connected, the potential on the capacitor does not change.(i) The capacitance of the capacitor becomes K times the original value i.e., C = KC0

(ii) Now new charge isQ = CV = KC0V = KQ0

(iii) The field between the plates becomes

E = i.e., no change.

(iv) The energy stored becomes

U = KC0V2 = KU0

100. Explain the underlying principle of working of a parallel plate capacitor. If two similar plates, each of area A having surface charge densities + and – are separated by a distance d in air, write expression for (i) the electric field at points between the two plates. (ii) the potential difference between the plates and (iii) the capacitance of the capacitor so formed.

Ans. Consider an insulated plate A. Let it be given a charge such that its potential becomes a maximum i.e., it can no more hold additional charge. Now, place another uncharged conductor B in its vicinity. Due to electrostatic induction –ve charge is induced on the inner face and +ve charge on the outer face. The –ve charge tends to decrease and the +ve charge tends to increase the potential of plate A. Since the –ve charge is close to plate A than the +ve charge therefore, there is an overall decrease in the potential on plate A. To bring the plate back to the same potential it must be given more charge. This means that the charge holding capacity of plate A has increased i.e., its capacitance increases. Now let us connected the conductor B to the earth i.e., ground it. Due to this the +ve charge on it “flows” to the ground, but the –ve charge is bound to the plate B due to the +ve charge on plate A. now, the potential of plate A decreases by a large value. Thus more charge will be required by it to come back to the same potential.

31


Therefore to store a large charge in a small space an uncharged earth connected conductor should be placed in the vicinity of a charged conductor. This forms the principle of a capacitor.

(i) The electric field between the two plates is given by E =

(ii) The potential difference between the plates is given by V = Ed =

(iii) The capacitance of the capacitor is given by C =

101. Using Gauss’s law show that no electric field intensity exists inside a hollow charged conductor.

Ans. Consider a hollow charged conductor (say a thin conducting shell) having charge +Q, which spreads on its outer surface. Consider a Gaussian surface for calculating the electric field intensity E at a point P inside the hollow conductor. Then, total flux over this Gaussian surface is

= = 0 as qin = 0 or E = 0.

Thus, electric field intensity at all points inside a hollow charged conductor is zero.102. Two plane sheets of charge densities + and – are kept in air as shown in the Fig. What are

the electric field intensities at points A and B?

Ans. (i) At point A the electric field intensities due to either sheet has magnitude E = / 20 but have a direction opposite to each other. Hence, net electric field at A.

E = E1 + E2 = 0

(ii) At point B the two fields are in the same direction. Also between the plates the electric field is uniform and has a value E = / 20. Therefore the total electric field between the plates at point B is

E = E1 + E2 = towards the second plate as shown.

103. Capacitors P, Q and R have each a capacity C, A battery can charge the capacitor P to a potential difference V. if after charging P the battery is disconnected from it and the charged capacitor P is connected in the following separate instances to Q and in the following separate instances to Q and R (i) to Q in parallel and (ii) to R in series, then, what will be the potential differences between the plates of P in the two instances.

Ans. (i) When the capacitor P is connected to Q in parallel, the two will share charge till they attain a common potential. Since the two capacitors have the same capacity, therefore both will have half the original charge. Hence their potential will also become half.

Thus V’ = V/2(ii) When capacitor P is connected in series to capacitor R, R will also be charged with the same charge as that on capacitor P. Thus there is no loss of charge of capacitor P, hence its potential does not change and remains V.

32

A

B+

–

A

+

–

E1

E21

E2E1+

2


104. Define electric line of force and give its two important properties.Ans. It is defined as a path straight or curved, a tangent to which at any point gives the direction of

the electric field at that point.(i) No two field lines can cross, because at the point of intersection two tangents can be drawn

giving two directions of electric field which is not possible.(ii) The field lines are always perpendicular to the surface of a charged conductor.

105. (a) Why does the electric field inside a dielectric decrease when it is placed in an external electric field?(b) A parallel plate capacitor with air between the plates has a capacitance of a pF. What will be the capacitance if the distance between the plates be reduced by half and the space between them is filled with a substance of dielectric constant K = 6?

Ans. (a) When a dielectric is placed in an electric field (E0), it gets polarize, i.e., within the dielectric an electric field (E) is induced in a direction opposite to that of the external field. Therefore the net field within the dielectric decreases to

(b) The capacitance of a parallel plate capacitor of plate area A and separation d is given by

C0 =

Now C0 = 8 pF, When filled with a dielectric of constant K and distance reduced by half, the capacitance C is given by

C = = 2 6 8 = 96 pF

106. The given graph shows the variation of charge Q versus potential difference V for two capacitors C1 and C2. The two capacitors have same plat separation but the plate area of C2 is double than that of C1. Which of the lines in the graph correspond to C1 and C2 and why?

Ans. The capacitance of a parallel plate capacitor is given by the expression C = In other

words C A that is, for the separation being the same, the capacitance is directly proportional to the plate area. Since the plate area of capacitor C2 is double than that of C1, therefore its capacitance will also be double than that of C1, therefore its capacitance will also be double compared as compared to that of C1. The slope of the Q-V graph gives the value of capacitance. The slope is greater for A than for B. Therefore graph A corresponds to C2 and graph B correspond to C1.

107. Define electric field E at a point in space due to a distribution of charges. Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance.

Ans. Electric field at a point is defined as the net force experienced by a unit charge placed at that point.(i) Electric field due to two similar charges.

33

Q

A

B

VO


(ii) Electric field due to two opposite charges

108. Define electric flux. Write its SI unit. Using Gauss’ theorem, derive an expression for the electric field intensity at any point outside a charged spherical shell.

Ans. Electric flux is defined as the scalar product of the electric field and the area vector.

Mathematically it is represented by the equation = In SI it is measured in Nm2C–1.

Consider a thin spherical shell of radius R and centre at O. Let Q be the total charge on it. The charge distribution is spherically symmetric. In order to find the electric field at a point outside the shell let us consider a Gaussian surface in the form of a sphere of radius r (r >> R). By symmetry we find that the electric field acts radially outwards and has a normal component at all points on the Gaussian sphere. Therefore by definition of electric flux we have

= E 4r2 ….(1)But by Gauss’s law

= ….(2)

from equations (1) and (2) it follows that

E 4r2 = or E =

This is the same as for a point charge.109. Define dielectric constant of a medium. Briefly explain why the capacitance of a parallel plate

capacitor increases, on introducing a dielectric medium between the plates.Ans. It is defined as the ratio of the force acting between two charges placed a certain distance

apart in vacuum to the force acting between the same two charges placed the same distance apart in the medium.Introduction of the dielectric between the plates of the capacitor reduces the electric field difference between the plates and hence the potential between the plates. As C = Q/V, therefore a decrease in V means an increase in C.

110. Two identical plane metallic surfaces A and B are kept parallel to each other in air, separated by a distance of 1 cm as shown in the figure.

A is given a positive potential of 10V and the outer surface of B is earthed.

34


(i) What is the magnitude and direction of the uniform electric field between Y and Z?(ii) What is the work done in moving charge of 20 C from X to Y?

Ans. (i) Electric field between the plates is

E = = 103 V m–1

directed from plate A at higher potential to plate B at lower potential i.e., from Y to Z(ii) Since X and Y are on the same plate A, which as an equipotential surface, so work done in moving a charge of 20 C from X to Y on equipotential surface is zero.

111. An electric dipole with moment is placed in a uniform electric field of intensity Write an expression for one torque experienced by the dipole. Identify two pairs of perpendicular vectors in the expression. Show diagrammatically the orientation of the dipole in the field for which the torque is (i) maximum, (ii) half the maximum value, (iii) zero.

Ans. Expression for torque on dipole in uniform electric field is = pE sin In vector form this is written as So, two pairs of perpendicular vectors are:(a) torque and dipole moments (b) torque and electric field intensity.(i) For maximum torque = 90o. The orientation is as shown below.

(ii) Half the maximum value. For this the angle is 30o or 150o. The orientations are as shown below.

(iii) When torque is zero: Torque is zero when = 0o or 180o. The orientation are as shown below.

112. Define capacitance of a capacitor. Give its S.I. unit. For a parallel plate capacitor, prove that the total energy stored in a capacitor is ½ CV2 and hence derive expression for the energy density of the capacitor.

