ing Charges and Magnetism · 2018. 10. 2. · Ampere's circuital law to calculate the magnetic...

- CHAPTER 04

,

Moving Charges"and Magnetism

Chapter Analysis w.r.t, Lost 3 Yeor's Boord ExomsThe analysis given here gives you an analytical picture of this chapter and will help you toidentify the concepts of the chapter that are to befocussed more from exam point of view.

Number of Questions asked in last 3 years----------

2015 2016 2017Delhi I All India Delhi ·All India Delhi All India

Very Short Answer (1 mark) lQ lQ lQShort Type I Answer (2 marks) - lQShort Type II Answer (3 marks) lQ lQ lQ

Long Answer (5 marks) lQ, Value Based Questions (4 marks) I

I

• In 2015, only one question of 5 marks based on Biot-Savarts Law was asked in Delhi set.• In 2016, in Delhi set, one question of 1mark based on Moving coil Galvanometer and

10 ..rs one question of 3 marks based on Biot-Savart's Law were asked.(. A iL ~' In 2017, in Delhi set, one question of 1 marks based on Induced Current, one question of

3 marks based on Magnetic Field were asked. In All India set, one question of 1markbased on Magnetic Field Lines, one question of 2 marks based on Magnetic LorentzForce and one question of 3 marks based on Magnetic Field were asked.

On the basis of above analysis, it can be said that from exam point of view Biot-Savarts Law,Moving Coil Galvanometer, Magnetic Field and Magnetic Lorentz Force are mostimportant concepts of the chapter.

[TOPIC1] Magnetic Field Laws andtheir Applications

1.1 Magnetic FieldThe space in the surroundings of a magnet or acurrent carrying conductor in which its magneticinfluence can be experienced is called magneticfield. Its SI unit is tesla (T).

Oersted ExperimentOersted experimentdemonstrated thatthe current carryingconductor carryingmoving chargesproduces magneticfield around it.When key Kisclosed, thendeflection occurs in the compass needle andvice-versa.

~~K

A- 0-~ Conducting\V wire

Composs needle

Magnetic Field due to a CurrentElement (Biot-Sovort's Low)According to this law,the magnetic field dueto small current carryingelement dl at any nearbypoint P is given by

dB = h.Idlsin941t r-

or dB =h. Idl r41t Irl2 X

and direction is given by Current carrying conductorAmpere's swimmingrule or right hand thumb rule.

where, h= 1O-7T-m/A=1O-7H/m41t

and Ilo = permeability of free space or vacuum andr = distance of point P from current carryingelement.

Permittivity and PermeabilityThe relation between Ilo' Eo and c is

1 2--=c1l0Eo

where, c is velocity of light, Eo is permittivity of freespace and Ilo is magnetic permeability.

Some Important Points Related toMagnetic Field .

(i) Magnetic field at thecentre of a circularcurrent carryingconductor/coil.

B=lloI2r

where, r is the radius of a circular loop.

For N turns of coil, B = lloN12r

(ii) Magnetic field at the centre of semi-circularcurrent carrying conductor

B = 1101

4r

I

I

(ill) Magnetic field at the centre of an arc ofcircular current carrying conductor whichsubtends an angle 9 at the centre

B=hJ941t r

124

(iv) Magnetic field at any point lies on the axisof circular current carrying conductor

2B = 1l0Ia

2 (r2 + a2)3/2

rx-, "r2 + a2, ---a' <,, ---, ---_ P

M

(v) Magnetic field due to straight currentcarrying conductor at any point P at adistance r from the wire is given by

l--~------- PLong straight wire

110 21 1B=-·- => Boc-41t r r

1.2 Ampere's Circuital LawThe line integral of themagnetic field B aroundany closed path in vacuumis equal to 110 times of thetotal current I threadingthrough the closed circuiti.e. loop. Mathematically,f B· dl= 1101

I

tMagneticfield lines

Important PointsRelated to Magnetic Field

I

• Magnitude of magnetic field of a straight wire

using Ampere's law B = 110121tT

(i) Magnetic field due to a straight solenoid

i):-~:-:[-]-j

o ehopterwise eBSE Solved Papers PHYSICS

(a) At any point inside the solenoid,B = 1l0nI

where, n = number of turns per unitlength.

(b) At points near the ends of air closedsolenoid,

1B = -llonI

2(ii) Magnetic field due to a toroidal sole~

~,'(ffflA415 ~//-"V~ (/ ::s~ ::fv1----r~

(a) Inside the toroidal solenoid, B = 11onINHere, n =-

21tr

and N = total number of turns(b) In the open space, interior or exterior of

toroidal solenoid, B = 0

B

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions

1. A bar magnet is moved in the directionindicated by the arrow between two coilsPQ and CD. Predict the direction of theinduced current in each coil. All India 2017PCJQ ---+ CCJONc==JS

A A

2. Draw the magnetic field lines due to acurrent carrying loop. Delhi 2013C

CHAPTER 4 : Moving Chorges and Magnetism

o 2 Marks Questions3. Two identical

circular loops Pand Q, each ofradius randcarrying equalcurrents are keptin the parallel planes having a commonaxis passing through O. The direction ofcurrent in P is clockwise and in Q isanti-clockwise as seen from 0 which isequidistant from the loops P and Q.Find the magnitude of the net magneticfield at O. Delhi 2D12

4. A long solenoid of length L havingN turns carries a current I. Deduce theexpression for the magnetic field in theinterior of the solenoid. All India 2011C

5. Obtain with the help of a necessarydiagram, the expression for the magneticfield in the interior of a toroid carryingcurrent. HOTS;All India 2011C

6. A straight wire of length L is bent into asemi-circular loop. Use Biot-Savart's lawto deduce an expression for the magneticfield at its centre due to the current Ipassing through it. Delhi 2011C

7. State Ampere's circuital law. Show throughan example, how this law enables an easyevaluation of the magnetic field when thereis a symmetry in the system? All India 2010

8. State Biot-Savart's law. A Zcurrent I flows in aconductor placed dlperpendicular to the plane O'r---.-"!:"""- yof the paper. Indicate the Pdirection of the magnetic Xfield due to a small elementdl at a point P situated at a distance r fromthe element as shown in the figure. Delhi 2009

9. A wire of length L is bent round in theform of a coil having N turns of sameradius. If a steady current I flows throughit in clockwise direction, then find themagnitude and direction of the magneticfield produced at its centre. Foreign 2009

125

y10. An element M = Sxlis placed at theorigin (as shown infigure) and carries acurrent I = 2 A.Find out themagnetic field at apoint P on theY -axis at a distanceof 1.0 m due to theelement tu = w em. Also, give thedirection of the field produced. Delhi 2009C

p

.0~/tJ

II

Zl

o 3 Marks Questions11. (i) State Biot - Savart's law and express

this law in the vector form.(ii) Two identical circular coils, P and Q

each of radius R, carrying currents1 A and 13A respectively, are placedconcentrically and perpendicular toeach other lying in the XY and YZplanes. Find the magnitude anddirection of the net magnetic field atthe centre of the coils. All India 2017

12. Two identical loops P and Q each of radius5 cm are lying in perpendicular planessuch that they have a common centre asshown in the figure. Find the magnitudeand direction of the net magnetic field atthe common centre of the two coils, if theycarry currents equal to 3A and 4A,respectively.

Q

P •

A r·/ •

Delhi 2017

13. Use Biot-Savart's law to derive theexpression for the magnetic field on theaxis of a current carrying circular loop ofradius R.Draw the magnetic field lines due to acircular wire carrying current (1).Delhi 2016

126

14. (i) State Ampere's circuital lawexpressing it in the integral form.

(ii) Two long co-axial insulated solenoids81 and 8 2 of equal length are woundone over the other as shown in thefigure. A steady current 1flowsthrough the inner solenoid 81 to theother end B which is connected to theouter solenoid 82 through which thesome current I flows in the oppositedirection so, as to come out at end A.If n1 and n2 are the number of turnsper unit length, find the magnitudeand direction of the net magneticfield at a point

(a) inside on the axis and(b) outside the combined system.

Dllhl2014

15. (i) How is a toroid different from asolenoid?

(ii) Use Ampere's circuital law to obtainthe magnetic field inside a toroid.All India 2014C

16. Figure shows a long straight wire of acircular cross-section of radius a carryingsteady current I.The current I isuniformly distributed across thiscross-section. Derive the expressions forthe magnetic field in the region (i) r < aand (ii) r > a. All India 2011C

1-

I I \I I ,I I II I II I I\ , I

~ •••••••••• ~~~~~~ ••••• ~.I'... .•..... _-_ ....

[2J ehapterwise eSSE Solved Papers PHYSICS

17. A long straight wire of a circularcross-section of radius a carries a steadycurrent I. The current is uniformlydistributed across the cross-section. ApplyAmpere's circuital law to calculate themagnetic field at a point in the region for(i) r < a and (ii) r > a. Deihl 2010

[ZJ 5 Marks Questions18.

19.

(i) Write using Biot-Savart law, theexpression for the magnetic field Bdue to an element dl carryingcurrent I at a distance r from it in avector form.Hence, derive the expression for themagnetic field due to a currentcarrying loop of radius R at a point Pand distance x from its centre alongthe axis of the loop.

(ii) Explain how Biot-Savart law enablesone to express the Ampere's circuitallaw in the integral form, uiz,

fB.dl =~oIwhere, I is the total current passingthrough the surface. Oelhl 2015

(i) State Ampere's circuital law. Usethis law to obtain the expression forthe magnetic field inside an air coredtoroid of average radius r, having nturns per unit length and carrying asteady current I.

(ii) An observer to the left of a solenoidof N turns each of cross-section areasA observes that a steady current Iflows in the clockwise direction.Depict the magnetic field lines due tothe solenoid specifying its polarityand show that it acts as a barmagnet of magnetic momentm = NIA All India 2015

A 1"---"I II I

t :\ \

N\ \'--'w' '

I I II I II I II I I\ \ \\ \ \, , \

CHAPTER 4 : Moving Charges and Magnetism

20. Twovery small identical circularloop (1) and (2)carrying equal current I areplaced vertically (with respect to the planeof the paper) with their geometrical axesperpendicular to each other as shown inthe figure. Find the magnitude anddirection of the net magnetic field producedat the point O.IG~2-----------____~o

II

: xIII

(2)

I

Delhl2D14

21. State Biot-Savart's law expressing it in thevector form. Use it to obtain the expressionfor the magnetic field at an axial pointdistance d from the centre of a circular coilof radius a carrying current 1. Also, findthe ratio of the magnitudes of the magneticfield of this coil at the centre and at anaxial point for which d = a.Ji Delhl2013C

22. State Biot-Savart's law and give themathematical expression for it.Use this law to derive the expression forthe magnetic field due to a circular coilcarrying current at a point along its axis.How does a circular loop carrying currentbehave as a magnet? Deihl 2011

23. (i) Using Ampere's circuital law, obtainthe expression for the magnetic fielddue to a long solenoid at a point insidethe solenoid on its axis.

(ii) In what respect, is a toroid.differentfrom a solenoid? Draw and comparethe pattern of the magnetic field linesin the two cases.

127

(iii) How is the magnetic field inside agiven solenoid made strong?

. All Indio 2011

24. (i) State Ampere's circuital law.(ii) Use it to derive an expression for

magnetic field inside along the axisof an air cored solenoid.

(iii) Sketch the magnetic field lines for afinite solenoid. How are these fieldlines different from the electric fieldlines from an electric dipole?Foreign 2010

25. (i) Using Biot-Savart's law, deduce anexpression for the magnetic field onthe axis of a circular currentcarrying loop.

(ii) Draw the magnetic field lines due to.a current carrying loop.

(iii) A straight wire carrying a current of12 A is bent into a semi-circular arcof radius 2.0 em as shown in thefigure. What is the magnetic field Bat 0 due to

(a) straight segments,(b) the semi-circular arc?

Foreign 2010

26. (i) State Ampere's circuital law. Showthrough an example, how this lawenables an easy evaluation of thismagnetic field when there is asymmetry in the system?

(ii) What does a toroid consist of? Showthat for an ideal toroid of closelywound turns, the magnetic field

(a) inside the toroid is constant.(b) in the open space inside an

exterior to the toroid is zero.All Indio 2010C

o Explanations1. By applying Lenz's law, we can find out direction

of current in the coil. On the right hand side coil,South pole is approaching towards the coil, so atend C, South pole will be produced and on theleft hand side, North pole is moving away, so at

. end Q coil, South pole will be produced. (1)

2. Magnetic field lines due to a current carrying loopare given by

s3. To calculate net magnetic field at point 0, first of

all, calculate the magnetic field at point 0 due toboth coils with direction. By vector addition ofthese two magnetic fields, net magnetic field canbe obtained.Magnetic field at 0 due to two rings will be insame direction (Q~ P, along the axis) and ofequal magnitude. (1/2)

B = Bl + B2 but B2 = Bl

B = 281 = 2 [ 1101r2 ]2 (r2 + r2)3/2

2 2B - 110Ir _ 110Ir- (2r2)3/2 - i!2r3

(1/2)

(112)

(1/2)

4. Figure shows the longitudinal sectional view oflong current carrying solenoid. The current comesout of the plane of paper at points marked.

oQI+-- I ----+l

dj-----------"jcI t---/1i.~..:::...~..·~f·.~... ~.• :::••. ~•••~•• ~•.• g'.

a L. -----. b

~B -----------/---~~mmmm~~1

The Bis the magnetic field at any point inside thesolenoid.Con;idering the rectangular closed path abcda.Applying Ampere's circuital law over loop abcda.(1Jf B· dl= 110x (Total current passing through

loop abcda)

J>.dl+ J:B.dl+ fcdB.dl+ J:B.dl=llo(~li)

rn

where, !!.. = number of turns per unit lengthL

ab = cd = I = length of rectangle.J: Ed! cos 0° + J: Bd! cos 90° + 0

+1: Bd! cos 90°=110 (~)li

B J: d!=llo (~)li => Bl=llo (~)li

=> B=llo (~}

M B=llo~ mwhere, n = number of turns per unit length.This is required expression for magnetic fieldinside the long current carrying solenoid.

