ONE MARK QUESTIONS - KV No 1 Jaipur Two similarly and equally charged identical metal spheres A and...
Transcript of ONE MARK QUESTIONS - KV No 1 Jaipur Two similarly and equally charged identical metal spheres A and...
ONE MARK QUESTIONS
Q.1 What is the angle between the directions of electric field and electric dipole moment at any
(i) axial and (ii) equatorial point due to an electric dipole?
Ans . axial point is 00 2. equatorial point is 1800
Q.2 What is the net force on a dipole in a uniform electric field?
Ans. The net fore is Zero
Q.3 Write S.I. unit of electrostatic flux, Is it a scalar or a vector quantity?
Ans. S.I. unit Volt x Meter. It is a scalar quantity.
Q.4 Write the S.I. unit of (i) e lectric field intensity and (ii) electric dipole moment.
Ans (i) SI unit of electric field intensity is Newton/ Coulomb (Nc-1)
(ii) SI unit of electric dipole moment is Coulomb x Meter (C-m)
Q.5 A charge 'q' is placed at the centre of a cube. What is the electric flux passing through the
cube?
Ans. Electric flux passing through whole cube is 0
q so
0
q
Q.6 An electric dipole of dipole moment 20x10-6 Cm is enclosed by a closed surface. What is
the net flux coming out of the surface?
Ans. Zero
Q.7 Why does the electric field inside a dielectric decrease when it is placed in an external
electric field?
Ans. Electric filed inside the plates decreases because of dielectric which sets up on electric filed
opposite to the applied filed.
Q.8 How does the coulomb force between two point charges depend upon the dielectric
constant of the intervening medium.
Ans. Coulomb’s force F 1/K, where K is the dielectric constant of the intervening medium
Q.9 If the radius of the Gaussian surface enclosing a charge is halved, how does the electric
flux through the Gaussian surface change?
Ans. Electric flux through a Gaussian surface enclosing the change q is
0
qE
This is independent of radius of Gaussian surface, so if radius is halved , the electric flux
through the surface will remain unchanged
Q.10 What is the electrostatic potential due to an electric dipole at an equatorial point?
Ans. Zero
TWO MARKS QUESTIONS Q.1 Define the term 'electric dipole moment'. Give its unit. Derive an expression for the
maximum torque acting on an electric dipole, when held in a uniform electric field.
Ans. The electric dipole moment is defined as the numerical product of either charge and the
distance b/w the charges. Its direction is from negative to positive charge
p = q l2
Its Unit is Coulomb meter and
ii) When dipole is placed perpendicular to the electric field.
Force on A and B Constitute couple = (qE) x BC = qE2a sin
= qx 2a Esin = P Esin
P x
E
Direction of torque is given by Right handed screw rule. Direction of torque is to
the plane of paper directed inward.
Q.2 S1 and S2 are two hollow concentric spheres enclosing charges Q and 2Q respectively as
shown in the figure.
1. What is the ratio of the electric flux through S1 and S2?
2. How will the electric flux through the sphere S1 change, if a medium
of Dielectric constant 5 is introduced in the space inside S1 in place of
air?
Ans. 0
Q for S1 ,
000
2
323.
QqqqdsE
for S2
Radio 2
1
=
0
0
/3
/
Q
Q
3
1 ::Q1:Q2::1:3
(ii) If a medium of dielectric constant 5 is introduced then flux through S, will become 5
1
time of flux in air
Q.3 What is an electric line of force? Sketch lines of force due to two equal positive charges
placed at a small distance apart in air.
Ans. Electric line :- Electric line is defined as the path followed by a unit positive charge when
it is free to more in an electric filed .
Q.4 Two similarly and equally charged identical metal spheres A and B repel to each other with
a force 2x10–5 N. A third identical uncharged sphere C is touched with A and then placed
at the mid-point between A and B. Calculate the net electric force on c.
Ans. 2.0x10-5 N←-----------------------------------→ C
Q.5 State Gauss' theorem in electrostatics. Using this theorem, derive the expression for the
electric field intensity at any point outside a uniformly charged thin spherical shell.
Ans. Gauss’s theorem in electrostics :- the total electric flux through any closed surface in free
space is 0
1
times the total electric charge enclosed by the surface.
Ǿ=
dsE . = 0
q
Consider a Gaussian surface at radius r outside the hollow sphere
0
24..
q
rEdsE
E= 04
1
2r
q
Q.6 Plot a graph showing the variation of coulomb force (F) versus
2
1
r, where r is the
distance between the two charges of each pair of charges: (1C, 2C) and (2C, -3C).
Interpret the graphs obtained.
Ans. F=04
1
2
2..1
r
The group between F and 2
1
r is straight
line of slope 04
1
q1 q2 passing through
originsince magnitude of the slope is more for
fattractive> frepulsive
Q.7 Two fixed point charges +4e and +e units are separated by a distance 'a'. Where should the
third point charge be placed for it to be in equilibrium?
Ans. For equilibrium 2
)4(
x
eKq=
2)(
)(
xa
eKq
2
4
x=
2)(
1
xa
x
2=
xa
1
xxa 22
3
2ax from A.
Q.8 A system has two charges qA=2.5x10–7 C and qB= –2.5x10–7 C located at points A(0, 0, –
15cm) and B (0, 0, + 15cm) respectively. Calculate the electric dipole moment of the
system. What is its direction?
Ans. 7.5x10-8 cm along –z axis.