Ans. It is the ability of a capacitor to store charge and energy.Suppose the capacitor is charged fully, its final charge is Q and final potential difference is V. These are related as

Q = CV ….(1)Let q and v be the charge and potential difference, respectively, at an intermediate stage during charging process then q = Cv. At this stage the small work done dW required to transfer an additional charge dq is

dW = vdq = ….(2)


W = ….(3)

or W = ….(4)


35


U = ….(5)

substituting Q = CV in equation 5 we have

U = CV2 ….(6)

which is the required expression.Consider a parallel plate capacitor with plate area A and distance between plates d. Now the total energy stored in a capacitor is

U = CV2 ….(7)

Now volume between the plates is v = Ad and the value of capacitance C is C =

Substituting these values in equation 6 we have u =

here V = Ed, hence energy density u is

u = 0E2 ….(8)

113. Derive an expression for the capacitance of a parallel plate capacitor when the space between the plates is partially filled with a dielectric medium of dielectric constant ‘K’. Explain why the capacitance decreases when the dielectric medium is removed from between the plates.

Ans. Consider a parallel plate capacitor having each plate of area A and separated by a distance d. When there is vacuum between the two plates, the capacitance of

the parallel plate capacitor is given by C =

Suppose that when the capacitor is connected to a battery, electric field of strength E0 is produced between the two plates of the capacitor. Further, suppose that when dielectric slab of thickness t (t < d) is introduced between the two plates of the capacitor as shown in figure, the electric field reduces to E due to polarization of the dielectric. Therefore, between the two plates of the capacitor; over a distance t, the strength of the electric field is E and over the remaining distance (d – t) the strength is E0. If V is the potential between the plates of the capacitor, then

V = Et + E0 (d – t) ….(1)Since E = E0/K where K is the dielectric constant, therefore the above equation becomes

V = + E0 (d – t) = E0 ….(2)

The electric field between the plates of the capacitor is given by

E0 = ….(3)

Hence the potential between the two plates becomes

V = E0 ….(4)

Hence the capacitance of the parallel plate capacitor is given by

C =

36


=

The capacitance decreases because K > 1. Therefore the removal of the dielectric decreases the capacitance.

114. An electric dipole is held in a uniform electric field (i) Prove that no translatory force acts on it. (ii) Derive an expression for the torque acting on the dipole. (iii) Calculate the work done on the dipole when it is rotated through 180o from its equilibrium position

Ans. (i) Consider an electric dipole consisting of charges –q and +q and dipole length d placed in a uniform electric field as shown figure below. Let the dipole moment make an angle with the direction of the electric field.The two charges experience force qE each. These forces are equal, parallel and opposite. Therefore the net force acting on the dipole is

Fn = qE – qE = 0Thus the net force acting on the dipole is zero.(ii) But these two forces constitute a couple. This applies a torque on the dipole given by

= Either force arm of the couple. = qE d sin where d sin is the arm of the couple. = qdF sin

where p = qd, dipole moment = pE sin .The work done in rotating a dipole in an electric field is given by

W = sin d = pE

= –pE [cos – cos ]W = pE [1 – cos ]

when = 180o = –1 W = pE [1 + 1] = 2pEW = –pE cos , since = 180o therefore the work done is W = –pE cos 180o = + pE

115. Define dipole moment of an electric dipole. Show mathematically that the electric field intensity due to a short dipole at a distance ‘d’ along its axis is twice the intensity at the same distance along the equatorial line.

Ans. It is defined as the product of the magnitude of either charge and the distance between them.Consider an electric dipole consisting of –q and +q charges separated by a distance 2a as shown in figure. Let P be the point of observation on the axial line where the electric field has to be found. Let it be at a distance r from the centre O of the dipole. Let us suppose that the dipole is placed in vacuum.Let EA and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

EA =

and EA =

The two fields at p are in opposite directions thus the resultant electric field at p is given by

37


E =

= = EB – EA

since = 180o Therefore the resultant electric field is

E =

=

=

Solving we have

E = ….(1)

where p = q 2aIf the dipole is short then r >> a, therefore a is neglected as compared to r, hence

E = ….(2)

Consider an electric dipole consisting of charges –q and +q separated by a distance 2a as shown in figure below. Let the point of observation P lie on the right bisector of the dipole AB at a distance r from its mid point O. Let EA and EB be the electric field intensities at point P due to charges at A and B respectively.The two electric fields have magnitudes.

EA = ….(3)

In the direction of AP

EB = ….(4)

in the direction of PBThe two fields are equal in magnitude, but have different directions. Resolving the two fields EA & EB into their rectangular components i.e. perpendicular to and parallel to AB. The components perpendicular to AB i.e. EA sin and EB sin being equal and opposite cancel out each other while the components parallel to AB i.e. EA cos and EB cos being in the same direction add up as shown in the Fig. Hence the resultant electric field at point P is given by

E = EA cos + EB cos

= 2EA cos = ….(5)

38


E =

∵ cos = and q 2a = p

For a short dipole r2 >> a2 therefore E =

Thus for a short dipole (ideal dipole) the electric field on the axial line is E a = where

as on the equatorial line is E = , thus we find that the electric field intensity due to a

short dipole on the axial line is twice that on the equatorial line.116. Derive an expression for the electrostatic energy stored in a parallel plate capacitor. Assuming

that the capacitor is disconnected from the charging battery, explain how the (i) capacitance (ii) potential difference and (iii) energy stored, in the parallel plate capacitor change, when a medium of dielectric constant ‘K’ is introduced between the plates.

Ans. Suppose the capacitor is charged fully, its final charge is Q and final potential difference is V. These are related as

Q = CV ….(1)Let q and v be the charge and potential difference, respectively, at an intermediate stage during charging process then q = Cv. At this stage the small work done dW required to transfer an additional charge dq is

dW = vdq = ….(2)


W = ….(3)

or W = ….(4)


U = ….(5)

39


substituting Q = CV in equation (5) we have

U = CV2 ….(6)

which is the required expressionConsider a parallel plate capacitor with plate area A and distance between plates d. Now the total energy stored in a capacitor is

U = CV2 ….(7)

When the charging battery is removed the charge on the capacitor remains the same. The capacitance increases and becomes K times its original value. The potential difference between the plates decreases and becomes 1/K times the previous value. Energy decreases and becomes 1/K times its original value.

117. Give the principle of working of a Van-de-Graff generator. With the help of a labeled diagram, describe its construction and working. How is the leakage of charge minimized from the generator?

Ans. The Van-de-graff generator is based on the following two principles.(i) The corona discharge.(ii) When a charged conductor is placed in contact with the inside of a hollow conductor, all

of the charge of the first conductor is transferred to the hollow conductor. The charge on the hollow conductor can be increased by repeating the process.

Construction: It consists of a large and highly polished hollow metal sphere called dome. It is insulated from the ground with the help of insulating supports. Two, comb like, metallic needles are provided at point C1and C2 as shown in figure below. The comb at point A is called the spray comb and that at point B is called the collector comb. There is a provision of two pulleys around which a belt of insulating material e.g. Silk, rayon, reinforced rubber, moves.Working: Due to high density of positive charge at the sharp point of the teeth of the combs C1, positive charge is sprayed on the belt (Corona Discharge). As the belt is rotated by the electric motor, these positive irons move upward along with the belt. A negative charge is induced on the sharp teeth of the collecting comb C2

and an equal positive charge is induced on the farther end of the comb C2. This positive charge shifts to the outer surface of the conducting shell S. Due to Corona discharge at the sharp teeth of comb C2 a negatively charged electric wind is setup. This neutralizes the positive charge on the belt. This process continues and sphere goes on getting more and more of positive charge.The leakage is minimized by housing the whole generator inside a steel chamber filled with N2 or methane sulphur hexafluoride or some other inert gas at high pressure.

MAGNETICS

1. Under what condition is the force acting on a charge moving through a uniform magnetic field minimum?

Ans. When the charge moves parallel to the direction of the magnetic field.2. What is the nature of the magnetic field in a moving coil galvanometer?

40


Ans. Radial magnetic field.3. A certain proton moving through a magnetic field experiences maximum force. When does this

occur?Ans. When it moves perpendicular to the direction of the magnetic field.4. In a certain arrangement a proton does not get deflected while passing through a magnetic field

region. Under what condition is it possible?Ans. When it moves parallel to the direction of the magnetic field.5. An electron and a proton moving with the same speed enter the same magnetic field region at

right angles to the direction of the field. For which of the two particles will the radius of the circular path be smaller.