5. Toroid is an endless solenoid to calculate themagnetic field in the interior of toroid, Ampere'scircuital law can be obtained.Toroid is a hollow circular ring on which a largenumber of insulated turns of a metallic wire areclosely wound.The direction of the magnetic field at a point isgiven by tangent to the magnetic field line at thatpoint. (1)

B/--- ......•..,I (J;>{\\\V)" I••..... ./--

....(i)

as fB.dl = 110I x Number ofturns ....(ii)

If n be the number of turns/Unit length, then totalnumber of turns = n x 2ltr

So, fS dl = 110n x 2ltrl ....(ill)

CHAPTER 4 : Moving Charges ond Magnetism

From Eqs. (i) and (ill), we getB 21tT = ~o n21tTI ~ B= ~o nl

Q.

Applying Ampere's circuital law over loop,we havef B· dl = ~o x Current passing through the loop (1)

6. When a straight wire is bent into semi-circularloop, then there are two parts which can producethe magnetic field at the centre, one is circularpart and other is straight part due to which fieldis zero.

II

r/I__ -.L-----.-----.l..- __-

r C r

Length L is bent into semi-circular loop.Length of wire = Circumference

of semi -equal circular wire~ L = 1tT

LT = - ... (i)

1t

Considering a small element dl on current loop.The magnetic field dB due to small current elementldl at centre C. Using Biot-Savart's law, we have

dB = &.. Idl sin 90° ['r:' . .1 r, :.e = 90°)41t ,2 (112)..

dB=&'. Idl41t ,2

:. Net magnetic field at C due to semi-circular loop,B = f ~o Idl

semicircle 41t ,2

B = ~o I: f dl41t,2 semicircle (1/2)

B = ~o . .!...- L41t ,2

But, r = ~1t

B = ~o . ....!!::....- = ~o x IL X 1t2 ~ B = ~o I1t41t (L/1t)2 41t L2 4L

This is the required expression.

129

7. As, Ampere's circuital law states that the lineintegral of magnetic field B around any closedloop is equal to ~o times the total currentthreading through the loop. (1)

i.e. fB.dl=~oI

I I

Bounda~ Surtace

To explain the Ampere's circuital law consider aninfinitely long conductor wire carrying a steady

':~:~:-"-r:-::~:::'(tMg,mlal to--------r-------- circumference)

In order to determine the magnetic field at point Pwhich is situated at a distance R from the centreof the circular loop around the conductor wire, B(magnetic field) is tangential to circumference ofthe loop. (1)

Now,fB.dl= fBdl=B21tR=~oI

~ B = ~o I [From Ampere's circuital law)21tR

The direction of magnetic field will be determinedby right hand rule.

8. Biot-Savart's law This law states that themagnetic field (dB) at point P due to small currentelement Idl of current carrying conductor is(i) directly proportional to the Idl (current)

element of the conductor.dB oc Idly

p

x(1) (ii) directly proportional to sin e dB oc sin e

where, e is the angle between dl and r.

130

(ill) inversely proportional to the square of thedistance of point P from the current element.

dB<x~~ rnCombining all the inequalities

dB <X Idl sin 9 = 110 ,Idl sin 9,2 47t ,2where, 110 = 10-7 T-m/A for free space.

47tThe direction of magnetic field can be obtainedusing right hand thumb rule.In v~ctor form, . •

dB = 110 ,Idlx r47t Ir 12 (1/2)

zCurrentelement

)4-~-dB: p

~~x (112)

The direction of magnetic field will beperpendicular to Y-axis along upward in the planeof paper.

9. When a straight wire is bent inthe form of a circular coil ofN turns, then the length of thewire is equal to circumference ofthe coil multiplied by thenumber of turns. Let the radiusof coil be r,As, the wire is bent round in the form of a coilhaving N turns.:. N x circumference of the coil

=Length of the wire(27t') x N = L

L,=--27tN

Magnetic field at the centre due to N turns of a coilis given by

B = 110 (NI) ~ B = 110 (NI) [From Eq. (i)]

2r 2L~)B = 1107tN21

L (1Ya)

The direction of magnetic field is perpendicular tothe plane of loop and entering into it. (112)

o... (i)

Of 1

<jfn

o Chapterwise eSSE Solved Papers PHYSICS

10. Biot-Savart's law states thatdB = 110 ,Idl x r

47t Irl2 (1)

Here, Idl = 2x wi

Sx = wcmill = illx1= 2A,' =1 m

dB = 110 • (2wi x j)47t (1)2

r = j

Irl=1 m

dB = 110w k ~ IdBI = 110w27t 27t

and direction along +Z-axis.

(112)

(1/2)

11. (i) Biot-Savart's Law This law deals with themagnetic field induction at a point due to asmall current element (a part of anyconductor carrying current).

y___ Current- element

-::- X

According to Biot-Savarr's law, the magnitudeof magnetic field intensity (dB) at a point Pdue to a current element is given by, (1/2)

dB <X idlsin9,2

This relation is called Biot-Savart's law.If conductor is placed in air or vacuum, then

dB = 110 idlsin947t ,2

where, ~ is a proportionality constant and 11047t

is the permeability of free space.110 = 47t X 10-7 Tm/A or weber/ ampere-metre.

Thus, in vector notationdB <X idl x r = 110 . idl x r

,3 47t,3

The above expression holds when the mediumis vacuum. The magnitude of this field is

IdBI = 110 . Idlsin947t ,2 (112)


(ii)

p

12.

Magnetic field due to circular wire P,

Bp = ~o X 2reI1 (along vertically upwards)4re R

= ~OIl2R

Magnetic field due to circular wire 0..BQ = ~o X 21tl2

4re R

(along horizontal towards left)= ~oI2

2R (112)Net magnetic field at the common centre ofthe two coils,

B=~

~ B= (~)\(~rB= (~)\II2+Ii)

~ B=~O~II2+Ii2R

B = 4re X 10-7 ~(l)2 + (.,[3)22xR

B = 4re X 10-7 TeslaR

Resultant magnetic field makes an angle awith direction of BQ, which is given by

tana= Bp =~BQ .,[3

~ a = 30° (1/2)

Magnetic field due to circular loop P,

s, = Iloip2r

Q

p

131

Magnetic field due to circular loop Q,

B - ~OiQQ---

2T (1)

13.

So, net magnetic field at the common centre ofthe loop is,

Ene, = ~B; + B~ = (~;:pr + (~;;Qr- ~o ~ _ 4re X 10-7 5- 2 10-5 T--"zP+ZQ- X - rex

2r zx 5xlO-2 (1)

Resultant magnetic field makes an angle ewithBQ which is given by, J01-'51[ t:'

tancp= Bp = ip =}.BQ iQ. 4 (1)

Let us consider a circular loop of radius a withcentre C. Let the plane of the coil beperpendicular to the plane of the paper andcurrent I be flowing in the direction as shown inthe figure. Suppose P is any point on the axis at adirection r from the centre.

Lm Q

I

""'" ~ (2 + a2.•.••••........

dB

dB cos odB

'------" Q'

Now, consider a current element Idl on top (L )where current comes out of paper normally,whereas at bottom (M) enters into the plane ofpaper normally.

LP .l IdlAlso, MP .lIdl

LP = MP = ~r2 + a2

The magnetic field at point P due to currentelement Idl. According to Biot-Savart's law,

dB = ~o . Idl sin 90°4re (r2 + a2)

where, a = radius of circular loop,r = distance of point P from centre along the axis.

The direction of dB is perpendicular to LP andalong PO.. where PQ.l LP. Similarly, the samemagnitude of magnetic field is obtained due tocurrent element Idl at the bottom and direction isalong PQ', where PQ' .l MP.

132 o ehopterwise eBSE Solved Papers PHYSICS

J

Now, resolving dB due to current element at LandM dB cos cp components balance each other and netmagnetic field is given by

B = fdB sin cP = f Ilo (~). a41t r2 + a2 ~r2 + a2

[':In MeL, sin cP = a 1

~r2 + a2

= Ilo fa .!dl = Ilo fa (21tQ)41t (r2 + a2)3/2 j 41t (r2 + a2)3/2

B = llofa2

2(r2 + a2)3/2

_ lloN fa2For N turns, B - 2 2 3/2 Tesla.

2(r + a ) (2)

The diagram of magnetic field lines due to a circularwire carrying current I is.

or

(1)

14. (i) Ampere's circuital law states that the lineintegral of magnetic field (B) around any closedpath in vacuum is Ilo times the net current (1)threading the area enclosed by the curve.Mathematically, f B·dl = Ilo I

Ampere's law is applicable only for an Amperianloop as the Gauss's law is used for Gaussiansurface in electrostatics. (1)

(il) According to Ampere's circuital .law, the netmagnetic field is given by B = Ilo n i (1)

(a) The net magnetic field is given byBnet = B2 -BI

=IlOn/2-llonl11=llo1(n2-nJl

The direction is from B to A. (1) .

(b) As the magnetic field due to SI is confinedsolely inside SI as the solenoids areassumed to be very long. So, there is nomagnetic field outside SI due to current inSI' similarly there is no field outside S2':. Bnet = 0 (1)

15. (i) A toroid can be viewed as a solenoid which hasbeen bent into circular shape to close on itselfrn

(il) Refer to Ans. 14 (il). (1)

."[':/2=11=1]

16. In these types of questions, first of all we haveto calculate the current per unit area ofcross-section, so that we can calculate thecurrent in each loop, then only we can find themagnetic field. The current is distributeduniformly across the cross-section of radius a.

Loop(r < a)

C .. I:. urrent passes per urut cross-section = --1ta2

:. Current passes through the cross-section ofradius r is

r = (~ x 1tr2) = 1r2

... (i)_ 1tQ2 a2 (112)

(i) cCder a loop of radius r whose centre liesat the axis of wire where, r < a as shown infigure inside the wire.Applying Ampere's circuital law,

fB.dl=llo1'

f Bdl cosO° = Ilo ( ~2) [From Eq. (i)]

1r2B.!dl=llo -t a2

" Ir2" IrB x 21tr = _•..._0_ ~ B = _•..._0_

a2 21ta2

(1/2)

(1/2)

(ii)~ B cc r

Considering a loop of radius r whose centrelies at the axis of wire and (r > a) as shown inouter dotted line.:. Current I threads the loops.Applying Ampere's circuital law,

fB.dl=llo1(1/2)

f Bdlcosoo=llo1 ~ Bf dl=llo1

B x 21tr =llo1 ~ B = llo121tr (1)


Boc~r

Thus, the field B is proportional to r as we movefrom the axis of cylinder towards its surface andthen decreases as ~ .

r

17. Refer to Ans. 16.18. (i) According to Biot-Savart's law,

placed in air or vacuum,Currentelement

(3)

conductor is

6Id I H--, -'----+lB,

~

then dB = llo idlsin64n r2

In vector form, Biot-Savart's law can be written as,dB oc Idl x r

r3

_lloIdlxr- 4n-r-3-

Let us consider a circular loop of radius a withcentre C. Let, the plane of the coil beperpendicular to the plane of the paper andcurrent I be flowing in the direction shown in thefigure.

I

---..,.QdB

dB cos edB

~_---2llQ'

Suppose P is any point on the axis at a direction rfrom the centre.Now, consider a current element Idlon top L,where current comes out of paper normallywhereas at bottom (M) enters into the planepaper normally... LP 1. dlAlso, MP 1. dl

.. MP = LP = ~r2 + a2

The magnetic field at P due to current element ldl,

133

According to Biot-Savart's law,dB= llo .Idlsin90°.

4n (r2 + a2)

where, a = radius of circular loopr = distance of point P from centre along the axis.The direction of dB is perpendicular to LP andalong po, where PQ 1. LP. Similarly, the samemagnitude of magnetic field is obtained due tocurrent element Idl at bottom and direction isalong PQ', where PQ 1. MP.Now, resolving dB due to current element at L andMdBcos$ components balance each other and netmagnetic field is given by

B = fdBSin$= f!:Q.( /dL 2)' a4n r + a ~r2 + 12

[

..:In MC/.:, a 1sm$=-;===

~r2 + a2

B = llo . Ia I.dl4n (r2 + a2)3/2 j

B = llo . Ia (2na) = lloIa2

4n (r2 + a2)3/2 2 (r2 + a2)3/22

For N turns, B = lloNla2(r2 + a2)3/2

The direction is along the axis and away from theloop. (2'12)

(ii) When current in the coil is in anti-clockwisedirection. y

134 [2] Chopterwise C8SE Solved Popers PHYSICS

Consider any arbitrary dosed path perpendicularto the plane of paper around a long straightconductor XY carrying current from X to Y, lyingin the plane of paper.Let, the closed path be made of large number ofsmall elements, whereAB = dll' BC = dl2, CD = dl}Let del' de2, de}, be the angles subtended by thevarious elements at point 0 through whichconductor is passing. Then

de I + de 2 + de J + ... = 21tSuppose these small elements AB, BC, CD, ... aresmall circular arcs ofradii '1' '2' ,} ,... respectively.

dll di2 dl}Then, del = -, de2 = -, de} = -

1j '2 ,}

If BI, B2, B} are the magnetic field induction at apoint along the small elements dl" dl2 ,dl3 ••••••

then from Biot-Savart's law we know that for theconductor of infinite length, magnetic field isgiven by

BI = Ilo 21 ; B2 = Ilo 21; B} = Ilo 21 ...41t '1 41t '2 41t ,}

In case of each elements, the magnetic fieldinduction B and current element vector dl are inthe same direction. Line integral of B aroundclosed path isfB.dl=BI·~ +B2·dl2+B}·dl}+ ...