Q.9 Define electric flux. Write its S.I. units. A spherical rubber ballon carries a charge that is
uniformly distributed over its surface. As the ballon is blown up and increases in size, how
does the total electric flux coming out of the surface change? Give reason.
Ans. Electric flux:- Electric flux is defined as the total number of electric lines of force passing
through a surface .
SI unit →Nm2/c
(ii) Total electric flux through the surface =0
qas charge remains unchanged when size of
balloon increases, electric flux through the surface remains unchanged
Q.10 Define the dipole moment of an electric dipole. How does the electric potential due to a
dipole vary on the dipole axis as a function of 'r'- distance of the field point from the mid-
point of the dipole- at large distance?
Ans Electric dipole moment:- electric dipole moment of an electric dipole is the product of the
magnitude of either charge and the dipole length.
Potential due to charge +q at A = 1vlr
Kq
Potential due to charge -q at B = 2vlr
Kq
Total Potential=V1+V2 =K
lr
q
lr
q
= Kq
22 lr
lrlr
= Kq 2222
)2(
lr
Kp
lr
l
H here p=2ql dipole moment V=2r
Kp r>>l
04
1
K
2
04 r
PV
Q.11 A uniformly charged conducting sphere of 2.8 m diameter has a surface charge density of
100 C/m2.(a) Find the charge on the sphere.(b) What is the total electric flux leaving the
surface of the sphere?
Ans. 19.6x10-14c , 0
qCNm /102.2
10854.8
1096.1 28
12
4
THREE MARKS QUESTIONS
Q.1 An electric dipole of dipole moment
P is placed in a uniform electrical field
E . Write the
expression for the torque
experienced by the dipole. Identify two pairs of perpendicular
vectors in the expression. Show diagrammatically the orientation of the dipole in the field
for which the torque is (i) Maximum (ii) Half the maximum value (iii) Zero.
Ans. Toque
P x
E , pair torque and p, torque and E
(i) ( =PE sin 900 =PE)
(ii)( =PE sin 30 =2
PE)
(iii) zero torque occurs when is or 1800
Q.2 Using Gauss theorem, derive an expression for the electric field intensity due to an
infinitely long straight wire of linear charge density C/m
Ans.
= E x 2πrl
According to gauss’s law =0
q
E.2πrl =00
lq
E=r02
Q.3 An electric dipole is held in a uniform electric field. (1) Using
suitable diagram, show that it does not undergo any translatory motion, and (ii) derive an
expression for the torque acting on it and specify its direction.
Ans. Let a dipole AB be placed in uniform electric field E .
(i) +q charge will experience a force qE parallel to
E where –q charge experience a force
qE anti-parallel to
E . Since There forces are equal and opposite so no net force is
experienced.
(ii) Force on A and B Constitute couple = (qE) x BC = (qE)x2a sin
= qx 2a Esin = P Esin
P x
E
Direction of torque is given by Right handed screw rule. Direction of torque is to
the plane of paper directed inward.
Q.4 State Gauss's theorem in electrostatics. Using it, deduce an expression for electric field
intensity at a point near a thin infinite plane sheet of electric charge.
Ans. Gauss’s theorm:- The total electronic flux through any
closed surface in free space is 0
1
times the total
electric charge enclosed by the surface .
=
S
dsE . =0
q
=(Ex A)x2 =2EA , 2EA =0
q But q=σ A 2EA=
0
A or (E= ]
2 0
ELECTRIC POTENTIAL AND CAPACITOR
One mark questions Q.1 The graph shown here, shows the variation of the total energy (E) stored a capacitor against
the value of the capacitance (C) itself. Which of the two-the charge on the capacitor or the
potential used to charge it is kept constant for this graph?
Ans. The change (q) is kept Constant.
Q.2 Sketch a graph to show how the charge 'Q' acquired by a capacitor of capacitance 'C'
varies with increase in potential difference between its plates.
Ans. The graph charge (Q) versus potential difference (v) is straight line whose slope is equal to
capacitance ‘C’. Slope=tan CV
Q
Q.3 A hollow metal sphere of radius 6cm is charged such that the potential on its surface is
12V. What is the potential at the centre of the sphere?
Ans. Potential at Centre of Sphere = 12V, because potential remains same inside a
metal/conducting sphere.
Q.4 The electric potential is constant in a region. What can you say about electric field there?
Ans As dr
dvE and v is constant therefore 0E
Q.5 What is meant by capacitance? Give its S.I. unit.
Ans. Capacity to store charge of a body is K/a electric capacitance. The capacitance of a
Capacitor is equal to the ratio of magnitude of change (Q) on either plate or potential diff.
(V) across the plate
V
QC Unit of capacitance is
Volt
Coulomb or farad (F).
Q.6 Define the term' potential energy' of charge 'q' at a distance 'r' in an external electric field.
Ans. work done on charge q in bringing it from infinity to the given point in the electric field
against the electrical force is called potential energy of the charge.
Two marks questions Q.1 Two indentical plane metallic surfaces A and B are kept parallel to each other in air
separated by a distance of 1.0 cm as shown in the figure. Surface A given a positive
potential of 10V and outer surface of B is earthed.
i) What is the magnitude and direction of the uniform electric field between points y and
z?
ii) What is the work done in moving a charge of 20 mC from point X and point Y?
Ans. i) 13
210
101
10
Vm
r
VE Direction of E is from y to z
ii) No work is done because plate A is an equipotential surface 00 qVqW
Q.2 Two point charges 3x 10–8 and –2 x 10–8 C are located 15 cm part in air. Find at what point
on the line joining these charges the electric potential is zero. Take potential at infinity to
be zero.