Ans. The radius of the circular path is given by the expression r = Since v, B and q are same

for both, therefore the radius of the circular path depends upon its mass. Since electron’s mass is less than that of a proton, therefore the radius of the circular path of the electrons will be smaller.

6. How will the magnetic field intensity at the centre of a circular coil carrying current change, if the current through the coil is doubled and the radius of the coil halved?

Ans. The magnetic field at the centre of a circular current carrying coil is given by the expression

B = When current is doubled and the radius is halved the magnetic field becomes four

times.7. State two properties of the material of the wire used for suspension of the coil in a moving coil

galvanometer.Ans. (i) High tensile strength. (ii) Small value of torque per unit twist.8. An ammeter and a milli-ammeter are converted fro the same galvanometer. Out of the two,

which current measuring instrument has higher resistance?Ans. A milli ammeter has higher resistance.9. Define the term magnetic moment.Ans. It is defined as the product of either pole strength and the dipole length.10. How does the intensity of magnetization of a paramagnetic material vary with increasing

applied magnetic field?Ans. It increases with the increase in the applied magnetic field.11. Why do magnetic lines of force prefer to pass through ferromagnetic substances than through

air?Ans. It is because the permeability of ferromagnetic material is greater than that of air.12. A small magnetic needle pivoted at the centre is free to rotate in a magnetic meridian. At what

place will the needle be vertical?Ans. At the poles13. What is the angle of dip at a place where the horizontal and vertical components of the earth’s

magnetic field are equal?Ans. 45o

14. In which direction would a compass needle align if taken to geographic (i) North and (ii) South pole?

Ans. In perpendicular direction for both cases.15. How does the intensity of a paramagnetic sample vary with temperature?Ans. It decreases with the increase in temperature.16. What should be the orientation of a magnetic dipole in a uniform magnetic field so that its

potential energy is maximum?Ans. It should be anti parallel to the applied magnetic field.17. Write the SI unit of (i) Magnetic pole strength (ii) Magnetic dipole moment of a bar magnet.Ans. (i) Am and (ii) Am2

18. Name the SI unit of intensity of magnetization.41


Ans. Am–1

19. Two wires of equal length are bent in the form of two loops. One of the loops is square shaped and the other is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience a greater torque? Give reasons.

Ans. Torque experienced by a current carrying loop placed in a uniform magnetic field is given by the expression - BInA. In other words torque is directly proportional to the area of the loop. Since a circular wire has more area than a square wire for the same dimension, therefore the circular wire experiences more torque than the square wire.

20. State the reason why soft iron is used in making magnets.Ans. It has a high retentivity.21. Under what condition an electron moving through a magnetic field experience maximum force?Ans. When it is moving perpendicular to the direction of the magnetic field.22. Write one condition under which an electric charge does not experience a force in a magnetic

field.Ans. When it moves parallel to the direction of the magnetic field.23. Write SI unit of magnetic field.Ans. tesla.24. What is the value of the horizontal component of the earth’s magnetic field at the magnetic

poles?Ans. Zero25. What is the force acting on an electron when moving (i) parallel and (ii) antiparallel to a

magnetic field?Ans. (i) Zero (ii) Zero.26. How does the magnetic moment of an electron in a circular orbit of radius r and moving with a

speed v change when the frequency of revolution is doubled?

Ans. The magnetic moment of an electron is given by the expression M = r2 = .

When the frequency is doubled the magnetic moment also becomes double. 27. Permanent magnets whereas soft iron is preferred for making electromagnets. Give one reason.Ans. (i) Retentivity of steel is more than that of soft iron.

(ii) Soft iron has a smaller value of coercivity than steel.28. What is the value of angle of dip at a place on the surface of the earth, where the ratio of the

vertical component to the horizontal component of the earth’s magnetic field is

Ans. Using the expression tan = Therefore = 30o.

29. Consider the circuit shown here where APB and AQB are semi-circles. What will be the magnetic field at the centre C of the circular loop?

Ans. Zero, because magnetic fields due to APB and AQB are equal in magnitudes but opposite in directions.

30. A current is set up in a long copper pipe. Is there a magnetic field (i) inside, (ii) outside the pipe?

Ans. (i) There is no magnetic field inside the pipe.(ii) There is a magnetic field outside the pipe.

31. The force F experienced by a particle of charge e moving with velocity v in a magnetic field B is given by Of these, name the pairs of vectors which are always at right angles to each other.

Ans. F and V as well as F and B are the pairs which are always at right angles to each other.

42


32. Which one of the following will describe the smallest circle when projected with the same velocity v perpendicular to the magnetic field B: (i) – particle, and (ii) – particle?

Ans. –particle as mass of –particle is less than that of alpha particle.33. Equal currents I and I are flowing through two infinitely long parallel wires. What will be the

magnetic field at a point mid-way when the currents are flowing in the same direction?Ans. Zero, because fields due to two wires will be equal but opposite.34. The figure shows a circular loop carrying current I. Show the

direction of the magnetic field with the help of lines of force.

Ans. The magnetic field lines are as shown

35. Which one of the following will have minimum frequency of revolution, when projected with the same velocity v perpendicular to the magnetic field B: (i) -particle (ii) -particle.

Ans. Frequency of revolution v = and of -particle is less, hence -particle will have

minimum frequency of revolution.36. What is the effect on a vertical spring carrying a weight when a current is passed through the

spring?Ans. As the currents in the neighboring loops are in the same direction, hence they will attract each

other and as a result the spring will contract.37. A current is passed through a loop of a flexible wire. What shape will it take?Ans. It will acquire a circular shape because each small part of the loop will experience a repulsive

force due to the magnetic field from the opposite part of the loop.38. An electron and a proton moving parallel to each other in the same direction with equal

moment, enter into a uniform magnetic field which is at right angles to their velocities. Trace their trajectories in the magnetic field.

Ans. Because both electron and proton have the same charge and momentum, therefore they will describe circles of equal radii as shown.

39. Name the physical quantity which has its unit JT–1. Is it a scalar or a vector quantity?Ans. Magnetic dipole moment has the unit JR–1. It is a vector quantity. 40. How does the (i) pole strength, and (ii) magnetic moment of each part of a bar magnet change if

it is cut into two equal pieces transverse to its length?Ans. (i) Pole strength remains unchanged.

(ii) Magnetic moment m is half of the original value.41. How does the (i) pole strength, and (ii) magnetic moment of each part of a bar magnet change if

it is cut into two equal pieces along its length?

43


Ans. (i) Pole strength of each part becomes half.(ii) Magnetic moment also becomes half.

42. What is the source of magnetic field?Ans. Magnetism is of electrical origin. The electrons revolving in an atom behave as tiny current

loops, which give rise to magnetism.43. Compare the magnetic fields due to a straight solenoid and a bar magnet.Ans. Both the fields are identical. In fact a long straight solenoid behaves as a bar magnet having

dipole moment m = N I A.44. A short bar magnet placed with its axis making an angle with a uniform external field B

experiences a torque. What is the magnetic moment of the magnet?

Ans. Torque on a short bar magnet = MB sin , therefore M =

45. Can two magnetic lines of force intersect? Justify your answer.Ans. Two magnetic field lines cannot intersect at any point because if they do so then there will be

two possible directions of the magnetic field at the point of intersection, which is not possible.46. Which one of the following will experience maximum force, when projected with the same

velocity ‘v’ perpendicular to the magnetic field (i) alpha particle and (ii) beta particle?Ans. The force experienced by a charged particle is given by the expression F = Bq v. Since an

alpha particle has more charge than the beta particle, therefore alpha particle will experience more force.

47. Which one of the following will have minimum frequency of revolution, when projected with the same velocity v perpendicular to the magnetic field B; (i) alpha particle and (ii) beta particle?

Ans. The frequency of revolution of a charged particle in a magnetic field is given by v = .

The ratio of for an alpha particle is less than that for a beta particle; therefore alpha particle

will have minimum frequency of revolution.48. What is the effect on a vertical spring carrying a weight when a current is passed through the

spring?Ans. A spring will behave as combination of different current carrying conductors carrying current

in the same direction; these conductors will attract each other. This will in turn contract the spring.