= BI (d4) + B2(d~) + B}(d4) + ...= Ilo 21d4 + Ilo 21 d~ + Ilo 21 d1

3+...

41t '1 41t T2 41t ,}

= llo21[d4 + d~ + d4 + ... J41t TI T2 T}

= llo21 [del + de2 + de} + ... J= Ilo 21 x 21t = llo141t 41t

=> fB. dl = Ilol, which is an expression of

Ampere's circuital law. (2'12)

19. (i) Ampere's circuital law states that the lineintegral of magnetic induction B around aclosed path in vacuum is equal to Ilo times thetotal current I passing through the surface, i.e.fB. dl = Ilol. (1)

A toroid is a hollow circular ring on which alarge number of turns of a wire are closelywound.Now, consider an air-cored toroid with centreo and in order to determine the magnetic fieldinside the toroid, we consider three amperianloops(loop 1, loop 2 and loop 3).

loop 1loop 2

-O(Jh--1~ ...""-Ioop 3

For loop 1,According to Ampere's circuital law, we have,fB. dl = Ilo(total current)

Total current for loop 1 is zero because nocurrent is passing through this loop.So, for loop 1, fB. 81= 0

For loop 3,According to Ampere's circuital law, we have,

fB. dl = Ilo (total current)

Total current for loop 3 is zero because netcurrent coming out of this loop is equal to thenet current going inside the loop.For loop 2,The total current flowing through the toroid isNI. Where, N is the total number of turns.

fB.dl=llo(Nl) ... (i)

Now, Band dlare in the same direction.fB.dl=Bfdl=> fB.dl=B(21tT) .... (ii)

On comparing Eqs. (i) and (ii), we get=> B = lloN1

21tTNumber of turns per unit length is given by

Nn=-21tT

.. B=lloN1This is the expression for magnetic field insideair-cored toroid. (1'12)

(ii) Since, it is given that the current flows in theclockwise direction for an observer on the leftside of the solenoid. It means that the left faceof the solenoid acts as South pole and rightface acts as North pole. Inside a bar, themagnetic field lines are directed from South toNorth. (1'12)

Therefore, the magnetic field lines are directedfrom left to right in the solenoid. (1)

Magnetic moment of a single current carryingloop is given by, m' = lA .So, magnetic moment of the whole solenoid isgiven by m = Nm' = N(lA) . (1)


20. The magnetic field at a point due to a circularloop is given by

2B = ~O. 2n:la

4n: (a2 + r2)3/2

where, I = current through the loop" = radius of the loopand r = distance of 0 from the centre of the loop.Since I,a and r = x are the same for both theloops, the magnitude of B will be the same and isgiven by (1)

2Bl = B2 = !:.Q... 2n:la

4n: (a2 + x2)3/2

The direction of magnetic field due to loop (1 )will be away from 0 and that of the magneticfield due to loop (2) will be towards 0 as shown.The direction of the net magnetic field will be asshown below: (1)

~

Bnet :

0) :__________________________ 0

90°

x

(2)

The magnitude of the net magnetic field is givenby

_ ~ _ ~o 2.J2itIa2Bnet -" B1 + B2 ~ Bnet - 2 2 3 2

4n: (a + x ) 1 (1)

21. For the statement of Biot-Savart's law Referto Ans. 8. (1)

For field at axial point Refer to AIlS. 18 (i). (1)

In this answer, put r = d.

Magnetic field induction at the centre of thecircular coil carrying current is

B2=!:.Q... 2n:I

4n: a_ ~o 2n;a2I

Bl - 4n:' (a2 + d2)3/2

B1 _ a2xa a3

B2 (a2 + d2)3/2 (a2 + d2)3/2

Bt _ a3 _ a3 _ a3

B2 - (a2 + 3a2)3/2 -(4a2)3i2" - 8a3

[':d = a.J3]

135

22. For Blot-Savart/s law Refer to Ans. 8. (2)

For the magnetic field due to a circular coilcarrying current at a point along its axis. Refer toAns.13. (1)

As current carrying loop has the magnetic heldlines around it which exists a force on a movingcharge. Thus, it behaves as a magnet with twomutually opposite poles. (1)

(1)

The anti-clockwise flow of current behaves like aNorth pole, whereas clockwise flow as South pole.Hence, loop behaves as a magnet. (1)

23. (i) Refer to Ans. 4. (2)(ii) Solenoid is a hollow circular ring having large

number of turns of insulated copper wire on it.Therefore, we can assume that toroid is a bentsolenoid to close on itself.The magnetic fields due to solenoid and toroidis given in figures below: (1)

(1)

Field due to solenoid

,/,,--~------ ---------,3 ' ,, ,, ,, ,, ,I ,I \I \I \I II II II II I\ I, I, I, I, I, ,, ,, ,

"',,---------------,,/

(3)

Field due to toroid

Magnetic field inside the solenoid is a uniform,strong and along its axis also field lines are allmost parallel while inside the toroid field linemakes closed path.

(iii) The magnetic field in the solenoid can beincreased by inserting a soft iron core inside it. (2)

136

24. (i) For statement of Ampere's circuital IawRefer to Ans. 7. (1)

(ii) Refer to Ans. 4. (2)(iii) Magnetic field lines due to a finite solenoid

has been shown as below:Q

(2)

All the magnetic field lines are necessarily closedloops, whereas electric lines of force are not.

25. (i) Refer to Ans. 8 and 18. (1)(ii) Magnetic field lines due to a current carrying

loop is given as below: (2)

(ill) Magnetic field due to straight partB - f~O Idlx r

41t r3 (112)For point 0, dl and r for each element of thestraight segments AB and DE are parallel.Therefore, dl x r = O. Hence, magnetic field dueto straight segments is zero.Magnetic field at the centre due to circularpoint

Magnetic field at the centre of circular coil2

[":Here, coil is half 1

o Chapterwise CBSE Solved Papers PHYSICS

_ 1 (~oI) _ ~oI-- -- ---2 2r 4r

~ B = ~oI = (41t x 10-7) x 124r 4x2xlO-2

= 61t X 10-5 T(1'I2J

26. (i) For statement of Ampere's circuital lawRefer to Ans. 7.For the evaluation of magnetic field for asymmetrical system, we can consider theexample of a current carrying solenoid. Now,refer to Ans. 4. (1V.)

(ii) A solenoid bent into the form of closed loop iscalled toroid. The magnetic field B has aconstant magnitude everywhere inside thetoroid. (112)

Q.

I I

(a) Let magnetic field inside the toroid is B alongthe considered loop (1) as shown in figure.

3

///-------------""',9 B

, ,r ,r ,

I ,I ,I ,I II II II II II II I, I, I, I, ,, ,, ,, ,

-,<, --------_ ,;/' F

Applying Ampere's circuital law,

! B·dl=~o (NI)iloopl

Since, toroid of N turns, threads the loop 1,N times, each carrying current I inside the loop.Therefore, total current threading the loop 1 isNI.~! B·dl=~oNI ~ B! dl=~oNI

iloop1 iloop

B x 21tr = ~o NIB = ~oNI

21tror

(1)


(b) Magnetic field inside the open spaceinterior the toroid. Let the loop (2) is shownin figure experience magnetic field B.No current threads the loop 2 which lie in theopen space inside the toroid.:. Ampere's circuital law

1, B·dl=l.I.o(O)=O:::) B=O~~2 rn

Magnetic field in the open space exteriorof toroid Let us consider a coplanar loop (3)in the open space of exterior of toroid.

137

Here, each turn of toroid threads the loop twotimes in opposite directions.Therefore, net current threading the loop

= NI -Nl = 0:. By Ampere's circuital law,

1, B.dl=l.I.o(Nl-NI)=O"loop 3

:::) B=O

Thus, there is no magnetic field in the open spaceinterior and exterior of toroid. (1)

[TOPIC 2] Lorentz Force and Cyclotron2.1 Force on moving charge in a

uniform Magnetic andElectric Field (Lorentz Force)

Force experienced by a single charged particle qmoving with speed v in a uniform magnetic fieldat an angle a with it is known as magneticLorentz force.

F = q (v x B) [vector form]Magnitude of F = qvB sin a and direction of forceis given by right hand palm rule or Fleming's lefthand rule.

51Unit of Magnetic FieldMagnetic lorentz force, F = qvB sin a=> B= __ F_

qv sin aIf F = IN, q = I C, v = 1ms-1

sin a = 1= sin 90 ° => a = 90 °

.. SI unit of B = • 1 N I(lC)(lms-)

= 1 NA - lm- I = 1 T

SI unit of magnetic field, B = 1tesla (T).

Right Hand Palm RuleIf all four fingers of right hand together points inthe direction of magnetic field and thumb pointsthe direction of motion of positive charge, thenthe palm of right hand faces in the direction offorce.

Magnetic Force on a Charged Particle• Work done by magnetic Lorentz force on charge

particle is zero as F 1- v, hence F is perpendicularto displacement.

• Magnetic force cannot increase the kineticenergy of charge particle.

• The trajectory/path traversed by the chargedparticle is a(i) straight line when angle between vand B is

0° or 180°.

(ii) circle when angle between v and B is 90°.

(iii) helix when angle between v x B is an acuteangle.

• When charge particle enter in a magnetic fieldperpendicularly, then. mv' '" mv(z) -- = qvB (II) radius, r = -

r qB

(iii) T 21tm (. ) f qBIII = -- zv = --qB 21tmq2B2r2

(v) KE=--2m

• When angle between v and B is a, then

(z') di f h u 1 h mv sin a mv.lRa IUS 0 e lCa pat ,r = = --qB qB

where, v.l = vsina= perpendicular componentof velocity.

~Helixes

138

(ii) Time period,

T_ 21tr _ 21tr mvsinB _ 21tm----_. ---

v.L vsine qB qB

(iii) Frequency, f = qB21tm

(i ) P' h 21tm v cosa. 21tmvllIV itc = =--

qB qB

where, vII = vcose = parallel component ofvelocity .

• Net force on a charged particle when bothelectric and magnetic fields are present.

F=q [E+(vxB)]• If a charged particle q accelerated by potential

difference V and speed changes from 0 to v,then,Work done = Change in KE

1 2 ~2qvqV = - mv => v = --2 m

2.2 CyclotronA compact device used to accelerate the positivelycharged ion or particles (like protons, deuterons,a-particles) up to very high speed.

N

CyclotronBased on the principle of magnetic resonance, acharge particle can be accelerated to high speed bypassing it again and again through small region ofoscillating electrical field by making the use ofstrong normal magnetic field.

o ehopterwise eBSE Solved Papers PHYSICS

• KEmax of charge particle accelerated byq2B2R2

cyclotron is KE = ---max 2m

where, R = radius of circular track of chargedparticle.

• Cyclotron cannot accelerate electron andelectrically neutral particles.

PREVIOUS YEARS'.EXAMINATION QUESTIONSTOPIC 201 Mark Questions

1. Write the expression in a vector form forthe Lorentz magnetic force F due to acharge moving with velocity v in amagnetic field B. What is the direction ofthe. magnetic force? Delhi 2014

2. A narrow beam of protons and deuterons,each having the same momentum, enters

. a region of uniform magnetic fielddirected perpendicular to their directionof momentum. What would be the ratio ofthe radii of the circular path described bythem? Foreign 2011

3. Two particles A and B of masses m and2m have charges q and 2q respectively.They are moving with velocities VI and V2

respectively in the same direction, entersthe same magnetic field B acting normallyto their direction of motion. If the twoforces FA and FB acting on them are in theratio of 1 : 2, find the ratio of theirvelocities. Delhi 2D11C

4. A beam of a-particles projected along+ X-axis, experiences a force due to amagnetic field along the + Y -axis. What isthe direction of the magnetic field?All Indio 2010, 2008


5. Use the expression F = q (v x B)to definethe S1 unit of magnetic field. All India 2010C

6. An electron does not suffer any deflectionwhile passing through a region of auniform magnetic field. What is thedirection of the magnetic field? All India 2009

o 2 Marks Questions7. Find the condition under which the charged

particles moving with different speeds inthe presence of electric and magnetic fieldvectors can be used to select chargedpartie' ,of a particular speed. All India 2017

8. Define one tesla using the expression forthe magnetic force acting on a particle ofcharge q moving with velocity u in amagnetic field B. Foreign 2014

9. State the underlying principle of acyclotron. Write briefly how this machineis used to accelerate charged particles tohigh energies. Delhi 2014

10. A particle of charge q and mass m ismoving with velocity v. It is subjected to auniform magnetic field B directedperpendicular to its velocity. Show that itdescribes a circular path. Write theexpression for its radius. Foreign 2012

11. Write the expression for Lorentz magneticforce on a particle of charge q moving withvelocity v in a magnetic field B. Show thatno work is done by this force on thecharged particle. All India 2011

12. An electron and a proton moving with thesame speed enter the same magnetic fieldregion at right angles to the direction ofthe field. Show the trajectory followed bythe two particles in the magnetic field.Find the ratio of the radii of the circularpaths which the particles may describe.Foreign 2010

x x x x

x x x

ee- x x x x

x x x xB

139

13. A deuteron and a proton moving with thesame speed enter the same magnetic fieldregion at right angles to the direction ofthe field. Show the trajectories followedby the two particles in the magnetic field.Find the ratio of the radii of the circularpaths which the two particles maydescribe. Foreign 2010

BZ

14. A charge q movingalong the X-axis with avelocity v is subjectedto a 'uniform magneticfield B acting along theZ-axis as it crosses the ~origin O.