Ans. (3 x 10-8c) O x P 15cm A (-2 x 10-8c)
Let p be the point, where potential is zero and let OP be x cm.
Potential at infinity is taken zero Potential,
Vp 010)15(4
102
)10(4
1032
8
2
8
xExE oo
015
23
xx x=9cm
Q.3 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply
and is connected to another uncharged 600 Pf capacitor. How much electrostatic energy is
lost in the process?
Ans. Loss of Energy 2
21
21
21 ][2
1VV
CC
CCU
Energy lost 6106 U Joule
Q.4 Find the equivalent capacitance of the combination of capacitors between the points A and
B as shown in the figure. Also calculate the total charge that flows in the circuit when a
100 V battery is connected between the points A and B.
Ans. Ceq= F20)4040(
4040
Total charge q= CeqV.
q C36 1021001020
Q.5 The graph shows the variation of voltage, 'V' across the plates of two capacitors A and B
versus increase of charge, 'Q' stored on them, which of the two capacitors has higher
capacitance? Give reason for your answer.
Ans. eslopeoflinV
QC
1 As slope of A is smaller, capacitance of A is higher.
Q.6 Parallel plate capacitor with air between the plates has a capacitance of 8PF. The
separation between the plates is now reduced by half and the space between them is filled
with a medium of dielectric constant 5. Calculate the value of capacitance of the
capacitor in the second case.
Ans. pFd
AC o 81
in the second care d1=d/2, K=5
PFPFC
d
A
d
A
d
AKC ooo
80810
10
2/
5
2
12
Q.7 Define 'dielectric constant' of a medium. Briefly explain why the capacitance of a parallel
plate capacitor increases, on introducing a dielectric medium between the plates.
Ans. Dielectric Constant: - Dielectric constant K, is the ratio of the capacitance with dielectric
between the plates C1 to the capacitance with vacuum/air between the plates C
K=C1/C.
the electric field inside the dielectric is E = E0 - Ep
Outside the dielectric field, remains E0 only. Therefore, potential difference between
the two plates is V = E0 (d-t) + Et.
Therefore, Capacitance of the capacitor with dielectric between is
Ktd
A
K
ttd
A
V
QC
11
00
i.e.
Ktd
AC
11
0 Clearly, C>C0
Q.8 Explain the underlying principle of working of a parallel plate capacitor.
If two similar plates, each of area A having surface charge densities + and – are
separated by a distance d in air, write expressions for.
1. the electric field at points between the two plates.
2. the potential difference between the plates.
3. the capacitance of the capacitor so formed.
Ans. Principles of a Capacitor: - A capacitor works on the principle that the capacitance of a
conductor increases appreciably when an earthed conductor is brought near it. Thus a
Capacitor has two plates separated by a distance having equal and opposite charges.
Suppose A be the area of each plate, separation between the plates, K the
dielectric constant of medium between the plates. If is the magnitude of charge density
of plates. A
Q
i) electric filed strength between the plates oK
E
Eo = permittivity of free space.
ii) potential diff between the plates VAB=Ed
oK
d
VAB =
AK
d
K
dA
oo
iii) Capacitance of the capacitor so formed
d
AKC
AK
Qd
Q
V
QC
o
o
AB
Q.9 (i) Can two equipotential surfaces intersect each other? Give reasons.
(ii) Two charges - q and +q are located at points A (0,0,-a) and B (0,0,+a) respectively.
How much work is done in moving a test charges from point P (7,0,0) to Q ( -3, 0,0)?
Ans. i) No,. If they intersect there will be two normal at the point of intersection, hence two
directions of electric filed at the same point, which is impossible.
ii) W = 0 Since X-axis is an equipotential surface.
Three marks questions Q.1 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 u
C and - 2 uc (and no external field) placed at (-9 cm,0,0) and (9 cm, 0, 0) respectively.
(b) How much work is required to separate the two charges infinitely away from each
other?
(c) Suppose that the same system of charges is now placed in an external field E= A. 2
1
r,
A=9x 105 cm2. What would the electrostatic energy of the configuration be?
Ans. a) Using r
qqU
0
21
4
JU 7.010)99(
109)102(1072
966
b) work required to separate charges = U2 – U1 = 0- U = 0-(-0.7) = 0.7 J
c) Electrostatic energy of configuration
E = total potential energy
= 2211
0
21
4vqvq
r
dr
dvE
r
Adr
r
AEdrV
2
Putting the value of V, we get
J
r
Aq
r
Aq
r
qqE
3.4920707.0
09.0
)102(109
09.0
)107(109)07.(
4
6565
2
2
1
1
0
21
Q.2 Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C'
and having charge 'Q'.
How will the (i) energy stored and (ii) the electric field inside the capacitor be affected
when it is completely filled with a dielectric material of dielectric constant 'K'?
Ans Let a capacitor of capacitance C be charged to potential V. If q is the charge on the plate
then
C=q/V or V=q/C
Small amount of work done by battery in charging the capacitor to small
charge dq at constant voltage V is given by
dw = V dq = dqC
q
qqqq
Cqdq
Cdq
C
qW
0
2
002
11
222
2
1)(
2
1
2
1CV
C
CV
C
qW
This work done is stored inside the capacitor as potential energy.
2
2
1CVU
a) Battery disconnected:
i) Energy stored is decreased to 1/K times the initial energy.
ii) Electric field is decreased to 1/K times the initial field.
b) Battery connected:
i) Energy stored will increase to K times the initial energy.
ii) Electric field will remain same.