49. Name the material used in making the core of moving coil galvanometer.Ans. Soft iron.50. A small magnet is pivoted to move freely in the magnetic meridian. At what place on the

earth’s surface, will the magnet be vertical.Ans. At the magnetic poles of the earth.51. An electron beam projected along + X-axis, experiences a force due to a magnetic field along

the + Y-axis. What is the direction of the magnetic field?Ans. The direction of the magnetic field is along z-axis. This is because the direction of motion, the

magnetic field and the force are perpendicular to one other as an electron carries negative charge.

52. The vertical component of Earth’s magnetic field at a place is times the horizontal component. What is the value of angle of dip at this place?

Ans. Given: Bv = BH, we know that tan = or = 60o

Therefore angle of dip, = 60o

53. An electron is moving along the +ve X-axis in the presence of uniform magnetic field along the +ve Y-axis. What is the direction of force acting on it.

Ans. The force acts along the –ve Z-axis.

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54. Why should the material used for making permanent magnets have high coercivity?Ans. It is because a permanent magnet should not easily de-magnetised.55. An electron and a proton, having equal momenta, enter a uniform magnetic field at right angels

to the field lines. What will be the ratio of the radii of curvature of the their trajectories?

Ans. As radius of charged particle in magnetic field is r = Since q, mv and B are same

therefore the ratio of their radii is 1.56. A compass needle, pivoted about the horizontal axis and free to move in the magnetic meridian

is observed to point along the (i) vertical direction at a place A.(ii) horizontal direction at a place BGive the value of the angel of dip at these two places.

Ans. (i) 90o and (ii) 0o

57. An electron is moving with velocity v along the axis of a long straight solenoid carrying current I. What will be the force acting on the electron due to the magnetic field of the solenoid?

Ans. Zero, as force on charged particle moving in magnetic field is F = Bqv sin Here both v and B are along the axis of solenoid, so = 0o between them. Hence F = qvB sin = 0.

58. Derive an expression for the torque on a rectangular coil of area A, carrying a current I and placed in a magnetic field B, the angle between the direction of B and the vector perpendicular to the plane of the coil is .

Ans. Current carrying conductors usually form closed loops. This prompts us to find the total force and torque acting on the loop. Consider a current carrying loop placed in a uniform magnetic field as shown in figure. The loop can be considered to be consisting of a series of straight-line segments. We will find that the total force acting on the loop is zero but there is net torque acting on it.Figure below shows a rectangular loop of wire with length ‘a’ and breadth ‘b’. A line perpendicular to the plane of the loop (i.e. a normal to the plane) makes an angle with the direction of the magnetic field B, and the loop carries a current I as shown. Let the forces acting on the various sides of the loop be as shown. It follows from the expression for the force experienced by a conductor in a magnetic field that force on arm AB is

….(1)

Here I is a vector in the direction of the current. In accordance with Fleming’s left hand

rule this force acts in the place of the paper and is directed upwards as shown.The force on arm CD is given by

….(2)

Here I is a vector in the direction of the current. In accordance with Fleming’s left hand

rule this force acts in the plane of the paper and is directed downwards as shown.The sides with length ‘b’ i.e. AB and CD make an angle (90o – ) with the direction of the magnetic field. Therefore the forces acting on these two sides given by equations (1) and (2) are equal and opposite.Since these two forces are equal and opposite and have the same line of action therefore they cancel out each others effect and their resultant effect on the coil is zero.

Now the force on arm BC is

….(3)

Here is a vector in the direction of the current. In accordance with Fleming’s left hand

rule this force acts perpendicular to the plane of the paper and is directed outwards as shown.

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Finally the force on arm DA is

….(4)

Here is a vector in the direction of the current. In accordance with Fleming’s left hand

rule this force acts perpendicular to the plane of the paper and is directed inwards as shown in figure below.Both forces F3 and F4 make an angle of 90o with the direction of the magnetic field. Therefore in magnitude these forces are given by

F3 = F4 = I l Bsin 90o = I l B ….(5)The lines of action of both these force are perpendicular to the plane of the patper.The two forces F3 and F4 lie along, different lines and each gives rise to a torque about the X-axis. The two torques, produce a resultant torque in +X direction. The arm of the couple (perpendicular distance between the lines of action of the two forces) from the figure below is given byArm of couple = b sin ….(6)Therefore by the definition of torque we have

Torque = either force arm of coupleUsing equation 5 and 6 we have

Torque = I B a b sin But a b = A, area of the coil, therefore

= I BA sin

59. Derive a formula for the force between two parallel straight conductors carrying current in the opposite directions and write the nature of the force. Hence define an ampere.

Ans. Consider two long, straight parallel wires separated by a distance ‘a’ and carrying I1 and I2 in the same direction as shown. We can easily determine the force on one wire due to the magnetic field set-up by the other wire. Wire 2, which carries a current I2, sets-up a magnetic field B2 at the position of wire 1.The direction of B2 is perpendicular to the wire, as shown in figure. Now the magnetic force on length L of wire 1 is

Since L is perpendicular to B2, the magnitude of F1 is given byF1 = I1LB2

….(1)

But the field due to wire 2 is given by the relation B =

….(2)Therefore from equations 1 and 2 we have

F1 = I1LB2

=

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We can rewrite it in terms of force per unit length as

….(3)

The direction of F1 is downward, towards wire 2, since is downwards. If one considers the field set-up at wire 2 due to wire 1, force F2 is found to be equal and opposite to F1, which is in accordance with Newton’s third law of motion. When the currents are in opposite directions, the forces are reversed and the wires repel each other. Hence we find that the force per unit length of wire between two parallel current carrying wires is given by

F =

Therefore two conductors carrying current in the same direction attract each other whereas two conductors carrying current in the opposite directions repel each other.The above expression can be used to define ampere, the SI unit of current. Let I 1 = I2 = 1 ampere, a = 1m then F = 2 10–7 Nm–1 Thus we haveOne ampere is that much current which when flowing through each of the two infinitely long straight conductors held 1 metre apart in space, produce a force of F = 2 10–7 N per metre of their length.

60. Explain how will you convert a galvanometer into a voltmeter to read a maximum potential of ‘V’ volt. Can one use a voltmeter to measure the emf of a cell? Justify your answer?

Ans. Suppose a galvanometer having resistance G is to be converted into a voltmeter, which can measure the potential difference from 0 to V volt. Let a high resistance R be joined in series with the galvanometer for this purpose. Its value is so chosen that when the galvanometer with the resistance is connected between two points having a potential difference of V volt, the galvanometer gives full-scale deflection. It is clear from the figure below that

V = Ig (R + G)

or R =

On connecting the above high resistance in series with a galvanometer, the galvanometer is converted into a voltmeter of range V volt.A voltmeter cannot measure the emf of the cell as it draws current from the cell while measuring potential difference.

61. Explain how will you convert a galvanometer into an ammeter to read a maximum current of ‘I’ ampere. An ammeter is always connected in series with a circuit. Why?

Ans. Suppose a galvanometer of resistance G is to be converted into an ammeter having range 0 to I ampere. Let Ig be the current, which gives full-scale deflection in the galvanometer. Suppose S is the appropriate shunt required for this purpose i.e. when shunt S is used, current Ig passes through the galvanometer and the remaining (I – Ig) passes through the shunt as shown in figure below. Since the shunt and the galvanometer are connected in parallel therefore the potential differences across both will be same. Hence

Ig G =(I – Ig) S ….(1)

S = ….(2)

An ammeter is used to measure current; therefore it is connected in series so that the entire current passes through it. Moreover an ammeter is a low resistance device.

62. Using Biot Savart’s law, derive an expression for the magnetic field at any point on the axis of a circular coil of radius r and having N number of turns. Indicate the direction of the magnetic field.

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Ans. Consider a circular loop of wire of radius R located in the YZ plane and carrying a steady current I as shown in figure below. Let us calculate the magnetic field at an axial point P a distance x from the centre of the loop. From the figure it is clear that any element dL is perpendicular to furthermore all the elements around the loop are at the same distance r from P, where r2 = x2 + R2. Hence by Biot Savart’s law the magnetic field at point P due to the current element dL is given by

dB = ….(1)

The direction of the magnetic field dB due to the element dL is perpendicular to the palen formed by and dL as shown in figure above. The vector dB can be resolved into components dBx along the X-axis and dBy which is perpendicular to the X-axis. When the components perpendicular to the X-axis are added over the whole loop, the result is zero. That is, by symmetry any element on one side of the loop will set up a perpendicular component that cancels the component set up by an element diametrically opposite it. Therefore it is obvious that the resultant magnetic field at P will be along the X-axis. This result can be obtained by integrating the components dBx = dB cos. Therefore, we have

B = ….(2)

Where the integral is to be taken over the entire loop since , x and R are constants for all elements of the loop and since

cos = therefore, we have

B =

The direction of magnetic field is as shown below.