(i) Trace trajectory(ii) Does the charged particle gain

kinetic energy as it enters themagnetic field? Justify your answer.HOTS; Delhi 2009

o'}------_y

x

y15. A point charge ismoving with aconstant velocityperpendicular to auniform magneticfield as shown in thefigure. What shouldbe magnitude anddirection of theelectric field, so that the particle movesundeviated along the same path? Foreign 2009

16. A cyclotron when being used to acceleratepositive ions? (Mass = 6.7 x 1O-27kg,charge = 3.2 x 1O-19q has a magnetic fieldof(1t/2)T. What must be the value of thefrequency of the applied alternatingelectric field to be used in it? All India 2009

x x xB

x x

x x x x x-+q

x X Vx x )(

x x x x x

x x x x x~-----------x

o 3 Marks Questions17. (i) Write the expression for the force

F acting on a particle of mass m andcharge q moving with velocity u in amagnetic field B. Under whatconditions will it move in(a) a circular path and(b) a helical path?

140

(ii) Show that the kinetic energy of theparticle moving in magnetic fieldremains constant. Deihl 2017

18. (i) Write the expression for themagnetic force acting on a chargedparticle moving with velocity v in thepresence of magnetic field B.

(ii) A neutron, an electron and an alphaparticle moving with equal velocities,enter a uniform magnetic field goinginto the plane of the paper as shownin the figure. Trace their paths inthe field and justify your answer.

x x x x x X<X-

X X X X X xn_

x x x x x xe e--------+

x x x x x x

CBSE Delhi 2016

19. A uniform magnetic field B is set up alongthe positive X-axis. A particle of charge qand mass m moving with a velocity ventersthe field at the origin in XY-plane such thatit has velocity components both along andperpendicular to the magnetic field B.Trace, giving reason, the trajectory followedby the particle. Find out the expression forthe distance moved by the particle alongthe magnetic field in one rotation.All India 2011

20. Draw a schematic sketch of the cyclotron.State its working principle. Show that thecyclotron frequency is independent of thevelocity of the charged particle. Delhi 2D11C

21. Explain the principle and working of acyclotron with the help of a schematicdiagram. Write the expression forcyclotron frequency. Delhi 2009

22. Show that the frequency of revolution of acharged particle (In the XY-plane), in auniform magnetic field B (B = Bok), isindependent of its speed. Which practicalmachine makes the use of this fact? Whatis the frequency of the alternating electricfield used in this machine? Delhi 2009C


23. (i) Write an expression for the forceexperienced by a charge q movingwith a velocity v in a magneticfield B. Use this expression to definethe unit of magnetic field.

(ii) Obtain an expression for the forceexperienced by a current carryingwire in a magnetic field. All India 2D09C

o 5 Marks Questions24. (i) Deduce an expression for the

frequency of revolution of a chargedparticle in a magnetic field and showthat it is independent of velocity orenergy of the particle.

(ii) Draw a schematic sketch of acyclotron. Explain the essentialdetails of its construction how it isused to accelerate the chargedparticles? Delhi 2014

25. (i) Draw a schematic sketch of acyclotron. Explain clearly the role ofcrossed electric and magnetic field inaccelerating the charge. Hence, derivethe expression for the kinetic energyacquired by the particles.

(ii) An a-particle and a proton arereleased from the centre of thecyclotron and made to accelerate.

(a) Can both be accelerated at thesame cyclotron frequency? Givereason to justify your answer.

(b) When they are accelerated inturn, which of the two will havehigher velocity at the exit slit ofthe does? Delhi 2012

26. Write the expression for the force F, actingon a charged particle of charge q movingwith a velocity v in the presence of bothelectric field E and magnetic field B.Obtain the condition under which theparticle moves undeflected through thefields. All India 2012C

27. With the help of a labelled diagram, statethe underlying principle of a cyclotron.Explain clearly how it works to acceleratethe charged particles? Delhi 2011


Show that cyclotron frequency isindependent of energy of the particle. Isthere an upper limit on the energyacquired by the particle? Give reason.

28. (i) Draw a schematic sketch of acyclotron, explain its workingprinciple and deduce the expressionfor the kinetic energy of the ionsaccelerated.

(ii) Two long and parallel straight wirescarrying currents of 2A and 5A in theopposite directions are separated bya distance of 1 cm. Find the natureand magnitude of the magnetic forcebetween them. Foreign 2011

29. Draw a schematic sketch of a cyclotron.State its working principle. Describebriefly, how it is used to acceleratecharged particles? Show that the period ofa revolution of an ion is independent of itsspeed or radius of the orbit. Write twoimportant uses of a cyclotron. All India 2011C

30. Draw a schematic sketch of cyclotron.Explain briefly how it works and how it isused to accelerate the charge particles?

(i) Show the time period of ions in acyclotron is independent of both thespeed and radius of circular path.

(ii) What is resonance condition? How isit used to accelerate the chargedparticles? Foreign 2009; All India 2009

o Explanations1. The expression in vector form is given by

F = q(v x B) (1/2)

where, q is the charged particle.The direction of the magnetic force is in thedirection of vx B, i.e. perpendicular to the planecontaining v and B. (112)

2. For the given momentum of charge particle,radius of circular paths depends on charge andmagnetic field as

mvr=-

qB1

~ roc-qB

[.: qvB = m:]

For constant momentum,:. r proton : r deuteron = 1 : 1 = q deuteron : q proton

141

As they have same momentum and chargemoving in a small magnetic field. (1)

Hence, rp :rd = 1 :13. Ratio of forces acting on the two particles

FA = qVI B sin 90° 1FB (24) v2 Bsin 90° 2

[BI = B2, same magnetic field)According to the question,

~ ~ = 1 ~ VI : v2 = 1 :1v2 (1)

4.. : Velocity of a:particlesv = vi [Projected along X-axis)

.: Magnetic force on a-particlesFm = q (v x B) = q (vi x B)

Fm = Fm j [Oriented along Y-axis)

~ Fm j = q (vi x B)~ B = - Bk = B(-k)

The direction of magnetic field must be alongZ-direction. (1)

5. Given, F= q (v x B)~ F = qvB sin ewhere, e is the angle between v and B.~ B= __F_

qv sin e

i.e. ifq =1 C v=1 ms-',e= 90°, thenB=FThe magnetic field at any point is given byB=_F_= IN =IN/A-m=l T

qv sin e [(1 C) (1ms ') sin 90°) (1)

:. S1unit of magnetic field = 1TThus, the magnetic field induction at a point is saidto be one tesla if a charge of one coulomb whilemoving at right angle to a magnetic field with avelocity of 1 ms " experiences a force of 1 N at thatpoint.

6. An electron does not suffer any deflection in themagnetic field, it means that the electron ismoving parallel or antiparallel to the magneticfield, i.e. along or opposite to the direction ofmagnetic field. (1)

7. A diagram in which particle moves in magneticand electric field is shown below (1)

x x x B+ + + +

vx

x

142 o Chapterwise CSSE Solved Papers PHYSICS

Forces on a charged particle areF. = electric force = qE

Fm = magnetic force = Bqv

For a particle to go straight without any deflection;EF = F ~ qE = Bqv ~ v = -e m B

separated.

EIn this way, particles having speed, v = - areB

8. One tesla is defined as the field which produces aforce of I newton when a charge of I coulombmoves perpendicularly in the region of themagnetic field at a velocity of! ms'" .

F INF=qvB ~ B=- ~ IT=----,-

qv uc) (I ms") (2)

9. In cyclotron, Lorentz force concept is utilised.Force is acting perpendicular to both velocity andmagnetic field.Principle A charged particle can be acceleratedto very high energies by making it pass through amoderate electric field a number of times. Thiscan be done with the help of a perpendicularmagnetic field which throws the charged particleinto a circular motion, the frequency of whichdoes not depend on the speed of the particle andthe radius of the circular orbit.The combination of crossed electric and magneticfields is used to increase the energy of the chargedparticle. Cyclotron uses the fact that the frequencyof revolution of the charged particle in a magneticfield is independent of its energy.

Magnetic field outof the paper Deflection plate

Oscillator (1)

Inside the dees the particle is shielded fromthe electric field and magnetic field acts on theparticle and makes it to go round in a circularpath.

(1)

Every time, the particle moves from one dee tothe other it comes under the influence ofelectric field which ensures to increase theenergy of the particle as the sign of the electricfield changed alternately.The increased energy increases the radius ofthe circular path, so the accelerated particlemoves in a spiral path.Since, radius of trajectory

[as we know, m: = BqV]

vmr=-

qBrqB

V=-m

Hence, the kinetic energy of ionsI I r2q2B2 I r2q2B2=-m;=-m--~KE =---22m2 2m (1)

10. A charge q projectedperpendicular to the x

uniform magnetic field B x

with velocity v. Theperpendicular force,F = qvB, acts like acentripetal forceperpendicular to themagnetic field. Then, the x

path followed by charge iscircular as shown in the figure.

m; mvqvB = -- or r = -

r qB

where, r = radius of the circular path followed bycharge projected perpendicular to a uniformmagnetic field.

B

• v

(1)

(1)

11. Lorentz force always acts along the directionperpendicular to the direction of velocity of theparticle.Magnetic Lorentz force,

Fm = q (v x B) (1J

F 1. v~ Force is perpendicular to displacement made bycharge particle... W = Fd cos 900 = 0

[:. Force F and displacement dareperpendicular to each other]

~ W=ONo work is done by magnetic Lorentz force on thecharged particle. (1)


12. When a charged panicle enters iri the magneticfield at right angle, then the particle follows acircular path.

x x x xCircular

pathx x

p-

a-- x x x x

~ectronx Circular x x x

path

Radius of the circular path, r = mvqB

For same speed v, magnitude of charge andmagnetic field

roc m!!. = m,rp rrIp (1)

where, m, and. mp are masses of electron andproton, respectrvely

m, <mp(Proton is much heavier than electron)

~ r,<rp

The curvature of path of electron is much morethan curvature of path of proton. (1)

13. Refer to Ans. 12. (1)

rd _ md---,rp mp

~or

md = 2rr1p

rd = 2rp

rd:rp=2:1l( l( l(l(

Deuteron

l( l(

l( l( l( l(

l( l( l( l(

NOTE Smaller the radius, greater the curvature andvice-versa. This is why, because proton's pathhas got greater curvature.

143

14. When a force acts along the direction perpendicularto the direction'of the velocity of a particle, then nowork is done on the particle by this force.(i) As the charged particle is moving perpendicularly

to the magnetic field. So, it will perform circularmotion in XY -plane.

Z

}----- ...•.y/ ,".--.q I"'" ""'\

I I\ ," ....._-.....--, ... '"X m

(ii) No, the charged particle does not gain any KE asLorentz force acting on it does not perform anywork as Fm .1 v. (1)

W = Fdcos90o= 0

15. v=- vi[': The particl~ is moving along X-direction]

B=-Bk

[.: The magnetic field is perpendicular tothe plane of the paper directed inwards, i.e.

l -direction]

(1)

:. Force acting due to magnetic field, .Fm = q (v x B) = q [- vi x (- B~]

Fm=-qvBJ [.: ixk=-1](1)=> Magnitude of Fm = IFmI = qvBThe direction is along Y-direction.For the undeflected motion of particle,Force due to electric field = Force due to magnetic field

qE= q(v x B).. E=vxB

Magnitude of electric field, lEI = [v x B1 and directionof magnetic field will be perpendicular to both v andB, i.e. along Y-axis. (1)

.: Frequency of alternating electric field cyclotron isgiven by f = qB

2nmHere, q = 3.2 X 10-19 C m = 6.7 x 10-27 kg

and B= ~T2

f = (3.2 x 10-19) x (nI2)

2 x n x 6.7 x 10-27

16.