Q.3 A parallel Plate capacitor of plate separation 'd' is charged to a potential difference V. A
dielectric slab of thickness 'd' and dielectric constant 'k' is introduced between the plates
while the battery remains connected to the plates.
1. Find the ratio of energy stored in the capacitor after and before the dielectric is
introduced. Give the physical explanation for this change in stored energy.
2. What happens to the charge on the capacitor?
Ans. d
ACair
0 , d
AKCair
0 (when battery remains connected)
i) Potential air = Potential Medium
ii) Capacity air = K capacity air (Capacity Increases)
iii) Charge air = K x Charge air = Charge also increases
iv) (Electric field)air = K
rElectricai=Electric field decreases
a) K
KCV
CV
afterU
beforeU 1
2
12
1
)(
)(
2
2
b) KKQ
Q
dielectriccontinuouseafterCh
beforeeairCh 1
)arg(
)arg(
Q.4 The two plates of a parallel plate capacitor are 5mm apart. A slab of a dielectric of
thickness 4 mm is introduced between the plates with its faces parallel to them. The
distance between the plates is adjusted so the capacitance of the capacitor become equal
to its original value. If the new distance between the plates equals 8 mm, what is the
dielectric constant of the dielectric used?
Ans. Ca=Cm
K
ttd
A
d
A oo , K=1/4
Q.5 Two parallel plate capacitor, X and Y have the same area of plates and same
separation between them X has air between the plates while Y contains a dielectric
medium of r = 4
(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination
is 4 F.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) What is the ratio of electrostatic energy stored in X and Y?
Ans.
i)Cx=d
Ao Cy=d
Ao
Cx=d
Ao Cy= Xo Cd
A44
,use parallel , CX=5, CY=20
ii) Total Charge =Total Capacitor/Total Potential= 66
103
1
12
104
iii) 5:1
)5(2
1
)1(2
1
y
X
U
U
.FIVE MARKS QUESTIONS
Q.1 A dielectric slab of thickness 't' is kept in between the plates, each of area 'A', of a parallel
plate capacitor separated by a distance 'd'. Derive an expression for the capacitance of this
capacitor for t << d.
Ans. Capacitance with dielectric slab between the plates:- Capacitance of a parallel plate
capacitor with air as the medium between two plates of area A lying at a distance d apart
is
d
AC 0
Let a dielectric slab of thickness t be put between the plates such that t<d. Due to
polarization, electric field will reduce from Eo to E. Potential different across the capacitor
is given by V= E0 (d-t) + E t
Dielectric Constant E
EK 0 Or
K
EE 0
V = E0 (d-t) + K
tE0
00
A
qE
K
ttd
A
qV
0
Capacitance of the arrangement = V
qC
Ktd
AC
11
0
Q.2 (i). Explain briefly how a capacitor stores energy on charging. Obtain
an expression for the energy thus stored.
(ii)A battery of 10 V is connected to a capacitor of 0.1 F. The battery is now removed and
the capacitor is then connected to a second uncharged capacitor of same capacitance.
Calculate the total energy stored in the system.
Ans. i) Capacitors stores energy on charging: - When battery is connected to the
plates of capacitor them charge is transferred to the plates of capacitor from battery. To
transfer the charge battery has to do work and this work done stored in form of electric
energy of capacitor plates.
Energy stored in a Capacitor: -
W= 222
2
1)(
2
1
2
1CV
C
CV
C
q
This work done is stored inside the capacitor as potential energy given by
Energy stored in a capacitor with air as dielectric = 20
2
1V
d
A
And Energy stored in a capacitor with dielectric medium= 20
2
1V
d
KA
ii) When battery connected
JU
U
in 5
10101.02
1
When uncharged capacitor connected with charged capacitor
JU
U
f
f
5.2
52.02
1 2
UNIT - 2
QUESTIONS
1. How does the drift velocity of electrons in a metallic conductor vary with increase in
temperature?
Ans remains the same
2. Two different wires X and Y of same diameter but different materials are joined in series cross
a battery. If the number density of electrons in X is twice that of Y, find the ratio of drift velocity
of electrons in the two wires.
Ans : Vdx/Vdy = ny/nx = ½
3. A 4Ω non insulated wire is bent in the middle by 1800 and boththe halvesare twisted with each
other. Find its new resistance?
Ans 1Ω
4. The resistance in the left gap of a metre bridge is 10Ω and the balance point is 45cm from the left end.
Calculate the value of the unknown resistance.
Ans S = 12.5Ω
5. Two wires of equal length one of copper and the other of manganin have the same resistance.
Which wire is thicker?
Ans Manganin.
6. The V-I graph for a conductor makes angle Ѳ with V- axis, what is the resistance of the
conductor?
Ans R =CotѲ
7. It is found that 1020 electrons pass from point X towards another point Y in 0.1s. What
are the current & its direction?
Ans 160A
8. Two square metal plates A and B are of the same thickness and material. The side B is twice
that of side A. If the resistance of A and B are denoted by RA and RB, find RA/ RB.
Ans : 1
9.The V-I graph of two resistors in their series combination is shown. Which one of these graphs
shows the series combinations of the other two? Give reason for your answer.
Ans : 1
10. Plot the graph showing the variation of conductive with the temperature in a metallic conductor.
Ans:
T
θ
V
I
2
3
(1)
(1)
(1)
(1)
(1)
(1)
(2)
(1)
(1)
11. Draw a graph to show the variation of resistance of the metallic wire as a function of its diameter
keeping the other factor constant.