63. Using Biot Savart’s law, derive an expression for the magnetic field intensity at the centre of a current carrying circular coil.

Ans. Consider a circular loop of radius r carrying I and having centre at O as shown in figure below.

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Consider a small current element dL on the loop. Then by Biot Savart’s law the magnitude of the magnetic field at the centre of the loop due to the current element we have

dB = ….(1)

In this case the angel between the current element dL and the radius vector is 90o therefore equation 1 can be written as

dB = ….(2)

The circular loop can be considered to be consisting of such small elements placed side by side, and then the magnetic intensities of these elements will be in the same direction.Thus, net intensity of B at the centre of the loop is given by

B =

= ….(3)

But = 2r Therefore, we have

B = ….(4)

64. A charge ‘q’ moving in a straight line is accelerated by a potential difference ‘V’. It enters a uniform magnetic field ‘B’ perpendicular to its path. Deduce in terms of V an expression for the radius of the circular path in which it travels.

Ans. When the electron moves in a magnetic field, it is acted upon by the magnetic force given by . Since the angle between v and B is 90o then the force is given by F = qvB. This

force is perpendicular to both the velocity vector and the magnetic field vector, hence the particle will move in a circular path with a constant speed v. The necessary centripetal force for circular motion will be provided by the Lorentz magnetic force.Let m be the mass of the charged particle and r the radius of the circular path, then the

necessary centripetal force is given by Hence for the circular motion of the charged

particle we have

= qvB or r =

Since mv = p = where V is the potential through which the electrons has been accelerated. Therefore we have

r =

65. A particle with charge ‘q’ moving with velocity ‘v’ in the plane of the paper enters a uniform magnetic field ‘B’, acting perpendicular to the plane of the paper. Deduce an expression for the time period of the charge, as it moves in a circular path in the field. Why does the kinetic energy of the charge not charge, while moving in the magnetic field?

Ans. Due to the presence of the perpendicular magnetic field the particle moves in a circle of radius r given by

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r = ....(1)

The charged particle moves in a circle of radius r. The time taken to complete one revolution is therefore

T =

=

The magnetic force acts an angle of 90o. Therefore it does not do any work. Hence the kinetic energy of the particle does not change.

66. Using Ampere’s circuital law, derive an expression for the magnetic field along the axis of a toroidal solenoid.

Ans. A toroid is a solenoid of finite length bent in the form of a circle. Let N be the number of turns and I be the current passed through it. If the coils are closely spaced, the field inside the toroidal coil is tangent to the dotted circular path as shown in figure below and is same at all points lying on the dotted line.

Therefore we have

(1)

By Ampere’s circuital law we have

(2)

From equations 1 and 2 we haveB(2r) = 0NI

or B = but = n i.e. number of turns per unit length

Therefore we have B = 0nIThis gives the field inside a toroidal solenoid.

67. Name the elements of the earth’s magnetic field at a place. Explain their meaning?Ans. The elements required to completely specific the earth’s magnetic field at a plane are called

magnetic elements. These elements are(i) Declination. ()(ii) Dip or inclination. ()(iii) Horizontal component of earth’s

magnetic field. (BH)(i) Declination: Declination at a place is defined as the acute angle between the magnetic meridian and the geographic meridian.In figure above angle is called the angel of declination. There are several periodic variation. There are several periodic variations in the value of declination for a given location.(ii) Dip: Dip at a place is defined as the angle between the direction of total intensity of earth’s magnetic field

50


and a horizontal line in the magnetic field and a horizontal line in the magnetic meridian at that place. It is denoted by .Suppose a freely suspended magnet aligns itself along the direction AO as shown in figure, then, this gives the direction of earth’s total magnetic field at that place. Then the angle BAO = is called the angle of dip or simply dip at that point.(iii) Horizontal Component of Earth’s Magnetic Field: It is the component of total intensity of earth’s magnetic field along a horizontal line in the magnetic meridian. It is denoted by BH or H.

68. State and illustrate Curie’s law in magnetism.Ans. Experimentally one finds that the intensity of magnetization (I) of a paramagnetic material is

1. directly proportional to magnetic induction (B) or applied field, and2. inversely proportional to the temperature (T) of the material. i.e.

I B ….(1)

and I = ….(2)

combining the above two equations we have

I = ….(3)

As B H therefore we have

I =

or we have

But = m therefore the above equation becomes m or m = ….(4)

where C is a constant called Curie constant. The above equation is called Curie law. 69. You are given two identical looking bars A and B. one of them is a bar magnet and the other is

an ordinary piece of iron. Give an experiment to identify which one of the two is a bar magnet. You are not to use any additional material for the experiment.

Ans. Take one of the rods in your hand and touch the other by (lying on the table) at the ends and also in the centre. If there is no attraction at the centre but there is attraction at the ends then the rod on the table is a magnet.

70. Write the expression for the magnetic dipole moment for a closed current loop. Give its SI unit. Derive an expression for the torque experienced by a magnetic dipole in a uniform magnetic field.

Ans. The required expression is = MB sin.It is measured in Am2

Consider a uniform magnetic field of strength B. Let a magnetic dipole be suspended in it such that its axis makes an angle with the field as shown in figure below. If ‘m’ is the strength of each pole, the two poles experience two equal and opposite forces ‘mB’ each. These forces constitute a couple which tends to rotate the dipole. Suppose the couple exerts a torque of magnitude .Then = either force arm of the couple = mB AN = mB 2 L sin orSince m 2L is the magnetic dipole moment of the magnet.Therefore = MB sin in vector form we have

71. Sketch the lines of force of a magnetic field around a bar magnet, placed along the magnetic meridian, with its North Pole pointing towards the geographical south. Indicate the position of the neutral points.

Ans. The required figure is as shown below.

51


Here P and Q are the neutral points.72. A magnetized needle suspended freely in a uniform magnetic field experiences torque but no

net force. An iron nail near a bar magnet experiences a force of attraction in addition to torque. Why?

Ans. A bar magnet produces a non-uniform magnetic field. Therefore in this field the nail experiences both a translatory force and a torque. Therefore in this field the nail experiences both a translatory force and a torque.

73. The susceptibility of a magnetic material is – 0.085. Identify the magnetic type of the material. A specimen of this material is kept in a non-uniform magnetic field. Draw the modified field pattern.

Ans. The material is a diamagnetic material as diamagnetic materials have negative susceptibility. The modified field pattern is as shown below.

74. Distinguish between a diamagnetic substance and a paramagnetic substance stating two points of difference.

Ans.

Diamagnetic ParamagneticsThese are the substances, which are feebly repelled by a magnet. E.g. Bi, Zn, Cu Ag, Au, diamond C, NaCl, H2O, Hg, N2, H2 etc Exhibited by solids liquids and gases.When a diamagnetic substance is placed in a magnetizing field, the lines of force prefer not to pass through the substance.

These are the substances, which are feebly attracted by a magnet. E.g. Al, Na, Pt, Mn, CuCl2, O2 etc. Exhibited by solids, liquids and gases.

When a paramagnetic substance is placed in a magnetizing field, the lines of force prefer to pass through the substance.

75. A magnetic needle free to rotate in a vertical position orients itself with its axis vertical at a certain place on the earth. What are the values of (i) angle of dip and (ii) horizontal component of earth’s magnetic field at this place? Where will this place be on the earth?

Ans. The angle of dip is 90o and the horizontal component of earth’s magnetic field is zero. This place is the magnetic pole of the earth.

76. Define neutral point. Draw lines of force when two identical magnets are placed at finite distance apart with their N-poles facing each other. Locate the neutral points.

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Ans. It is a point near a magnet where the magnetic field of the earth is completely balanced by the magnetic field of the magnet. The figure is as shown below.The cross indicates the neutral point.

77. A diamagnetic material and a paramagnetic material of the same shape and size are in turn kept in an external uniform magnetic field. Draw the modification of the magnetic field lines of force in the two cases. How does the intensity of magnetization of a paramagnetic material vary with temperature?