(1/2)

(1)

f = 1.2X 107 cycle/s (112)

144 o Chapterwise C8SE Solved Papers PHYSICS

17. (i) Force acting on the particle, F = BqvIn vector form, F =q(vx B)where, B is uniform magnetic field and v isvelocity with particle which is moving.From this equation, it is clear that direction offorce is perpendicular to the plane containingboth v and B . In other words, force actsperpendicular to both v and B. When velocitybecomes perpendicular to force, the path of theobject becomes circular. (1)"~'L

v sin 9 B

In this case, B is assumed to act perpendicularto v. In case, B is not perpendicular to v, acomponent of v remains perpendicular to v. Itcreates circular path. The component ofv parallel to B will create linear path.Here, circular path is due to vcos 9 and linearpath is due to vsin9. Both when combined giveshelical path. (1)

(ii) Since, force always adjusts itself in a directionwhich becomes perpendicular to velocity, soonly direction of velocity is changes not themagnitude. Hence, the kinetic energy of the-particle always remains constant. (1)

18. (i) When a charged particle (q) moves withvelocity (v) inside a uniform magnetic field B,then force acting on it is

F=q(vxB)From right hand thumb rule, the force F isperpendicular to the velocity (v) and magneticfield (B).Hence, it changes its path continuously. (1)

(ii) According to question, magnetic force on acharge F particle is given by

F =q (vxB)The direction of force on the charged particle isgiven by (v x B) with the sign of chargedparticle, i.e. for a-particle, charge is positiveand direction of v is + iand direction of B is -k.So, direction of force is + (i x - k) , i.e. + j.For a-particleIt describes a circle with anti-clockwisemotion.For neutronIt is a neutral particle so, it goes undeviated.As F = q (v x B) = 0

For electronForce is given by F = - e(v x B)So, direction = - (i x - k) =}- j(e- describes a circle with clockwise motion)

® '~ ®I

® 0 ®a..........-.:'

® ® ®® ® ®n_--- •.-----® ® ®

e---+--.,® ~ ®

Ie ,tiJ ®(2)

19. y

v-+

B-+

Vol =V sin 9t

The path of the charged particle will be helix. As,the charge moves linearly in the direction of themagnetic field with velocity vcos9 and alsodescribe the circular path due to velocity vsin9.Time taken by the charge to complete one circularrotation, T = 27tr (1)

vol

f= qVolB

mvi-- = qVolB

r

From Eqs. (i) and (ii), we getvim--=qvolB

r

... (i)

and ... (ii)

=} Volm = r =} T = 27tvolm =} T = 27tmqB qB· Vol Bq (1)

Distance moved by the particle along themagnetic field in one rotation (pitch of the helixpath)

= VII X T t:VII = V parallel)

= V cos9x 27tmBq

p = 27tmvcos9 = 27tmvcos9qB qB (1)


20. (i) Schematic diagram of cyclotron is shown asbelow:

D1

Target

(ii) For Principle of Cyclotron Refer to Ans. 9.(1)

(ill) Radius of a charged particle,mv

r=-qB

~ Time period of charged particleT = 21tr = 21tm

v qB~ Frequency, f = qB =.!..

znm TFrom the above formula, it is clear that frequency ofcharged particle does not depend on the velocity ofcharged particle. (1)

21. Principle of Cyclotron The cyclotron works onthe principle that a positively charged ion can beaccelerated to high kinetic energy by making itpass again and again smaller value of sameoscillating electrical field by making use of strongperpendicular magnetic field. Also, the frequencyof charge particle must be equal to the frequencyof oscillating electrical field.Expression for cyclotron frequency,

f=~21tm (1)

For schematic diagram Refer to Ans. 20. (1)

Working Let initially positively charged isaccelerated towards D2 and enter into it.Now, the charged particle experiences magneticLorentz force due to a strong nonnal magneticfield. It performs circular motion. The time takenby the charge particle to complete half revolutionis equal to half of time period of AC oscillatorbetween two dees.

145

(1)

The charge d particle again accelerated towards Dlas D2 acquires positive and d negative polarity.Thus, the charge particle is brought again andagain in the small region of oscillating electricalfield by strong normal magnetic field.The charged particle repeatedly passes throughoscillatirlg electrical field. It traversed on spiralpath and finally having radius of its circular pathbecomes equal to the radius of dees and finallycomes out through window Wand strikes to thetarget. n)

22. For first part refer to Ans. 9. (1)Cyclotron is that machine which uses the abovefact. n)

The frequency of alternating electrical field isequal to the frequency of charge particle which is

B .given by f = -q-.

2~ m23. (i) The required expression for the face is

F = q (v x B) = q vB sin eNow, refer to Ans. 5. (1)Consider the segment of a conductor given inthe below figure.

(ii)

24. (i)

Let the number of electrons per unit volume ofthe conductor is n, the drift speed of electroninside the conductor is vd and the magneticfield is B.Then, Lorentz magnetic force, f = - e (vd x B):. Force on all the mobile electrons of theconductor F = f· nA/ = - nAZe (Vd x B)or F = /(1x B)= IlB sine (2)

This is the expression for the force experiencedby current carrying wire.

If the particle is performing circular motiondue to magnetic force, thenCentripetal force = Magnetic force

mv' . 0 mv-- = qvB Sill 90 ~ r = -r qB (1)

Time period is given byT = 21tr = 21t . mv = 21tm

v v qB qB

T = 21tmqB

1 qBThus, frequency, f = - = --T .21tm

(1)

(1)

146

25.

(ii) Essential details of construction ofcyclotron Cyclotron consists of(a) two semi-circular, hollow metallic D

shaped half cylinders known as dees.(b) high frequency oscilla ting electric field is

produced by AC oscillator. (having highfrequency of AC voltage of few kVs)

(c) strong normal magnetic field is producedin dees using field magnetic.

(d) whole system is enclosed in high vacuumchamber. (1)

For working and schematic sketch Refer toADs. 20. (1)

(i) Refer to ADs. 20(ii) (a) Let the mass of proton = m; charge of

proton = q, mass of a-particle = 4mCharge of a-particle = 2q

Cyclotron frequency, (1)

v=l!!L ~ .;».21tm m

For proton frequency, vp oc i..m

For a-particle,

Frequency, v oc~IX 4m

v oc-.!LIX 2m

Thus, particles will not accelerate with samecyclotron frequency. The frequency of proton istwice than the frequency of a-particle. (1)

(b) Velocity, v = Bqr ~ voc i..m m

For proton velocity, v cc i..p m

or

For a-particle,Velocity, VIX cc ~ or VIX cc -.!L

4m 2m

26.

Thus, particles will not exit the dees with samevelocity. The velocity of proton is twice than thevelocity of a-particles.Force on the charged particle due to electric andmagnetic field s, F= qE+ q(v x B)For undeflected motion,

F= 0

qE+q(vxB)=O

~ E+ (v x B) = 0

E = - (v x B) ~ I E I = I - v x B IE= vB sin I)

where, I) = 90°v=Em W

o Chopterwise CBSE Solved Papers PHYSICS

27. Refer to ADs. 20 (i) for diagram of cyclotron. (2)

Refer to ADs. 21 for working of cyclotron. (2)

Maximum KE of charged particle= q2B2r~

2m

where, ro = radius of dees., Radius of dees is limited, therefore, KGnax alsohave limited value. (1)

28. (i) For figure of cyclotronRefer to ADs. 20(i) and 21 for principle.For working of cyclotron Refer to ADs. 21..,'In case of the cyclotron, the particle moves ona circular path, the centripetal force required isprovided by magnetic force, so magneticLorentz force = centripetal force

m; mv qBrqvB=- ~ r=-~ V=-r qB m

KE = ~ mv: = ~m (qBr)2= q2B2r222m 2m (2)

For maximum KE, r = ro (radius of dees). (1)

Given, II = 2A, 12 = 5A, r = I cm = 1 x 10-2 m

Force per unit length between two wires.(il)

~ = ~ 21112 = !0-7 X 2 x 2 x 5= 20 X 10-5L 41t r lXl02

(3)

~ = 2x 10-4 Nm' 'L rn

Currents flowing in wires are in oppositedirections, so the force will be of repulsivenature. (1)

29. Refer to ADs. 20. (5)

30. For schematic sketch of cyclotron Refer toADs. 20 (i). (1)

For working of cyclotron Refer to ADs. 21.(i) Refer to ADs. 20. (1)

(ii) The frequencies of charge particle must beequal to the frequency of AC oscillator. This isknown as resonance condition. (1)

This makes time period of charged particle andoscillator equal. Therefore, the time taken bycharge particle to complete half revolution isequal to the time of change the polarity ofdees. This facilitate the acceleration of chargeparticle.If two frequencies do not match, then insteadof acceleration, charged particle mayaccelerate. (2)

[TOPIC3] Magnetic Force and Torque betweenTwo Parallel Currents

3.1 Force on a Current CarryingConductor in a UniformMagnetic Field

The magnetic force experienced by acurrent-carrying conductor placed in a uniform.magnetic field is given by

F = I (1x B) [vector form]where, 1= a vector whose magnitude is equal tolength of the conductor and has identical directionin the flow of electric current I and B = magneticfield.Magnetic force, F = IBI sin ewhere, e is the angle between current andmagnetic field.The direction of force is given by Fleming's lefthand rule.

Magnetic Force between TwoParallel Current Carrying ConductorMagnetic force between twostraight parallel current-carryingconductors is given by

~ = h.21112L 41t r

II

I, ~--~-!'1I II II II I

Force will be of attractive nature,if direction of flow of currents arein the same direction i.e. parallel.The force of repulsion will act when direction offlow of currents are in opposite directions.(antiparallel)If both wires are oflength I} and 12 (if II > 12), thenmagnetic force between two straight currentcarrying wires/conductors is given by

F=h21112 1241t r smaUlength

3.2 Torque Experienced by aCurrent Carrying Loop inUniform Magnetic Field(Magnetic Dipole)

Torque experienced by a current-carrying loopplaced in a uniform magnetic field B is given by

't = NIAB sin ewhere, e is the angle between the direction ofmagnetic field B and that of vector N drawnnormal to the plane of the coil.or 't= Mx Bwhere, M = NIA

F4 s

F3

P I

I B

B90°

F1 B

Q I F2

F3 = BII

p

F1 = BII

and M is known as magnetic dipole moment ofcoil. Its SI unit is A_m2.

148

3.3 Circular Current Loop as aMagnetic Dipole

A current loop behaves as a magnetic dipole. If welook at the upper face, current is anti-clockwise, soit has North polarity. If we look at lower face, thencurrent is clockwise. so it has South polarity so.

Qt5upper face

r IiI

Lower face

Magnetic dipole moment of the loop, M = IA

::::} IMI= I1trThe magnitude of magnetic field on axis of acircular loop of radius r carrying stedy current I isgiven by

for x» r

If loop has N turns, M = NI A

Magnetic Dipole Moment ofa Revolving ElectronAn electron being a charged particle, constitutes acurrent while moving in its circular orbit aroundthe nucleus, the magnetic dipole moment

M=_e_nh41tme

where, me = mass of electronh = Planck's constant

and n = 1,2,3, ...

3.4 Moving Coil GalvanometerIt is a device used to detect the current inelectrical circuit. It is based on the principle that acurrent carrying loop placed in a uniformmagnetic field experiences torque, the magnitudeof which depends on the strength of current.

o Chopterwise CBSE Solved Papers PHYSICS

• The current sensitivity and voltage sensitivity ofgalvanometer depends on number of turns ofcoil magnetic field B, area A of the coil andtorsion constant k of the' spring or suspensionwire.Restoring torque = Deflecting torque

ka=NIBA

[Torque 't = Force x perpendicular distance= NIbB x a sin 90° = NIBA]

Current sensitivity, I = ~ = NBAs I k

Its SI unit is rad/A or div/A,

1 . . . e e Is NBAYo tage SenSItIVIty,V = - = - = - = --Sv IRR kRIts SI unit is rad/V or div/V

Torque or moment of galvanometer

G =~= NBAI k

Clearly I~= ~ IConversion of Galvanometer into anAmmeter and VoltmeterA galvanometer can be converted into an ammeterby connecting a very low resistance (shunt S) inparallel with galvanometer whose value is given by

S= IsGI - Is

where, G = resistance of galvanometerIs = current through galvanometerI= total current in circuit and

S = resistance of the shunt (low resistance).Ammeterr--------------------------------~

S

(I-Ig)

~~------~ G r-----~~-I A Ig Ig B I---------------------------------


Conversion of a Galvanometer toVoltmeterA galvanometer can be converted into voltmeterby connecting a very high resistance R in serieswith galvanometer which is given by

R=~-G19

where, 19 = current through thegalvanometer

G = resistance of galvanometer andV = potentials difference across the

terminal A and B.

Voltmeterr--------------------------------, ,,,,,,,,r-Jyy"y'Vv-----r'-.BA-----{ ,,,,,,I V I:L !

NOTE The resistance of an ideal ammeter is zero and anideal voltmeter is infinite. Ammeter is alwaysconnected in series with electrical circuit, whereasvoltmeter is connected in parallel with the circuit.

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 3o 1 Mark Questions

1. Write the underlying principle of amoving coil galvanometer. CSSE Delhi 2016

2. Using the concept of force between twoinfinitely long parallel current carryingconductors define one ampere of current.All Indio 2014

3. Is the steady electric current the onlysource of magnetic field? Justify youranswer. Delhi 2013C

149

o 2 Marks Questions4. A square loop of side 20 cm carrying current

of lA kept near an infinite long straightwire carrying a current of 2A in the sameplane as shown in the figure.

2A

~tOi'·20cm

Calculate the magnitude and direction ofthe net force exerted on the loop due to thecurrent carrying conductor.All Indio 2015C

5. A rectangular coil of sider land b carrying acurrent I is subjected to a uniform magneticfield B acting perpendicular to its plane.Obtain the expression for the torque actingon it. Delhi 2014C

6. (i) Two long straight parallel conductors aand b carrying steady currents Ia andIb respectively are separated by adistance d. Write the magnitude anddirection, what is the nature andmagnitude of the force between the twoconductors?

(ii) Show with the help of a diagram, howthe force between the two conductorswould change when the currents inthem flow in the opposite directions.

Foreign 2014

7. A coil of N turns and radius R carries acurrent 1. It is unwound and rewound tomake a square coil of side a having samenumber of turns N. Keeping the current Isame, find the ratio of the magneticmoments of the square coil and the circularcoiL Delhi 2013C

8. A circular coil of closely wound N turnsand radius r carries a current I. Write theexpressions for the following:

(i) The magnetic field at its centre.(ii) The magnetic moment of this coil.