Ans:
12. Two cells each of emf E and internal resistances r1 and r2 are connected in series to an external
resistance R. Can a value of R be selected such that the potential difference of the first cell is 0.
Ans : I= 2E/(R + r1 + r2) Potential diff. for first cell V1 = E – I r1 = 0
E = (2E r1)/R + r1 + r2 Solving, R = r1 - r2
13. A battery has an emf E and internal resistance r. A variable resistance R is connected across the
terminals of the battery. Find the value of R such that
(a)The current in the circuit is maximum
(b)The potential difference across the terminal is maximum.
Ans : (a) I = E/ (r + R) I = Imax when R =0 Imax = E/r
(b)V = ER/(r + R) = E/(r/R + 1) V = Vmax when r/R + 1= minimum , r/R = o, V= E
14. A piece of silver has a resistance of 1Ω. What will be the resistance of the constantan wire of one
third length and one half diameter if the specific resistance of the constantan wire is 30 times that of the
silver.
Ans : 40Ω
15 .Calculate the current shown by the ammeter in the circuit shown
Ans : R = 2Ω and I = 5A
16. The plot shows the variation of current I through the cross section of a wire over a time interval of
10s. Find the amount of charge that flows through the wire over this time period.
Ans : Area under the I-t graph, q = 37.5C
17. Find the resistance between the points (i)A and B and (ii) A and C in the following network
Ans : (i) RAB = 27.5Ω (ii) RAC = 30Ω
10Ω
5Ω
10Ω
10Ω 10Ω 10Ω
+ -
10V
5Ω A
5 t(s) 10 0
I(A) 5
10Ω 10Ω 10Ω
A
D
B
C
10Ω 10Ω 10Ω
10Ω 10Ω
(2)
(2)
(2)
(2)
(2)
(2)
(2)
18. Two wires of the same material having lengths in the ratio 1:2 and diameter 2:3 are connected in
series with an accumulator. Compute the ratio of p.d across the two wires
Ans : R = ρl/A = 4ρl/πd2 RA/RB = 9/8 VA/VB = IARA/IBRB = 9/8
19. An infinite ladder network of resistances is constructed with 1Ω and 2Ω resistances shown
A 1Ω C 1Ω 1Ω 1Ω
6V 2Ω 2Ω 2Ω
B D
A 6V battery between A and B has negligible resistance.
(i)Find the effective resistance between A and B.
𝑅 =2𝑅
𝑅 + 2+ 1 𝑅 = 2Ω
20. The resistance of a tungsten filament at 150°C is 133Ω. What will be its resistance at 5000C ?
The temperature coefficient of tungsten is 0.00450C-1 at 00C.
Ans : Use Rt = R0 (1+ α t) R500 = 258Ω
21.The circuit shown in the diagram contains two identical lamps P and Q. What will happen to the
brightness of the lamps , if the resistance Rh is increased? Give reason.
ns : Brightness of P and Q decrease and increase respectively.
22.Using Kirchhoff’s laws, calculate I1,I2 andI3
Ans : I1 = 48/31A I2 = 18/31A I3 = 66/31A
23. In the circuit, find the current th rough the 4Ω
resistor.
2Ω
1Ω
2Ω
1Ω
R
1Ω
D 2Ω
B
5Ω
12V
I1 2Ω
I2
I3
3Ω
6V
3Ω 2Ω 2Ω
2Ω 2Ω 2Ω 9V
4Ω 8Ω 8Ω
I1 I2 I3
Ans : Since the circuit is infinitely
long, its total resistance remains
unaffected by removing one mesh
from it. Let the effective resistance
of the infinite network be R, the
circuit will be
Ans : Since the circuit is infinitely long, its total resistance remains unaffected by removing one
mesh from it. Let the effective resistance of the infinite network be R, the circuit will be
B
(3)
(3)
(3)
(3)
(3)
(3)
Ans : I = 1A
24. How will you compare emf of two cells using a potentiometer? Explain with a neat circuit diagram
25.Two cells of emfs E1 and E2 (E1> E2) are connected as shown
A B C
E1 E2
When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is
300cm. On connecting the same potentiometer between A and C, the balancing length is 100cm.
Calculate the ratio of E1 and E2.
Ans : E1 α 300 , E1 – E2 α 100, E1/ E2 = 3/2 (3)
26. Calculate the current drawn from the battery in the given network.
Ans : I = 2A
27.Find the value of X and current drawn from the batte ry of
emf 6V
Ans : X = 6Ω and I = 1A
28.Find the value of the unknown resistance X and the current drawn by the circuit from the battery if
no current flows through the galvanometer. Assume the resistance per unit length of the wire is
0.01Ωcm-1.
Ans X = 3Ω
29. In the circuit shown, AB is a resistance wire of uniform cross – section in which a potential gradient
of 0.01V cm-1 exists.
(a)If the galvanometer G shows zero deflection, what is the emf E1 of the cell used?
(b)If the internal resistance of the driver cell increases on some account, how will it change the
balance point in the expt.
2Ω
4Ω
1Ω 5Ω
2Ω
4V
2Ω 4Ω 10Ω
3Ω X
6V 2.4Ω
X 2Ω
G
4V
120 80
(3)
(3)
Ans : (a) PD VAB = 1.8 V (b)Balance pt. will shift towards B since V/l decreases.