Ans. The modified lines of force are as shown below.

The intensity of a magnetization of a paramagnetic materials depends inversely on its absolute temperature.

78. A uniform magnetic field gets modified as shown below, when two specimens X and Y are placed in it.

(i) Identify the two specimens X and Y.(ii) State the reason for the behaviour of the field lines in X and Y.

Ans. (i) X is a diamagnetic substance and Y is a paramagnetic substance.(ii) This is because the permeability of a diamagnetic substance is less than one and that of a paramagnetic substance is greater than one.

79. A charged particle having a charge q, is moving with a speed of v along the X-axis. It enters a region of space where the electric field is and a magnetic field are both present. The particle, on emerging from the region, is observed to be moving, along the X-axis only. Obtain an expression for the magnitude of B in terms of v and E. Give the direction of B.

Ans. Since the particle continues to move along the X axis therefore the magnetic force acting on it should be completely balanced by the electric force. Since the electric force acts along the Y-axis therefore the magnetic field must be along the Z-axis. Thus is equilibrium

qE = Bqv or v = E/B80. A straight wire of length L carrying a current I, stays suspended horizontally in mid air in a

region suspended horizontally in mid air in a region where there is a uniform magnetic field. The linear mass density of the wire is . Obtain the magnitude and direction of the magnetic field.

Ans. The magnetic fore acting on the straight wire balances the weight of the wire. Therefore in equilibrium we haveMg = BIL, here M = L, therefore we haveLlg = BIL or B = /IgThis field acts horizontally to the length of the wire.

81. Name the three types of magnetic materials which behave differently when placed in a non uniform magnetic field. Give two properties for each of them.

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Ans. The three types of magnetic materials are(i) Diamagnetic (ii) Paramagnetic and (iii) Ferromagnetic

Diamagnetic Paramagnetic FerromagneticThese are the substances, which are feebly repelled by a magnet. e.g. Bi, Zn, Cu Ag, Au, diamond C, NaCl, H2O, Hg, N2, H2

e.t.c. Exhibited by solids liquids and gases.When a diamag-netic substance is placed in a magnetizing field, the lines of force prefer not to pass through the substance.

These are the substances, which are feebly attracted by a magnet. e.g. Al, Na, Pt, Mn, CuCl2, O2 e.t.c. Exhibited by solids, liquids and gases.

When a paramagnetic substance is placed in a magnetizing field, the lines of force prefer to pass through the substance.

These are the substances, which are strongly attracted by a magnet. e.g. Fe, Ni, CO, Fe3O4 etc. Exhibited by solids only, that too crystalline

When a ferromagnetic substance is placed in a magnetizing field, the lines of force prefer to pass through the substance.

82. Show mathematically that the cyclotron frequency does not depend upon the speed of the particle.

Ans. Due to the presence of the perpendicular magnetic field the particle moves in a circle of radius r given by

r = ….(1)

The path of the particles in the dees is a semicircle and the time the particle spends in each dee is

t = = ….(2)

using equation (1)The above time is independent of the radius of the path and the velocity of the charged particle. Now the time period of the cyclotron is twice the time spent by the particle in each dee. Thus

T = 2t = ….(3)

Hence cyclotron frequency or the magnetic resonance frequency is given by

v = ….(4)

which is independent of the speed of the particle.83. Why does a paramagnetic substance display greater magnetization for the same magnetizing

field when cooled? How does a diamagnetic substance respond to similar temperature charges?Ans. (i) The tendency to disrupt the alignment of dipoles (with the magnetizing field) arising from

random thermal motion is reduced at lower temperature, hence, greater magnetization of the given paramagnetic sample.(ii) The induced dipole moment is a diamagnetic simple is always opposite to the magnetizing field, no matter what the internal motion of atoms is. Thus it is independent of the temperature.

84. Is any work done by a magnetic field on a moving charge? Why?Ans. No work is being done by a magnetic field on a moving charge because force due to magnetic

field acts perpendicular to the direction of motion of the charged particle. Consequently, work done W = = FS cos 90o = 0.

85. Define the SI unit of magnetic field. “A charge moving at right angles to a uniform magnetic field does not undergo change in kinetic energy,” why?

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Ans. The SI unit of magnetic field is tesla (T). Magnetic field is said to be 1 tesla if a charge of 1 C, while moving with a velocity of 1 ms–1 at right angles to the magnetic field, experiences a force of 1 N.When a charge is moving at right angle to a uniform magnetic field, no work is being done on the charge by the magnetic field because force due to magnetic field behaves as the centripetal force. Hence, in accordance with work-energy theorem, the kinetic energy of the charge remains unchanged.

86. A stream of electrons traveling with speed v ms–1 at right angles to a uniform magnetic field ‘B’

is deflected in a circular path of radius ‘r’. Prove that

Ans. Let a stream of electrons be traveling with speed v at right angles to a uniform magnetic field B then force due to magnetic field provides the requisite centripetal force which deflects the electron beam along a circular path of radius ‘r’ such that

Bev = or

where e = electronic charge and m = mass of the electron.87. Why do two long parallel conductors carrying current exert force on each other?Ans. We know that a magnetic field is set up around a current carrying straight conductor. Under

the influence of this magnetic field conduction electrons moving in the other conductor (also carrying a current) experience Lorentz force. As the electrons are confined to the conductor, force on electrons show its effect as the force on the conductor.

88. Why do two straight parallel metallic wires carrying current in the opposite directions repel each other?

Ans. Consider two long, parallel, straight wires carrying currents I1

and I2 respectively. Due to current I1, the magnetic field developed at a point P on second wire, as per right hand rule, points into the plane of paper. Now in accordance with Fleming’s left rule second conductor carrying current I2

experiences a force due to magnetic field and the

direction of force is normally away. Thus, it experiences a repulsive force.

89. In Fig. below, the straight wire AB is fixed while the loop is free to move under the influence of the electric currents flowing in them. In which direction does the loop begin to move? Give reason for your answer.

Ans. The loop moves towards the straight wire AB. In the loop in the side nearer to wire AB current I2 is in same direction as I1 and hence attractive force acts. However, in the side farther away from wire AB current I2 is in the opposite direction and the force is repulsive. But as the magnitude of attractive force is attractive in nature and hence, the loop moves towards the wire AB.90. Define the terms ‘Magnetic Dip’ and ‘Magnetic declination’ with the help of relevant diagrams.Ans. Magnetic dip: Magnetic dip at a place is defined as the angle between

the direction of total intensity of earth’s magnetic field and a horizontal line in the magnetic meridian at that place. It is denoted by .

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Magnetic declination: Magnetic Declination at a place is defined as the acute angle between the magnetic meridian and the geographic meridian.In figure below shows both the magnetic dip and the magnetic declination.

91. State the principle of working of a cyclotron. Write two uses of this machine.Ans. It is based on the principle that the positive ions can be accelerated to high energies with a

comparatively smaller alternating potential difference by making them to cross electric field again and again, by making use of a strong magnetic field.Uses: (i) It is used to accelerate charged particles like protons, deutrons, etc. which is used to induce nuclear reaction.(ii) It is used in hospitals to bombard tumour with nuclear radiations.

92. Which one of the two, an ammeter or a milliammeter, has a higher resistance and why?Ans. The shunt resistance connected to convert a galvanometer into an ammeter or a milliammeter

is given by the expression S = where S is shunt resistance, G galvanometer resistance,

I total current through G and S, and Ig galvanometer current. In case of milliammeter I is small.Therefore Smilliammeter > Sammeter. Hence resistance of a milliammeter is greater than that of an ammeter.

93. Derive an expression for the maximum force experienced by a straight conductor of length L, carrying current I and kept in a uniform magnetic field, B.

Ans. Consider a straight segment of a conducting wire; with length L and cross sectional area A, the current is from bottom to top as shown in figure below. The wire is in a uniform magnetic field B perpendicular to the plane of the diagram and directed into the plane. Let us assume that the moving charges are positive.The drift velocity is upward, perpendicular to B. The average force experienced by each charge is

….(1)

Directed to the left as shown in the figure

Since and are perpendicular, the magnitude of the force is given by

f = qvdB ….(2)Let n be the number density of charges i.e. number of charges per unit volume. A segment of the conductor with length L has volume V = A L and contains a number of charges N given by

N = nA L ….(3)Now the total force F on all the charges moving in this segment is

F = Nf = (n AL) qvd B

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= (nqvdA) (LB)But nqvd A = I, therefore the above equation becomes

F = BIL ….(4)

94. Three identical specimens of magnetic materials Nickel, Antimony, Aluminium are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.