All Indio 2012

150

9. A steady current 11 flows through a longstraight wire. Another wire carryingsteady current 12 in the same direction iskept close and parallel to the first wire.Show with the help of a diagram, howthe magnetic field due to the current 11exert a magnetic force on the secondwire. Deduce the expression for this force.All India 2011

10. How is a moving coil galvanometerconverted into a voltmeter? Explain givingthe necessary circuit diagram and therequired mathematical relation used.All India 2011C

11. A square coil of side 10 em has 20 turnsand carries a current of 12 A. The coil issuspended vertically and normal to theplane of the coil, makes an angle e withthe direction of a uniform horizontalmagnetic field of 0.80 T. If the torque,experienced by the coil equals 0.96 N-m,find the value ofe. Oelhl2010C

12. Define current sensitivity and voltagesensitivity of galvanometer. Increasingthe current sensitivity may notnecessarily increase the voltagesensitivity of a galvanometer, justify youranswer. All India 2009

13. Two long straight parallel conductor carrysteady current 11 and 12 separated by adistance d. If the currents are flowing inthe same direction, show how themagnetic field set up in one produces anattractive force on the other. Obtain theexpression for this force. Hence, defineone ampere. Deihl 2016

14. State the principle of working of agalvanometer. A galvanometer ofresistance G is converted into a voltmeterto measure upto V volts by connecting aresistance Rt in series with the coil. If aresistance R.z is connected in series withit, then it can measure upto V/2 volts.Find the resistance, in terms of Rt and R2,

required to be connected to convert it intoa voltmeter that can read upto 2V.Also,find the resistance G of the galvanometerin terms of B; and R2. All India 2015

(23 ehapterwise eBSE Solved Papers PHYSICS

15. A metallic rod oflength 1 is rotated with afrequency v with one end hinged at thecentre and the other end at thecircumference of a circular metallic ring ofradius r about an axis passing through thecentre and perpendicular to the plane ofthe ring. A constant uniform magnetic fieldB parallel to the axis is presenteverywhere. Using Lorentz force, explainhow emf is induced between the centreand the metallic ring and hence obtainedthe expression for it. Deihl 2013

16. A wire ABis carrying a steady current of12Aand is lying on the table. Anotherwire CD carrying 5Ais held directly aboveAB at a height of 1 mm. Find the mass perunit length of the wire CD, so that itremains suspended at its position whenleft free. Give the direction of the currentflowing in CD with respect to that in AB.[Take, the value of g = 10ms-2] All India 2013

17. A rectangular loopofwire of size2.5 em x 4 ern carries steady current of1A. A straight wire carrying 2 A currentis kept near the loop as shown. If the loopand the wire are coplanar, find the (i)torque acting on the loop and (ii) themagnitude and direction of the force onthe loop due to the current carrying wire.Delhl2D12

1= 2A2.5 em

r4em

IlAlA

1 em

18. Draw a labelled diagram of a moving coilgalvanometer and explain its working.What is the function ofradial magneticfield inside the coil? Foreign 2012

19. Depict the magnetic field lines due to twostraight, long, parallel conductorscarrying currents 11 and 12 in the samedirection. Hence, deduce an expression forthe force per unit length acting on one ofthe conductors due to the other. Is thisforce attractive or repulsive? Delhl2011C


20. Find the expression for magnetic dipolemoment of a revolving electron. What isBohr magneton? Delhi 2011

21. State the underlying principle of workingof a moving coil galvanometer. Write tworeasons why a galvanometer cannot beused as such to measure the current in agiven circuit. Name any two factors onwhich the current sensitivity of agalvanometer depends. Deihl 2010

22. A moving coil galvanometer of resistanceG gives its full scale deflection when acurrent Ig flows through its coil. It can beconverted into a ammeter of range (0 to I)(I> I g) when a shunt of resistance S isconnected is converted into an ammeter ofrange 0 to 1, find the expression for theshunt required in terms of Ig and G.Delhl2010C

23. Write the expression forthe magnetic moment (m)due to a planar squareloop of side I carrying asteady current I in avector form. In the givenfigure, this loop is placedin a horizontal planenear a long straightconductor carrying a steady current 11 at adistance I as shown.Give reasons to explain that the loop willexperience a net force but no torque.Write the expression for this force actingon the loop. Deihl 2010

11

tJS R

24. Derive the expression for force per unitlength between two long straight parallelcurrent carrying conductors. Hence,define one ampere. Deihl 2009

25. Deduce the expression for the torqueexperienced by a rectangular loopcarrying a steady current I and placed ina uniform magnetic field B. Indicate thedirection of the torque acting on the loop.Foreign 2009

151

o 3 Marks Questions26. An electron of mass me revolves around a

nucleus of charge +Ze. Show that itbehaves like a tiny magnetic dipole. Hence,prove that the magnetic momentassociated with it is expressed asJ.I. = - _e_L, where L is the orbital angular

2m.momentum of the electron. Give thesignificance of negative sign. Deihl 2017

27. (i) Obtain the expression for thecyclotron frequency.

(ii) A deuteron and a proton areaccelerated by the cyclotron. Canboth be accelerated with the sameoscillator frequency? Give reason tojustify your answer. Oalhl2D17

o 5 Marks Questions28. Explain, using a labelled diagram, the

principle and working of a moving coilgalvanometer. What is the function of (i)uniform radial magnetic field (ii) soft ironcore?Define the terms (i) current sensitivityand (ii) voltage sensitivity of agalvanometer.Why does increasing the currentsensitivity not necessarily increasevoltage sensitivity? Deihl 2015. Foreign 2016

29. (i) Draw a labelled diagram of a movingcoil galvanometer. Describe brieflyits principle and working.

(ii) Answers the following questions.(a) Why is it necessary to introduce a

cylindrical soft iron core inside thecoil of a galvanometer?

(b) Increasing the current sensitivity ofa galvanometer may not necessarilyincrease its voltage sensitivity.Explain with giving reasons.All Indio 2014

152 o ehapterwise eBSE Solved Papers PHYSICS

30. (i) State using a suitable diagram, theworking principle of a moving coilgalvanometer. What is the functionof a radial magnetic field and the softiron core used in it?

(ii) For converting a galvanometer intoan ammeter, a shunt resistance ofsmall value is used in parallel,whereas in the case of a voltmeter aresistance of large value is used inseries. Explain, why? Deihl 2014C

31. (i) Explain giving reasons, the basicdifference in converting agalvanometer into (a) a voltmeterand (b) an ammeter.

(ii) Two long straight parallel conductorscarrying steady currents II and 12 areseparated by a distance d.Explain briefly with the help of asuitable diagram, how the magneticfield due to one conductor acts on theother? Hence, deduce the expressionfor the force acting between the twoconductors. Mention the nature ofthis force. All Indio 2012

32. A rectangular loop of size I x b carrying asteady current I is placed in a uniformmagnetic field B. Prove that the torque't acting on the loop is given by 't = m x B,where, m is the magnetic moment of theloop. All Indio 2012

33. (i) Show that a planer loop carrying acurrent I, having N closely woundturns and area of cross-section A,possesses a magnetic momentm=NIA.

(ii) When this loop is placed in amagnetic field B, find out theexpression for the torque acting on it.

(iii) A galvanometer coil of 50 n resistance)shows"full scale deflection for acurrent of 5 mA. How will you convertthis galvanometer into a voltmeter ofrange 0 to 15 V? Foreign 2011

34. (i) With the help of a diagram, explainthe principle and working of amoving coil galvanometer.

(r)

35.

(ii) What is the importance of radialmagnetic field and how is itproduced?

(iii) Why is it that while using a movingcoil galvanometer as a voltmeter, ahigh resistance in series is requiredwhereas in an ammeter a shunt isused? All Indio 2010; Delhi 2009

(i) Derive an expression for the forcebetween two long parallel currentcarrying conductors.

(ii) Use this expression to define SI unitof current.

(iii) A long straight wireAB carries a currentI. A proton P travels Iwith a speed v,parallel to the wire ata distance d from it ina direction opposite to Athe current as shownin the figure. What is the forceexperienced by the proton and what isits direction? All Indio 2010; Foreign 2008

B

p-d-rproton

v

36. (i) Two straight long parallel conductorscarry currents II and 12 in the samedirection. Deduce the expression forthe force per unit length betweenthem. Depict the pattern of magneticfield lines around them.

(ii) A rectangular current carrying loopEFGH is kept in a uniform magneticfield as shown in the figure.

(a) What is the direction of themagnetic moment of thecurrent loop?

(b) What is the torque acting onthe loop maximum and zero?

Foreign 2010; Delhi 2009

o Explanations1. The principle of moving coil galvanometer is based

on the fact that when a current carrying coil isplaced in a magnetic field, it experiences a torque.Torque produced is given by t = NIBAsina[Symbols have their usual meaning) (1)

2. Refer to Text.3. Yes, the net magnetic force acting on a wire

carrying a steady (constant) electric current I inan external magnetic field B and is given

F = IdlB (1)

4. According to the question, as the loop is square,its sides are parallel. So, force between twoparallel current carrying wires,

2A•1~ 1A

AO· ~o C

20 em

F = JloII 12121tr

Force on arm AB,F _Jlox2Xlx20xlO-2

AB - 21tX10 X10-2 (1)

= 2Jlo N (Attractive, towards the wire)1t

Force on arm CD,

F. _JloX2XlX20xlO-2

CD- 21tx30Xl02

= 2Jlo N (Repulsive, away from the wire)31t

Force on arms BC and DA are equal and opposite.So, they cancel out each other.Net force on the loop is F = F AB - FCD

= Jlo [2-~] = 4Jlo = 4x41txl0-7

1t 3 31t 31t

= 5.33 X 10-7 N (Attractive, towards the. wire) (1)

5. Equivalent magnetic moment of the coilm = IAn

.. m= Ilbnwhere, n = unit vector.!.. to the plane of the coil.

:.Torque =mxB=Ilb(nxB) =0 (2)

As nand B are parallel or anti-parallel to eachother.

a0816. (i) Let a and b be two longstraight parallel conductors. I.and Ib are the current flowingthrough them and separatedby a distance d.

Magnetic field induction at Qa point P on a conductor bdue to current I. passingthrough a is

BI = Jlo2I.47td

d

Now, unit length of b will experience a forceas F2 = ~Ib xl = Bllb f

•• F2 = Jlo 2I.Ib41t d

Conductor a also experiences the same amountof force directed towards b. Hence, a and battract each other. (1)

(ii) a®81 B2®b

t,

p---4QF1

Now, let the direction of current in conductor bbe reversed. The magnetic field B2 at point Pdue to current I. flowing through a will bedownwards. Similarly, the magnetic field BI atpoint Q due to current Ib passing through b willalso be downward as shown. The force on a willbe, therefore towards the left. Also, the force onb will be towards the right. Hence, the twoconductors will repel each other as shown. (1)

7. Ratio of the magnetic momentsM, = 2INA,Me INAe

_ 2(~r _1-~-2

(2)- ,8. (i) Magnetic field at centre due to circular current

carrying coil, B = Jlo NI2r

(ii) Magnetic moment, M = NIA = NI (1tr2)

M = 1tutr'

(1)

(1)

where, r is the radius of circular coil, Jlo ispermeability of free space and N is number ofturns.

154

9. In these types of questions, we are calculating forceon a wire in the field produced by the other currentcarrying wire.

To find expression of force between two parallelwires.Let two infinitely long straight current carryingconductor carry currents I) and 12 in the samedirection.

ALt Magnetic field B) due to first wire on seconds, i.e.(112)

B) = ~o 2l) = ~OI) ... (i)41t r 21tr (112)

":)f I

The magnetic field is perpendicular to the plane ofpaper and directed inwards i.e. (X) type.

•• ::I[2x x

F2 x xL

x x 1x x

x x

x x

• • r• •

[..• •

Now, magnetic force on length L of second wire isgiven by (1/2)

F2 = I~) L sin 90°

F2 = 12 (~O . 2l) ) L41t r

~ F2 = ~o . 2l) 12 = ~OI)I2L 41t r 21tr

By Fleming's left hand rule, the direction of forceF2 is perpendicular to the second wire in the planeof pape~ towards the first wire.

! 11 1 q J ""1 I , •

Similar y, magnetic force on 1st wire is given by,.F fi.=~o.2I)I2 ... (iii)

L. 41t r

F)=-F2The force Fi is directed towards the second wire.Thus, two straight parallel current carryingconductor have the same direction of flow ofcurrents attracting each other. (112)

... (ii)

o Chapterwise eSSE Solved Papers PHYSICS

10. The resistance of an ideal voltmeter is infinity or veryhigh in practical condition. So, to convert agalvanometer into voltmeter, its resistance needs tobe increased, which can be done by a highresistance in series connection with it.

A galvanometer can be converted into a voltmeterby connecting a very high resistance R in serieswith it. (112)

Let R is so chosen that current I9 gives fulldeflection in the galvanometer where I9 is therange of galvanometer.

,Voltmeter

r---------------------.I I

~:[g R:

I G I

A : : BI It 11+--- V -----+l (112)

Let galvanometer of resistance G, range I9 is to beconverted into voltmeter of range V (volt). Now,

V VV=Ig(G+R) ~R+G=- ~R=--GIg Ig

11.

The appropriate scale need to be graduated tomeasure potential difference.Here, Area (A) of coil

= 10 x 10 = 100 cm2 = 10-2 m2

Number of turns, N = 20 turnsCurrent, I = 12 A

Number of the coils make angle with magneticfield = a =?

Magnetic field, B = 0.8TTorque, 't = 0.96 N-m (112)

.: Torque ('t) experienced by current carrying coil,the magnetic field is

't = NIABsina

0.96= 20 x12xlO-2 x 0.8 x sin asin a = 0.96 = ~ ~ a = ~ rad

1.92 2 6

(1)

(1)

(1/2)

12. Current sensitivity The deflection produced inthe coil of galvanometer per unit flow of electriccurrent through it, i.e.Current sensitivity,

I = ~ = NBA9 I K

where abbreviations are as usual. (1/2)

Voltage sensitivity The deflection produced inthe galvanometer is per unit applied potentialdifference across it.


V I . .. v: e e NBA. . 0 tage sensinvny. = - = - = --

s V IR K.Rwhere abbreviations are as usual.Its unit is rad IAor div IA. (112)Increasing the current sensitivity may notnecessarily increase the voltage sensitivity,because the current sensitivity increases with theincrease of number of turns of the coil but theresistance of coil also increases which affectadversely on voltage sensitivity. (1)

13. To find the force on a current-carrying wire due to asecond current-carrying wire, first find the field dueto second wire at the site of first wire. Then, find theforce on the first wire due to that field.