30. In a potentiometer circuit of a battery of negligible internal resistance is set up as shown to develop a
constant potential gradient along the wire AB. Two cells of emfs E1 and E2 are connected in series as
shown in the combination (1) and (2). The balance points are obtained respectively at 400cm and 240cm
from the point A. Find (i) E1/ E2 and (ii) balancing length for the cell E1 only.
Ans : E1 + E2 α 400,E1-E2 α 240,Solving E1/ E2 = 4,
E1 α l1, (E1 + E2)/E1 = 400/l1 l1 = 320cm
31. A potentiometer wire of length 100cm has a resistance of 10Ω is connected in series with a resistance
and cell of emf 2V of negligible internal resistance. A source of emf of 10mV is balanced against a length
of 40cm of potentiometer wire. What is the value of the external resistance?
E = 2V
A
100cm
Ans : I = E/(R + 10) = (2/R + 10) Resistance of 40cm wire is 4Ω.
At J, (2/R +10) x 4 = 10 x 10-3 R = 790Ω
UNIT – III
Important Derivations
Three marks questions
Q. 1. State Biot-Savart law for magnetic field produced at a point due to a small current element.
How will you find the direction of magnetic field?
J 180cm
G
A B
E1
E2
G E1 E2
E1 E2
E = 10mV
(3)
(3)
R
(3)
G
40cm J B
Ans. Consider a current element ld
of a conductor carrying current I. let P be a point having
position vector r
with respect to current element ld
. Let 𝜃 be the angle between ld
and
r
According to Biot-Savart law, the magnetic field dB at point P due to current element is
(i) dB I (ii) dB∝dl (iii) dB sin 𝜃 (iv) dB∝ 2
1
r
Combining all the four factors, we get
dB 2
sin
r
dlI
dB = 2
sin
r
dlIK
The direction of Bd
is the direction of the vector rld
. It is given by right hand screw
rule. If we place a right handed screw at point P perpendicular to the plane of paper and
turn its handle from ld
to r
, then the direction in which the screw advances gives the
direction of Bd
. Thus the direction of Bd
is perpendicular to and into the plane of paper,
as has been shown by encircled cross at point P in the figure.
Q. 2. State the factors on which the force acting on a charge moving in a magnetic field
depends. Write the expression for this force. When is this force minimum and maximum?
Ans. Suppose a charge q moves with velocity v
in a magnetic field B
at angle 𝜃.
It is found that the charge q experiences a force F
such that
1. The force is proportional to the magnitude of the magnetic field, i.e., F B
2. The force is proportional to the charge q, i.e.,F q
3. The force is proportional to the component of the velocity v in the perpendicular direction
of the field B, i.e., F v sin𝜃
Combining the above factors, we get
F Bqv sin𝜃 or F = kqvB sin𝜃
The unit of magnetic field is so defined that the proportionality constant k becomes unity
in the above equation. Then
F = qvB sin𝜃
As the direction of F
is perpendicular to both v
and B
, so we can express F
as
F
= q( Bv
)
Q. 3. Derive a mathematical expression for the force acting on a current carrying straight
conductor kept in a magnetic field. Under what conditions is this force (i) zero and (ii)
maximum?
Ans. consider a conductor of length l, area of cross-section A, carrying current I lies
perpendicular to a magnetic field B
.
Each electron moving with drift velocity vdexperiences a magnetic Lorentz force
)( Bvef d
If n is the number of free electrons per unit volume, then total number of electrons in the
conductor is
N = n x volume = nAl
Total force on the conductor is
][)]([ BvlenABvenAlfNF dd
If lI
represents a current element vector in the direction of current, then vectors l
and
dv
will have opposite directions and we can take
l dv
= lvd
][)(
)(
IenAvBlIF
BlenAvF
d
d
The magnitude of the force is F = I l B sin𝜃
(i) If 𝜃=0 or 180 , then F = I l B(0) = 0
Thus a current carrying conductor placed parallel to the direction of the magnetic field
does not experience any force.
(ii) If 𝜃=90 , then F = I l B sin90o or Fmax = I l B
Thus a current carrying conductor placed perpendicular to the direction of a magnetic
field experiences a maximum force.
Q. 4. Define current sensitivity and voltage sensitivity of a galvanometer. State the factors on
which the sensitivity of a moving coil galvanometer depends.
Ans. Current sensitivity: It is the deflection produced in the galvanometer when a unit
current flows through it.
Current sensitivity, k
NBA
II s
Voltage sensitivity: It is the deflection produced in the galvanometer when a unit
potential difference is applied across its ends.
Voltage sensitivity, G
sRk
NBA
VV
Factors affecting the sensitivity:
1. Number of turns N in its coil, NI s
2. Magnetic field B BI s
3. Area A of the coil, AI s
4. Torsion constant k of the spring, k
I s
1
Q. 5. How will you convert a galvanometer into an ammeter of range 0 - I amperes? What is
the effective resistance of an ammeter?
Ans. An ammeter is connected in series to a circuit. So it must have very small resistance so
that it does not affect the current. Therefore to convert a galvanometer into an ammeter, a
low resistance, called shunt, is connected in parallel with the galvanometer coil.
Let RG be the resistance of galvanometer and Ig be the current with which galvanometer
gives full scale deflection. A shunt of resistance S is connected in parallel with it. To
measure maximum current I, the maximum current through galvanometer must be Ig and
hence rest current I - Ig should pass through the shunt.
As galvanometer and shunt are in parallel, the potential difference across them is equal.
So,
SIIGI gg
or g
g
II
GIS
The effective resistance of the ammeter becomes
SG
GSRA
Q. 6. How can a galvanometer be converted into a voltmeter to read a maximum potential
difference V? Discuss with related mathematical expression.