Ans. Nickel is ferromagnetic. Antimony is diamagnetic and Aluminium is paramagnetic. Therefore they will show the behaviour as shown in the figures below.

95. Give one example each of a diamagnetic and a paramagnetic material. In what way would each of these tend to move when placed in a non-uniform external magnetic field?

Ans. Diamagnetic Gold,Paramagnetic – Platinum.When a diamagnetic substance is placed in a non-uniform magnetic field it moves from stronger portion of the magnetic field towards the weaker portion of the magnetic field.When a paramagnetic substance is placed in a non-uniform magnetic field it moves from weaker portion of magnetic field towards the stronger portion of the magnetic field.

96. A charged particle enters a region of uniform magnetic field with its initial velocity directed (i) parallel to the field and (ii) perpendicular to the field. Show that there is no change in the kinetic energy of the particle in both the cases.

Ans. The force experienced by a charged particle in a magnetic field is given by the expression .

(i) When the initial velocity is parallel to the magnetic field = 0. F = qvB sin = 0 and no change in energy takes place.

(ii) In this expression the velocity and the force vectors are always perpendicular to each other. Since the direction of motion of the particle is given by the direction of velocity, therefore the displacement of the charged particle will also be perpendicular ( = 90o) to the direction of the force.

57


Now work = change in K.E.Since W = FS cos , and = 90o, therefore work done in zero.Hence change in kinetic energy is zero.

97. Write the relation for the force F acting on a charge q moving with a velocity v through a magnetic field B in vector notation. Using this relation deduce conditions under which this force will be (i) maximum (ii) minimum.

Ans. The expression for force in vector notation is

In scalar form this is written asF = Bqv sin , where is the angle between the velocity vector and the magnetic field vector.Force will be maximum if sin is maximum orSin = 1 or = 90o. This maximum value of force is F = BqvForce will be minimum if sin is minimum orSin = 0 or = 0o. This minimum value of force is F = 0

98. Obtain an expression for the magnetic moment of an electron in a circular orbit of radius ‘r’ and moving with a speed ‘v’. State the rule to find its direction. How does this magnetic moment change when(i) the frequency of revolution is doubled(ii) the orbital radius is halved?

Ans. Equivalent current of an electron revolving in an orbit of an atom is I = ve = where v is

frequency and T is time period of revolution given by

T =

or v =

therefore I = ve

=

So, the magnetic moment of a current loop isM = I A

or M = IA

= r2

or M =

This gives magnetic moment of an electron revolving with speed ‘v’ in a circular orbit of radius r.Direction of magnetic moment is given by right hand screw rule.If fingers are curled in the direction of current, then thumb gives the direction of magnetic field.(i) As M = IA = veA, so when frequency v is doubled, magnetic moment M is doubled.(ii) By principle of conservation of angular momentum mvr = mv‘r’

or vr = v’

or v’ = 2vSo when radius r is halved speed v gets doubled. Therefore magnetic moment remains same.

99. Write the expression for the magnitude of force per unit length, between tow infinitely long parallel straight current carrying conductors. Hence define the SI unit of current.

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Ans. The expression for force is F = The above expression can be used to define ampere,

the SI unit of current.Let I1 = I2 = 1 ampere a = 1m then F = 2 10–7 Nm–1. Thus we have one ampere is that much current which when flowing through each of the two infinitely long straight conductors held 1 metre apart in space, produce a force of F = 2 10–7 N per metre of their length.

100.Obtain an expression for the frequency of revolution of a charged particle moving in a uniform transverse magnetic field. How does the time period of the circulating ions in a cyclotron depend on (i) the speed, (ii) the radius of the path of ions.

Ans. The necessary centripetal force required by charged particle to revolve in a circular path in magnetic field is provided by force due to magnetic field i.e.,

= Bqv

or v =

So, the frequency of revolution is then given by

n =

=

=

Time period of the circulating ions in a cyclotron is

T =

(i) From the above expression we find that time period is independent of the speed v.(ii) Also it follows that time period is independent of the radius of the path of ions.

101.Explain with the help of a labeled diagram the underlying principle, construction and working of a moving coil galvanometer.

Ans. It is an instrument used to detect weak currents in a circuit.Principle: It is based on the principle that, whenever a loop carrying current is placed in a magnetic field, it experiences a torque, which tends to rotate it.Construction: It consists of a rectangular or circular coil made by winding a fine insulated copper wire on an aluminium frame. A thin phosphor bronze strip from a torsion head, which is connected to terminal screw S1, suspends this coil. To the lower portion of the screw is attached a small light mirror which rotates along with the strip and whose deflection can be measured by a lamp and scale arrangement. The lower end of the coil is connected to a fine spring which is connected to another terminal screw S2. The coil hangs in space between the pole pieces of a powerful horseshoe magnet NS as shown in figure below. The pole pieces are made concave cylindrical. This provides a radial magnetic field. Since the field is radial, therefore the plane of the coil remains parallel to the magnetic field in all the orientations of the coil. In between the pole pieces, within the coil, lies a soft iron cylindrical piece called ‘core’. The core does not touch the coil

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anywhere. The whole arrangement is enclosed in a non-magnetic box to protect it from air currents. Three leveling screws are provided at the base.Working: When the closed loop is suspended in the magnetic field, it experiences a torque, which tends to rotate it along a vertical axis. This torque, called the deflecting torque, is given by

= nBIA sin ….(1)where is the angle which the normal to the plane of the coil makes with the direction of the magnetic field. Since the magnetic field is radial therefore the normal to the plane of the coil will always be perpendicular to the applied magnetic field i.e. = 90o and sin = 1, Therefore equation 1 can be written as

d = BInA ….(2)As a result of this torque the coil gets deflected. This produces a twist in the suspension wire, due to which the coil is acted upon by another torque called restoring torque. This torque tends to take the coil back to its original position. If ‘C’ is the restoring torque per unit angular twist and is the angle through which the wire has been turned then the restoring torque is given by

= c …(3)The closed loop is twisted till the restoring torque becomes equal to the deflecting torque. Therefore, in equilibrium, from equations (2) and (3) we have

BInA = c

I = ….(4)

or I = K 102.Explain with the help of a labeled diagram the underlying principle, construction and working

of a cyclotron.Ans. Principle: It is based on the principle that the positive ions can be accelerated to high energies

with a comparatively smaller alternating potential difference by making them to cross electric field again and again, by making use of a strong magnetic field.Construction: It consists of two D-shaped, hollow metal dees D1 and D2. The dees are placed with a small gap in between them. The dees are connected to a source of high frequency alternating potential difference. The dees are evacuated in order to minimize energy losses resulting from collisions between the ions and air molecules. The whole apparatus is placed between the poles of an electromagnet as shown in figure below. The magnetic field is perpendicular to the plane of the dees.

Working: In a cyclotron, particles of mass m and charge q move inside an evacuated chamber in a uniform magnetic field that is perpendicular to the plane of their paths. The alternating potential difference applied between the hollow electrodes D1 and D2, creates an changes precisely twice in each revolution, so that the particles get a push each time they cross the gap. The pushes increase their speed and kinetic energy, boosting them into paths of larger radius. The electric field increases the speed and the magnetic field makes the particles move in circular paths. Due to the presence of the perpendicular magnetic field the particle moves in a circle of radius r given by

r = ….(1)

The path of the particles in the dees is a semicircle and the time the particle spends in each dees is

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t = [by using equation (1)]

The above time is independent of the radius of the path and the velocity of the charged particle. Hence the charged particle spends the same time in each dee and encounters a dee of opposite polarity whenever it comes into the gap between the dees and thus is continuously accelerated.

103.State Biot Sarart’s law for the magnetic field due to a current carrying element. Use this law to obtain a formula for magnetic field due to an infinitely long straight conductor.