Let us consider AI BI andA2B2 are two infinite longstraight conductors.II and 12 are the currentsflowing through them andthey are r distance apart.Magnetic field induction ata point P on the conductorA2B2 due to current IIpassing through AI BI isB] = ~o2l]

4~r n)The unit length of A2B2 will experience a force as

P2=BI12Xl=B]I2

or P2

= ~o . 2l]124~ r

Conductor AI BI also experiences the same amountof force directed towards the wire A2B2.

Therefore, force between two current-carryingparallel conductor per unit length is

P = h. 2l]124~ r

Two linear parallel conductors carrying-currents inthe same direction attract each other while inopposite direction they repel each other.Definition of One Ampere1 ampere is the current which flows through eachof the two parallel uniform long linear conductors,which are placed in free space at a distance of 1 mfrom each other and which attract or repel eachother with a force of 2 x 10-7 Nm-] of theirlengths. (1)

14. According to the principle of working of a movingcoil galvanometer, when a current carrying coil isplaced in a magnetic field, it experiences atorque.

B, B2

B2 B,

PF, F2

I, 12A, A2

155

A high resistance that is connected in series withthe galvanometer to convert into voltmeter. Thevalue of the resistance is given by (1)

R=~-GIg

where, V = potential difference across theterminals of the voltmeter, Ig = current throughthe galvanometer andG = resistance of the galvanometer.When resistance R] is connected in series vy'tf" thegalvanometer, then

VR] =- -G ...(i)

Ig

When resistance R2 is connected in series with thegalvanometer, then

VR2 = - - G ... (ii)»,From Eqs. (i) and (ii). we get

VR] - R2 = - and G = RI - 2 R22Ig

The resistance R3 required to convert the givengalvanometer into voltmeter of range 0 to 2V isgiven by R3 = 2V - G

Ig

~ R3 = 4 (R] - R2) - (2R1 - R2) = 3R] - 2R2G in terms of RI and R2 is given by

G = RI - 2R2 (1)

15. Suppose, the length of the rod is greater than theradius of the circle and rod rotates anti-clockwiseand suppose, the direction in the rod at anyinstant be along +Y-direction. Suppose, thedirection of the magnetic field is along+ Z-direction.Then, using Lorentz law, we get the followingF = - e (v x B) ~ F = - e (v j x B~ ~ F = - evBl (1)

Thus, the direction of force on the electrons isalong X-axis.Thus, the electrons will move towards the centre,i.e. the fixed end of the rod., c . - •

This movement of electrons wili'ieshff Ji\urrentand hence, it will produce emf in the ~od Between,the fixed end and the point touching the ring.Let e be the angle between the rod and radius of thecircle at any time t.Then, area swept by the rod inside the circle

= ~ ~ril2

156

Now, induced emf = B x :!... (~ 1tr1l)dt 2

= ~1tr2Bde2 dt

= ~ 1tr2Bw2

= ~1tr2B(21tv)2

= 1t2r2Bv (1)noi -o'J

NOTE There will be an induced emf between the two ends ofthe rods also.

16. Anti-parallel currents repel.The magnetic force of repulsion on the upper wireshould be balancing its own weight.For wire CD to remain suspended at its position inequilibrium,magnetic force on CD due to AB = weight of CD

.. ~Olj 121 = mg (g = 10m/s2)21tr

m=2xlO-71jI2 (forl=lm)rg

2xlO-7 xizx 5m=--,----10 3 xlO

= 1.2 X 10-3 kgm "Current in CD should be in opposite direction tothat in AB.

17. There will be force of attraction between thestraight wire and 4 cm long arm of loop nearer tothe straight conductor.

Fi =~o 2x2xl x(4XlO-2)

41t (2xlO 2)

[towards straight conductor]Fi = 8 x 1O-7N ... (i) (1)

Similarly, force on other 4 ern arm of loop, awayfrom the straight conductor

F2 = ~o x 2x2xl x (4 X 10-2)41t ("4.5x 10 2)

F2 = 3.55 x 1O-7N ... (ii)

1 [away from conductor](i) Since, Fi and F2 are of different magnitudes,

therefore, do not form couple and henceTorque, t = 0 (112)

o Chapterwise CSSE Solved Papers PHYSICS

18.

(ii) Net force on loop,F=Fi -F2

[towards straight conductor]F = 8 X10-7 - 3.55 X10-7

F= 4.45 x 1O-7N

The forces on two branches of loop are equal inmagnitude and opposite in the directions, hencethey balance each other. (112)

Moving Coil Galvanometer PrincipleIts working is based on the fact that when acurrent carrying coil is placed in a magnetic field,it experiences a net torque.

Scale

Permanent magnet

T

T1

Phosphorbronzestrip

T2

(1/2)


WorkingSuppose, the coil PQRS is suspended freely in themagnetic field.Let I =length PQ or RS of the coil

b = breadth QR or SP of the coiln = number of turns in the coil

Area of each turns of the coil, A = I x b

Let B = strength of the magnetic field in whichcoil is suspended.

I = current passing through the coil in thedirection of PQRS

Let at any instant of time, a be the angle which thenormal drawn on the plane of the coil makes withthe direction of magnetic field. The rectangularcurrent carrying coil when placed in the magneticfield experiences a torque whose magnitude isgiven by 't = NIBA sin a (112)

Due to this deflecting torque, the coil rotates andsuspended wire gets twisted. A restoring torque isset up in the suspension wire.Let 9 be the twist produced in the phosphor bronzestrip due to rotation of the coil andk be the restoring torque per unit twist of thephosphor bronze strip.Then, total restoring torque produced = k9In equilibrium position of the coil,Deflecting torque = Restoring torque

. . NIBA = k9 or I = _k_9 = G9NBA

where, _k_ = GNBA

[constant for a galvanometer]It is known as galvanometer constant.• Current sensitivity of the galvanometer is the

deflection per unit current.~=NABI k

• Voltage sensitivity is the deflection per unitvoltage.* = N:S (i) = N:S i

The uniform radial magnetic field keeps the planeof the coil always parallel to the direction of themagnetic field, i.e. the angle between the plane ofthe coil and the magnetic field is zero for all theorientations of the coil. (1)

[','V=IR]

157

19. For deduction of the force acting between twolong parallel conductors Refer to Ans. 8 (i). (1)

Magnetic field lines due to both conductors

Current-carrying conductors having same directionof flow of current, so the force between them ~be attractive. (1)

20. As electric current associated with the revolvingelectron

l=~=~T 21tr

where, time period T = 21trv

r = radius of orbitv = velocity of electron

The magnetic moment due to the current,

~ = LA = ~ x 1tr2 ~ ~ = evr21tr 2 (1)

If electron revolves in anti-clockwise sense, thecurrent will be in clockwise sense. Hence, accordingto right hand rule, the direction of magneticmoment must will be perpendicular to the plane oforbit and directed inwards to the plane.

evtm elSo, ~ =--=-On On

where, vrm = I = angular momentum orbital ofelectron and in vector form,

I~=-e-

OnNegative sign indicates !.l and I are in mutuallyopposite directions. From Bohr's postulates,

nh1= mvr = -, where n = 1, 2, 3, ...21t

e nh~ = 2m' 21t = n!.lInin

eh!.l . = -- called Bohr's magneton.mm 47tm

21. Principle The current carrying coil placed innormal magnetic field experiences a torque whichis given by

(1)

where,

't =NIAB

158

where, N = number of turnsI = current

A = area of coilB =magnetic field

The galvanometer cannot be used to measure thecurrent because(i) all the currents to be measured passes through

coil and it gets damaged easily as hair linespring or

(ii) its coil has considerable resistance because oflength and it may affect original current.

['12-2=1)

Current sensitivity of galvanometer can beincreased by(i) increasing the magnetic field

(ii) decreasing the value of torsional constant. (1)

22. The resistance of an ideal ammeter is zero or verylow in practical condition, so to convert agalvanometer into ammeter its resistance needs tobe decreased which can be done by connecting alow resistance in its parallel order.

A moving coil galvanometer of range I9 andresistance G can be converted into ammeter byconnected very low resistance shunt in parallelwith galvanometer..,' To convert a galvanometer into an ammeter,shunt resistance is connected in parallel with thegalvanometer, so the potential difference across thecombination is same.:. PD across galvanometer = PD across shunt S

IgG = Is S (1)

But Is + Ig = I:::) Is = I - 19:::) IgG=(I-1g)S

:::) S = IgG1- 19 (1)

The shunt resistance S to be connected to convertgalvanometer into an ammeter.

Ammeter

s

(I - Ig) (I - Ig)

1---;-4~~G~--~r---A Ig B

I

IZI ehapterwise eBSE Solved Papers PHYSICS

23. The magnetic moment of a current carrying loopm=IA

where, A = area of the loop (square).. A = /2 nHere, n is a unit vector normal to the direction ofarea vector.The forces acting on the arms QR and SP of given(in question figure) loop are equal, mutuallyopposite and collinear. Hence, they are balanced byone another. (1)

"I "I IForce on arm PQ, Fi = B II = _,.._0_I II = _,.._0_1-21t/ 21t

Obviously, Fi is of the attractive nature anddirected towards MN.Again, force on arm RS,

F2=B2Il=J.1.01IIl = J.1.oII I21t(2/) 41t (1)

F2 is perpendicular to wire RS and directed awayfrom the conductor MN.

,'.Net force on loop PQRS,:::) Fnet=~-F2

= J.1.0III _ J.1.0II I21t 41t

or F - J.1.0Il I [attractive]net-~

As, FI and F2 are collinear, hence does not producetorque on the loop PQRS.

24. Force between two straight parallel currentcarrying conductors

F = &. 2/11241t r

Let II = 12 = 1 A. r = 1rn,

then F = 10-7 X 2 (1) (1) = 2 x 10-7(1) (1)

One ampere It is the value of steady currentwhich flows through two straight, parallelinfinitely long current carrying conductors ofnegligible cross-section placed in air or vacuum ata distance of 1 m and they experience a force ofattractive or repulsive nature of magnitude2x 10-7 N/mon their unit length. (2)

25. Let a current carrying rectangular loop PQRScarrying a steady current I placed in a uniformmagnetic field B keeping axis of the coilperpendicular to field as shown in figure. Let atany instant the area vector A makes an angle ewith the direction of magnetic field B.


p

: QI F1 i

IIIIIII

B !IIII

! F2III

S t-:::::::: i I

b~R

F3Fig. (0)

l-!B

mIup I

IIIIIIL___________________ Idown

b sin e

Fig. (b)

Let length and breadth of coil are 1 and b,respectively.Now, magnetic force on PS arm of the coil is givenby 1) = IBI sin 90°

[.,' PS II axis of coil.:', 9= 90°]

Fj=IBI ... (i)By Fleming'S left hand rule, the direction of force isperpendicular to SP and B is along upwarddirection. Similarly, force of QR arm of the coil.

F2 = IBI sin 90° ... (il)The direction of force is perpendicular to QR and Bis along downward direction..,' 1) and F2 are equal in magnitude, opposite indirection, parallel to each other acting on the loopforms a couple which try to rotate the coil. (1)

Now, force on RS part of the coilF3 = IBb sin (90° + 9)F3 = IBb CDS a

and force on PQ part of the coilF4 = IBb sin (90° - 9)= IBb cos a

But Fleming'S left hand rule, F3 and F4 are equal inmagnitude and opposite in direction along the

159

same line of action. Therefore, they balance eachother. (cancel out)Now, torque due to Fj and F2 is given byt = force x perpendicular distance between lines ofaction of Fj and F2•

t=FxbsinaBut, Fj = F2 = F = IBI ~ t = (IBI) x (b sin 9)

t = IB (Ib) sin at = IBA sin a

where, A = Ib = area of coil for N turns of coilt = NIAB sin a (1)

26. When electron revolves around a nucleus, itcreates circular current around it. In this way, itis equivalent to a current carrying coil. So, itbehaves as a tiny magnetic dipole. (lJWe know that a current carrying coil behaves like amagnetic dipole having dipole moment equal toi A, where i is current and A is area of the coil.

i = :...= ero = eMer2ro

T 21t 21tmr2where, e is charge of electron.Angular momentum of electron is L = M,r2ro

Substituting the above in Eq. (I), we get,. eLZ=---

27tM,r2

jJ.= iA = eL1tr2

= ~27tM,r2 2M,-eLjJ.=-2M, rn

Here, negative sign indicates jJ.is directs away from L.

27. (i) In the presence of the magnetic field, theproton moves in the circular path whoseradius is r, which depends on its speed.Hence,magnetic force on q = centripetal force on q.

B . 900 mv2B m;~ qv sm =-- ~ qv =--

r r

... (i)

tn

Hence,

In vector form,

sin900= 1)

~ mvr=-qB m

Time period of proton is, T = 21tr = 21tmv qB

... (i)., Frequency, I=.!... = qBT 21tm

The frequency of oscillation of the chargedparticle from the above expression is

qBlost: = 21tm

It is also known as cyclotron frequency. (lJ

160

(ii) Let the mass of proton", mCharge of proton", q, Mass of deuteron= 2m

Charge of deuteron e q

Cyclotron frequency, v "'~ ~v oc!L21tm m

For proton frequency, vp oc!L ... (i)m

For deuteron frequency, Vd oc!L ... (ii)2m

From Eqs. (i) and (ii), we get, vp '" 2vd

Thus, frequency of proton is twice that ofdeuteron.No, both cannot be accelerated with sameoscillator frequency as they have different mass.