Ans. A voltmeter is connected in parallel with a circuit element. So it must draw a very small
current, otherwise the voltage to be measured would decrease. To insure it, a large
resistance is connected in series with the galvanometer.
Let RG be the resistance of galvanometer and Ig be the current with which galvanometer
gives full scale deflection. To measure a maximum potential difference V, a high
resistance R is connected in series with it.
Total resistance of the device = R + RG
Therefore by Ohm’s law
)( Gg RRIV
g
GI
VRR or G
g
RI
VR
Q. 7. How will you select materials for making permanent magnets, electromagnets and cores
of transformers?
Ans. A. Permanent magnets-The material used for making permanent magnets must have the
following characteristics:
1. High retentivity 2. High coercivity 3. High permeability.
B. Electromagnets-The material used for making cores of electromagnets must have the
following characteristics:
1. High initial permeability 2. Low retentivity
C. Transformer cores-The material used for making cores of transformers must have the
following characteristics:
1. High initial permeability 2. Low hysteresis loss 3. Low resistivity
Five marks questions
Q. 1. Using Biot-Savart law, deduce an expression for the magnetic field on the axis of a circular
current loop. Hence obtain the expression for the magnetic field at the centre of the loop.
Ans. Consider a circular loop of wire of radius a and carrying current I, as shown in figure.
Let the plane of the loop be perpendicular to the plane of paper. We wish to find field B
at an axial point P at a distance r from the centre C.
Consider a current element ld
at the top of the loop. It has an outward coming current.
If s
be the position vector of point P relative to the element ld
, then from Biot-Savart
law, the field at point P due to the current element is
2
0 sin
4 r
dlIdB
Since .,., eisld
𝜃 = 90 , therefore
2
0
4 s
IdldB
The field Bd
lies in the plane of paper and is perpendicular to s
, as shown by PQ.Let
be the angle between OP and CP. Then dB can be resolved into two rectangular
components.
1. dB sin along the axis, 2. dB cos perpendicular to the axis.
For any two diametrically opposite elements of the loop, the components perpendicular to
the axis of the loop will be equal and opposite and will cancel out. Their axial
components will be in the same direction, i.e., along CP and get added up.
Therefore, total magnetic field at the point P in the direction CP is
𝐵 = sin dB
𝐵𝑢𝑡 s
asin and
2
0
4 s
IdldB
B = s
a
s
Idl..
4 2
0
Since 0 and I are constant, and s and a are same for all points on the circular loop, we
have
𝐵 =
]2[2
2.44 3
2
0
3
0
3
0 anceCircumferedIs
aIa
s
IadI
s
Ia
21
22 )(, arsAs
23
22
2
0
)(2 ar
aIB
If the coil consists on N turns, then .)(2 2
322
2
0
ar
aINB
Magnetic field at the centre: For the field at the centre of the loop, r = 0. Therefore
a
INB
2
0
Q. 2. State ampere’s circuital law connecting the line integral of B
over a closed path to the net
current crossing the area bounded by the path. Use the law to derive the formula for the
magnetic field due to an infinitely long straight current carrying wire.
Ans. Ampere’s circuital law: This law states that the line integral of the magnetic field B
around any closed circuit is equal to 0 (permeability constant) times the total current I
threading or passing through this closed circuit.
Mathematically IIdB 0.
Application of Ampere’s law to an infinitely long straight conductor: Figure shows a circular loop of radius r around an infinitely long straight wire carrying a
current I.
As the field lines are circular, the field B
at any point of the circular loop is directed
along the tangent to the circle at that point. By symmetry, the magnitude of field B
is
same at every point of the circular loop. Therefore,
rBdlBlldB 2.0cos. dB
Form Ampere’s circuital law,
B.2 Ir 0
r
IB
2
0
Q. 3. A long solenoid with closely wound turns has n turns per unit of its length. A steady current
I flows through this solenoid. Use Ampere’s circuital law to obtain an expression for the
magnetic field at a point on its axis and close to its mid point.
Ans. The magnetic field inside a closely wound long solenoid is uniform everywhere and zero
outside it. Figure shows the sectional view of a long solenoid. At various turns of the
solenoid, current comes out of the plane of paper at points marked and enters the plane
of paper at points marked .
To determine the magnetic field B
at any inside point, consider a rectangular closed path
abcd as the Amperean loop. According to Ampere’s circuital law,
abcdloopthethroughcurrentTotalldB 0.
Now
b
a
c
b
d
c
a
d
ldBldBldBldBldB
.....
b
a
b
a
b
a
lBdIBdlBldBldB 0cos..
Where, l = length of the side ab of the rectangular loop abcd.
Let number of turns per unit length of the solenoid = n
Then number of turns in length l of the solenoid = nl
∴ Total current threading the loop abcd = nlI
Hence Bl = nlI0
nIBor 0
Q. 4. Discuss the motion of a charged particle in a uniform magnetic field with initial velocity
(i) parallel to the field, (ii) perpendicular to the magnetic field and (iii) at an arbitrary
angle with the field direction.
Ans. When a charged particle having charge q and velocity v
enters a magnetic field B
, it
experiences a force
)( BvqF
The direction of this force is perpendicular to both v
and B
. The magnitude of this force
is
F = qv B sin𝜃
Following three cases are possible:
1. When the initial velocity is parallel to the magnetic field:
Here 𝜃=0 , So F = qvB sin0 = 0.
Thus the parallel magnetic field does not exert any force on the moving charged particle.