Ans. Consider a thin, straight wire carrying a constant current I along the X-axis as shown in figure. Consider an element dL at a distance r from P where the magnetic field has to be found. The direction of the field due to this element is out of the paper. Since is out of the paper. All elements of the conductor give a contribution directly out of the paper at P. Therefore by Biot Savart’s law we have

dB = ….(1)

In order to integrate this expression we must relate , dL and r. From the figure below, from right angled triangle AOP we have

+ = 90o or = 90o – Thereforesin = sin (90o – ) = cos ….(2)

Again in right angled triangle AOP we have

cos = or r = ….(3)

Also tan = or L = a tan ,

DifferentiatingWe get dL = a sec2 d ….(4)Substituting equations 2, 3 and 4 in equation 1 we have

dB =

= ….(5)

The total intensity of the magnetic field can be determined by integrating equation 5 within limits = –1 to = 2 Therefore we have

B = cos d

=

or B = [sin (2) – sin (–1)]

or B = [sin 2 + sin 1] ….(6)

This gives the value of magnetic field at a distance ‘a’ form the conductor.

If the conductor is infinitely long, then 1 = 2 = then equation 6 becomes

B =

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or B = [1 + 1]

or B =

104.Figure below shows a long straight wire of circular cross section (radius a) carrying steady current I. The current is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a. Draw a graph showing the variation of magnetic field for the above two cases.

Ans. Consider a long circular wire of radius ‘a’ carrying a steady current (dc) that is uniformly distributed along the cross section of the wire as shown in figure. Let us calculate the magnetic field in the regions r a and r < a. In region 1 let us choose a circular path of radius r centred at the wire. From symmetry, we find that B is perpendicular of dL at every point on the circular path 1 is I0, therefore by Ampere’s law we have

∮ B.dL = B∮dL = B(2r) = 0I0 ….(1)

or B = ….(2)

for r aNow consider the interior of the wire i.e., region (2) where r < a. In this the current I enclosed by the path is less than I0. Since the current is assumed to be uniform over the area of the wire,Therefore

….(3)

Now applying Ampere’s circuital rule to region (2) we have∮B.dL = B(2r)

= 0I = 0 ….(4)

or B = r for r < a

The magnetic field versus r for this system is as shown in the figure below.

105. (a) With the help of a labeled diagram, explain the principle and working of a moving coil

galvanometer.(b) Two parallel coaxial circular coils of equal radius ‘R’ and equal number of turns ‘N’, carry equal currents ‘I’ in the same direction and are separated by a distance ‘2R’. Find the magnitude and direction of the net magnetic field produced at the midpoint of the line joining their centres.

Ans. (a) It is an instrument used to detect weak currents in a circuit.Principle: It is based on the principle that, whenever a loop carrying current is placed in a magnetic field, it experiences a torque, which tends to rotate it.

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Working: When the closed loop is suspended in the magnetic field, it experiences a torque, which tends to rotate it along a vertical axis. This torque, called the deflecting torque, is given by

= n BI A sin ….(1)

where is the angle which the normal to the plane of the coil makes with the direction of the magnetic field. Since the magnetic field is radial therefore the normal to the plane of the coil will always be perpendicular to the applied magnetic field i.e., = 90o

and sin = 1, therefore equation 1 can be written asd = BInA ….(2)

As a result of this torque the coil gets deflected. This produces a twist in the suspension wire, due to which the coil is acted upon by another torque called restoring torque. This torque tends to take the coil back to its original position. If ‘C’ is the restoring torque per unit angular twist and is the angle through which the wire has deflected then the restoring torque is given by

r = C ….(3)The closed loop is twisted till the restoring torque becomes equal to the deflecting torque. Therefore, in equilibrium, from equations (2) and (3) we have

Bin A = C = ….(4)

(b) The magnetic field at a distance R from a circular coil is given by the expression

B =

For coil 1 B1 =

For coil 2 B2 =

Both are directed in the same direction, therefore the resultant magnetic field at the centre is

B =

106. (a) State Biot-Savart’s law. Using this law, derive the expression for the magnetic field due to a current carrying circular loop of radius ‘R’, at a point which is at a distance ‘x’ from its centre along the axis of the loop.(b) Two small identical circular loops, marked (1) and (2), carrying equal currents, are placed with the geometrical axes perpendicular of each other as shown in the figure. Find the mag-nitude and direction of the net magnetic field produced at the point O.

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Ans. (a) It states that the magnetic field due to a small current element at a distance r from it is

given by dB =

Consider a circular loop of wire of radius R located in the YZ plane and carrying a steady current I as shown in figure below. Let us calculate the magnetic field at an axial point P a distance x from the centre of the loop. From the figure it is clear that any element dL is perpendicular to furthermore all the elements around the loop are at the same distance r from P, where r2 = x2 + R2. Hence by Biot Savart’s law the magnetic field at point P due to the current element dL is given by

dB = ….(1)

The direction of the magnetic field dB due to the element dL is perpendicular to the plane formed by and dL as shown in figure above. The vector dB can be resolved into components dBx along the X-axis and dBy which is perpendicular to the X-axis. When the components perpendicular to the X-axis are added over the whole loop, the result is zero. That is, by symmetry any element on one side of the loop will set up a perpendicular component that cancels the component set up by an element diametrically opposite it. Therefore it is obvious that the resultant magnetic field at P will be along the X-axis. This result can be obtained by integrating the components dBx = dB cos. Therefore, we have

B = ∮dB cos = ….(2)

where the integral is to be taken over the entire loop since , x and R are constants for all elements of the loop and since

cos = therefore, we have

B =

(b) The magnetic field at O due to the circular loop 1 is B1 = directed towards

left.

The magnetic field at O due to the circular loop is B1 = directed upwards

The net magnetic field is therefore

B =

The direction of the net magnetic field is 45o with the axis of the loop as shown in the figure below.

107.Derive an expression for the magnetic field along the axis of an air-cored solenoid, using Ampere’s circuital law.

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Sketch the magnetic field lines for a finite solenoid. Explain why the field at the exterior mid-point is weak while at the interior it is uniform and strong.

Ans. A solenoid is a coil of wire with a length, which is large as compared with its diameter. Consider an ideal solenoid carrying current I and having n turns per unit length.(i) Consider a rectangular path abcd of length L as shown in figure below with cd just

outside the solomoid along cd B = 0. Let us apply Ampere’s circuital law to this rectangular path, so that we have

= = 0 + 0 + 0 = BL ….(1)

since ∵ = 90o

∵ B = 0

But by Ampere’s circuital law we have∮ = 0NI = 0(nL)I ….(2)From equations 1 and 2 we haveBL = 0 nLI or B = 0nIThis gives the value of magnetic field inside a solenoid.

(ii) The sketch is as shown below.

The magnetic field gets added inside the solenoid whereas it is not added outside the solenoid

108.Distinguish the magnetic properties of dia-, para-and ferro-magnetic substances in terms of

(i) susceptibility, (ii) magnetic permeability and (iii) coercivity. Give one example of each of these materials.

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Draw the field lines due to an external magnetic field near a (i) diamagnetic, (ii) paramagnetic substance.

Ans. The distinct is shown in the table below.

External magnetic field near (i) diamagnetic and (ii) paramagnetic are as shown below.

109.How will a dia-, para- and a ferromagnetic material behave when kept in two examples of each of these materials name two main characteristics of a ferromagnetic material which help us to decide its suitability for making (a) a permanent magnet, (ii) an electromagnet. Which of these two characteristics should have high or low values for each of these two types of magnets?

Ans. When kept in a non-uniform magnetic field.(i) diamagnetic substances moves from stronger to weaker parts of the field.(ii) para-magnetic substances moves from weaker to stronger parts of the field.(iii) ferromagnetic substances have greater tendency to move from weaker to

stronger parts of the field.Example of(i) diamagnetic substances are Copper and Gold(ii) paramagnetic substances are Platinum and Aluminium(iii) ferromagnetic substances are Iron and Cobalt.Characteristics of ferromagnetic material deciding its suitability for making a (i) permanent magnet have very large permeability and high retentivity.(ii) Electromagnet have low retentivity and small hysteresis loss.For permanent magnet retentivity and coercivity should be large and for electromagnet retentivity and coercivity should be small.

Dia Para FerroSusceptibility Very small and

negativeSmall and positive High and positive

Permeability Less than one Greater than one Extremely large compared to one

Coercivity …. …. ExistsExample Bi, Cu, Pb, Ag,

AuAl, Na, Ca, Pt Fe, Ni, Co.

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C.B.S.E Study Material 2 (Unit-1)

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