(1)

28. PrincipleIts working is based on the fact that when acurrent carrying coil is placed in a magnetic field,it experiences a torque. (1/2)

For Fig. refer to Ans. 18.Let at any instant, a be the angle which normaldrawn on the plane of the coil makes with thedirection of magnetic field, the rectangular coilcarrying current when placed in the magneticfield experiences a torque. Whose magnitude isgiven by t '"NIBAsinaDue to deflecting torque the coil rotates andsuspension wire gets twisted. A restoring torqueis set up in the suspension wire. (1'10)

Let 0 be the twist produced in phosphor bronzestrip due to rotation of the coil and k be therestoring torque per unit twist of the phosphorbronze strip.Then, total restoring torque produced", kOIn equilibrium position of the coil,Deflecting torque", Restoring torque

NIBA '" kO ~I '" _k_O '" GONBA

where, _k_ '"GNBA

rr

t·JIl


It is known as galvanometer constant, i.e. I ocO.Itmeans that the deflection produced isproportional to the current flowing through thegalvanometer. (1)

As, we can see in the figure given thatCylindrical soft iron core which not only makesthe field radial but also increases the strength ofthe magnet.Radial magnetic fields is a field in which coil ofthe galvanometer always remain parallel to thefield even on large deflection.Current sensitivity It is defined as thedeflection produced per unit current flowingthrough the galvanometer. It is given by

o NABIs "'-"'--I k

Voltage sensitivity It is defined as thedeflection produced per unit voltage

~ Vs '" % '" (N~)i ~ Vs '" N~ . :R~ v'" NAB

kR (1)

29.

Current sensitivity does not depend uponresistance (I?), whereas voltage sensitivity does, asevident from their expression. Current sensitivitycan be increased by increasing the number ofturns of the coil. However, this increase theresistance of the coil, since voltage sensitivitydecreases with increase in the resistance of thecoil the effect of increase in number of turn isnullified in the case of voltage sensitivity (1)

(i) Refer to Ans. 18. (1)(ii) (a) It is necessary to introduce a cylindrical

soft iron core inside the coil of agalvanometer because magnetic field isincreased, so its sensitivity increases andmagnetic field becomes radial. So, anglebetween the plane of coil and magnetic lineof force is zero in all orientations of coil. (2)

(b) The relation between current sensitivityand voltage sensitivity is given byVoltage sensitivity Current sensitivity

Resistance of coilIsVs "'-Rc

If Rc is constant, then Vs ocIsThis means if Vs increases, Is alsoincreases.But, if Rc increases in the given ratio, thenVs is a constant. (1)

(1)


30. (i) Refer to Ans. 18. (2)(ii) Refer to Ans. 22 (i). (3)

31. (i) A galvanometer of range Is and resistance G1

can be converted into(a) a voltmeter ofrange Vby connecting a high

resistance R in series with it where value isgiven by R = !.-G

Is(b) an ammeter of range I by connecting a very

low resistance (shunt) in parallel withgalvanometer whose value is given by

S = IsG

I - Is (1)

(ii) Let two straight wires of infinite length arecarrying currents, II and 12 in the samedirection and separated by distance d apartfrom each other. (1)

The magnetic field due to wire 1 at any pointon wire 2,

B1

= Ilo 21J ••• (i)47t d

The distance of Bl is perpendicular to plane ofpaper and directed inward.

CD ®I II II I

• •

T• • x x

• •• • x x

L • • x x

111 12• • x x

• • d x x

• • F1 F2 x xB

• • x x

• • x x

(112)

Magnetic force on wire 2 in L length of it

F2 =12 Bl L sin 90°= 12 (IlO. 211)L x i47t d (1/2)

.. F2 = Ilo 21112 [towardswire] ... (ii)47t d

By Fleming'S left hand rule.Similarly, force on wire 1 due to wire 2 can beproved 1) = Ilo 21112 ... (ill)

47t dThus, the nature of force is attractive.When direction of flow of current gets in oppositedirection, the nature of force becomes repulsive. (2)

161

32.

[1)

(112)

(1)

--~----------------BA

The magnetic force on Be and DA part of wire areequal in magnitude, opposite in direction alongthe same line. Therefore, they balance each other.

(112)

Let at any instant area vector of coil made anangle a with the direction of magnetic field.:.1) and F2 form couple which try to rotate thecoil.From figure,:. Torque, t = force x arm of the couple

= (IBI) x MD= IBI x bsin a = IB (lb) sin a

t = IB Asin a (1)

where, A = Ib = area of coil

But,

t = IABsina

m=IA

.. t = mBsinaIn vector form, t = m x B (1)

33. (i) Torque on rectangular loop,

t = NIAB sin a ... (i)where, symbols are as usual.Also, torque experienced by magnetic dipole ofmoment m are placed in a uniform magneticfield.

(ii)

t = mBsinaComparing Eqs. (i) and (ii), we getThe magnetic dipole moment,

m=NIAAlso, m is along A. ~ m = NIARefer to Ans. 21.

[2)(1/2)

... (ii)

162

(ill) G=50o.,Is=5X1O-3AV=15V+__v-----+

•..----- -------------,I I

~ •..•~ •.~high resistance~yyyy+-

Ig i iI II I_______________ - I" 1

Voltmeter1 V=ls(G+R)

r/;i#, .,,',b ,,[ R=~-G I,'f -+<Jq+l.-) "!:~ _ 19 ,

~ _1_5__ 50 ~ R = 29500.5x 10-3

A resistance R = 29500. is to be connected in serieswith galvanometer to convert it into a desiredvoltmeter. (1'10)

34. (i) Refer to Ans. 18for principle.H

f!I

Phosphorbronzestrip

T2

.n .(

(112)

The coil remains suspended in radial magneticfield so that it always experiences maximumtorque.When current passes through the coil,deflection torque 't (9) is produced given by

'tdeflection = NIAB sin 90° ... (i)As a result, coil rotates and phosphor bronzestrip gets twisted. As a result restoring torquegiven by

't restoring = k9 ... (ii)where, k = torsional restoring constant. . In equilibrium,

(deflecting = 't restoring ~ NIAB = k9

l = (:AB)9~ I ac9

greater the current, greater the deflection. (1)

o ehopterwise eBSE Solved Popers PHYSICS

(1)

(ii) In radial magnetic field, the plane of the coilis always parallel to the plane of the magneticfield and area vector of coil is perpendicular tomagnetic field. It is always exerts maximumtorque on the coil. (1)

(ill) The voltmeter connected in parallel with theelectrical circuit elements to measurepotential difference. For exact measurementof PD voltmeter must draw minimum currentwhich is possible only when it has highresistance. Ammeter is connected in serieswith the electrical circuit and current to bemeasured passes through it.In order to protect the galvanometer, a feeblecurrent must pass through the galvanometer,it is possible only when a low resistance(shunt) is connected in parallel withgalvanometer to allow the major part of thecurrent to pass through it. . (2)

35. (i) Refer to Ans. 19 (ii). (2)

(ii) As, ~ =!-Lo 2l, l2L 41t r

IJ = [2 = 1 A, r = Irn

~ = 2x1O-7 Nm-'L

For definition Refer to Ans. 20. (1)(ill) Here, magnetic field due to the current

carrying conductor at a distance d from it isgiven by

(112)

B =!-Lo 2l41t d (1/2)

. . Force on proton,F = (e) (0 B sin 90°

~ F= evB

F = ev (!-Lo 2l)41t d

F = !-Lo . 2lev41t d (1)

The proton is directed perpendicular tostraight conductor and away from it. (1/2)

36. (i) Refer to Ans. 19. (3)(ii) (a) Perpendicular to the plane of the paper

and directed inward. (1)

(b) When angle between area vector of coiland magnetic field is 90°, then maximumtorque experienced by the coil.When 9 = 0° or 180°, then torque will beminimum, i.e. zero. (1)

Value Based Questions (From Complete Chapter)04 Marks Questions

1. Kamal's uncle was advised by his doctorto undergo an MR1 scan test of his chestand gave him an estimate of the cost. Notknowing much about the significance ofthis test and finding it to be too expensivehe first hesitated. When Kamal learntabout this, he decided to take help of hisfamily, friends and neighbours andarranged for the cost. He convinced hisuncle to undergo this test so as to enablethe doctor to diagnose the disease, he gotthe test done and resulting informationgreatly helped the doctor to give himproper treatment.(a) What according to you, are the values

displayed by Kamal?(b) Assuming that the MR1 scan test

involved a magnetic field of 0.1 T,find the maximum and minimumvalues of the force that this fieldcould exert on a proton moving witha speed of104 ms-I. State thecondition under which the force canbe minimum. Foreign 2013

Ans. (a) Values displayed by Kamal:(i) Being educated, he knows about MRI

(magnetic resonance imaging}.(ii) Took prompt decisions to take the help of

his family, friends and neighbours toarranged the cost of MRI.

(iii) He showed his empathy, helping attitudeand caring nature for his uncle. (2)

(b) Magnetic force on moving charge particle inuniform magnetic field 8 can be given as

F = q(v x 8) or IFI= qvBsine(i) Maximum force at e = 90°

F = qvB= 1.6 X 10-19 X 104 x 0.1= 1.6 x 1O-16N [1]

(ii) Minimum force at e = 0° and 180°F=O

i.e. force is minimum when the chargeparticle either move paraliel oranti-parallel to the magnetic field lines. [1]

2. Mr. Sharma a 65 year old person oftencomplained of neck pain. One day hisgrandson Mridul, suggested that magnetictherapy is very effective in reducing pain.He said that the permanent magnet!electromagnet, used in the device willhelp to produce joule's heating effects inthe blood stream, which helps the bloodflow better. He immediately contacted hisfriend in Chennai who was running amagnetic therapy clinic.Read the above passage and answer thefollowing questions.

(i) What values did Mridul exhibittowards grandfather?

(ii) What is the S1 unit of magnetic field?(iii) Give the magnitude of magnetic field

in 4A current carrying circular coil orradius 2 cm.

Ans. (i) He is of responsible behaviour, concern andawareness. (1)

(ii) The 51 unit of magnetic field is Tesla. (1)

( ... ) B _ ~oI _ 41t xur ? X 4 _ 50.24 10-5III --- ---x

]r ·2 x 2 x 10- 2 4

= 12.56 x 10-5 T (2)

3. Ms. Kanchan is a student of PG course innanotechnology lab in lIT Kanpur. Thefirst day, when she went to the lab shemet Mr. Cobra, the lab assistant. Hegreeted her and advised her not to touchthe wires, which were suspended from theroof at every part of the lab as they werefrom high voltage lines. He also told hernot to bring any of the two wires closer toeach other during any experimentalapplications. He helped her inunderstanding about precautions that hasto be taken in the lab.Read the above passage and answer thefollowing questions.

(i) What values did Mr. Cobra exhibit?Give any two.

(ii) Why should not the two high voltagewires be brought close to each other?

164

Ans. (i) He is of responsible behaviour, sensitivity,concern for others. [2]

(ii) Because the two wires have high magneticfield around them. [2]

4. Alka and her sister were watching amovie in which the phenomena of auroraborealis was shown. Alka could notbelieve her eyes that, such a colourfuldisplay like the one during commonwealthgames could be created by the nature. Shewent to the library, but could not find theright book. So, she consulted her teacherwho guided her. Hence, Alka understoodthat during a solar flare, a large numberof electrons and protons are ejected fromthe sun. Some of these get trapped in theearth's magnetic field and move in ahelical path along the field lines. As thedensity of the field lines increases nearthe poles, these particles collide withatoms and molecules of the atmosphereemitting green and pink light. Alkashared this knowledge with her classwhen they studied the chapter of movingcharges in magnetic field.Read the above passage and answer thefollowing questions

(i) What values did Alka have?(ii) What is the radius of the path of

an electron moving at a speed of3 x 107 mls in a magnetic field of6 gauss perpendicular to it? What isits frequency? Calculate its energy inkilo electron volt.

Ans. (i) She has analytical knowledge, nature ofappreciations, diligence, curiosity, presence ofmind. [1]

(ii) Using equation,mv2

qvB=-r

mvr=-

qB9 x 10-31 X 3 X 107

1.6x 10 19X 6x 10 4

= 2.8125X10-1

= 28.125 ern

o ehopterwise eSSE Solved Papers PHYSICS

vV=-

27tr= 2x106s-1

= 2x106 Hz

=2MHz

Energy, (E) = ~m;2

= ~ x 9 X 10-31 X 9 X1014

2= 40.5 X10-17 J= 4 x1O-6J

= 25keV

[1]

[1]

5. In the birthday party of Kamal, hisparents gave big slinkie to all his friendsas a return gift. The very next day, duringthe physics class Mr. Mohan, the teacherexplained them about the production ofmagnetic field using current carrying coiland also said that they can makepermanent magnet, using such coils bypassing high currents through them. Thatnight Priyanshu, a friend of Bharat, askedhis father about the coils, and their shape.His father asked him to bring the slinky,that his friend gave and explained the useof toroid and solenoid.Read the above passage and answer thefollowing questions

(i) What value did Priyanshu's fatherhave?

(ii) What is the difference betweensolenoid and toroid?

(iii) Give the value or magnitude ofmagnetic field in solenoid.

Ans. (i) Priyanshu's father is responsible andknowledgeable, makes his child understand theconcepts of solenoid and toroid. [1]

(ii) A solenoid has magnetic field straight withinthe turns and in toroid, it is in form ofconcentric circles. [1]

(iii) The magnitude of magnetic field in solenoid isgiven by

[1]

B = lJ.onI[where, n = number of turns/length,I = current in coil] [2]

ing Charges and Magnetism · 2018. 10. 2. · Ampere's circuital law to calculate the magnetic...

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Transcript of ing Charges and Magnetism · 2018. 10. 2. · Ampere's circuital law to calculate the magnetic...