The charged particle will continue to move along the line of force.
2. When the initial velocity is perpendicular to the magnetic field:
Figure shows a magnetic field B
directed normally into the plane of paper, as shown by
small crosses.
A charge +q is projected with a speed v in the plane of the paper. The velocity is
perpendicular to the magnetic field. A force F = qvB acts on the particle perpendicular to
both v
and B
. This force continuously deflects the particle sideways without changing
its speed and the particle will move along a circle perpendicular to the field. Thus the
magnetic force provides the centripetal force. Let r be the radius of the circular path.
Centripetal force, r
mv2
= Magnetic force, qvB
r = qB
mv
Period of revolution, T=qB
m
qB
mv
vv
r 2.
22
The frequency of revolution is m
qB
Tf c
2
1
This frequency is independent of v and r and is called cyclotron frequency.
3. When the initial velocity makes an arbitrary angle with the field direction:
Consider a charged particle q entering a uniform magnetic field B
with velocity v
inclined at an angle 𝜃 with the direction of B
,as shown in figure.
The perpendicular component, v = v sin𝜃 of the initial velocity makes the charge move
along a circular path of radius,
r = 𝑚
v
𝑞𝐵=
𝑚𝑣 sin 𝜃
𝑞𝐵
The period of revolution is
qB
m
qB
mv
vv
rT
2sin
sin
22
The parallel component, llv = v cos 𝜃 of the initial velocity makes it move along the
direction of the magnetic field. Hence the resultant path of the charged particle will be a
helix, with its axis along the direction of B
, as shown in figure.
The distance moved along the magnetic field in one rotation is called pitch of the helical
path.
Therefore
Pitch = qB
vm
qB
mvTvII
cos22cos
Q. 5. With the help of a labeled diagram, explain the principle, construction, theory and working
of a cyclotron.
Ans. It is a device used to accelerate charged particles like protons, deuterons, 𝛼 - particles,
etc., to very high energies.
Principle: A charged particle can be accelerated to very high energies by making it pass
though a moderate electric field a number of times. This can be done with the help of a
perpendicular magnetic field which throws the charged particle into a circular motion, the
frequency of which does not depend on the speed of the particle and the radius of the
circular orbit.
Construction: As shown in figure, a cyclotron consists of two small, hollow, metallic
half- cylinders D1 and D2, called dees. An alternating voltage is applied across the gap
between the two dees. The dees are placed between the poles of a strong electromagnet.
A source of charged particles is placed near the centre of the dees. These ions move on a
circular path in the dees, D1 and D2, on account of the uniform perpendicular magnetic
field B. The whole arrangement is evacuated to minimize collisions between the ions and
the air molecules.
Theory: As a particle of charge q and mass m follows a circular path under the effect of
perpendicular magnetic field B, so
Magnetic force on charge = Centripetal force
Or qB
mvror
r
mvBqv
2
90sin
Period of revolution of the charged particle is given by
T= qB
m
qB
mv
vv
r 2.
22
Hence frequency of revolution of the particle will be m
qB
Tfc
2
1
Clearly, this frequency is independent of both the velocity of the particle and the radius of
the orbit and is called cyclotron frequency of magnetic resonance frequency.
Working: Suppose a positive ion enters the gap between the two dees and finds dee D1 to
be negative. It gets accelerated toward dee D1. As it enters the dee D1, it does not
experience any electric field due to shielding effect of the metallic dee. The perpendicular
magnetic field throws it into a circular path. At the instant the ion comes out of dee D1, it
finds dee D2 negative. It now gets accelerated towards dee D2 . It moves faster through D2
describing a larger semicircle than before. Thus if the frequency of the applied voltage is
kept exactly the same as the frequency of revolution of the ion, then every time the ion
reaches the gap between the two dees, the electric field is reversed and ion receives a
push and finally it acquires very high energy.
Q. 6. Derive an expression for the torque on a rectangular coil of area A, carrying a current I and
placed in a magnetic field B. The angle between the direction of B and vector perpendicular
to the plane of the coil is 𝜃.
Ans. Consider a rectangular coil PQRS suspended in a uniform magnetic field B
, with its axis
perpendicular to the field.
Let I be the current flowing through the coil PQRS, a and b be the sides of the coil
PQRS,
A = ab = area of the coil and 𝜃 is the angle between the direction of B
and normal to the
plane of the coil.
According to Fleming’s left hand rule, the magnetic forces on sides PS and QR are equal,
opposite and collinear (along the axis of the loop),so their resultant is zero.
The side PQ experiences a normal inward force equal to IbB while the side RS
experiences an equal normal outward force. These two forces form a couple which exerts
a torque given by
= Force x perpendicular distance
= IbB x a sin𝜃 = IBA sin𝜃
If the rectangular loop has N turns, the torque increases N times i.e.,
= NIBA sin𝜃
But NIA = m, the magnetic moment of the loop, so
= mB sin𝜃
In vector notation, the torque
is given by
= Bxm
The direction of the torque t
is such that it rotates the loop clockwise about the axis of
suspension.
Q. 7. With the help of a neat and labeled diagram, explain the underlying principle, construction
and working of a moving coil galvanometer. What is the function of (i) uniform radial field
(ii) soft iron core in such a device?
Ans. A galvanometer is a device to detect current in a circuit, the magnitude of which depends
on the strength of current.
Construction: A pivoted-type galvanometer consists of a rectangular coil of fine
insulated copper wire wound on a light aluminium frame. The motion of the coil is
controlled by a pair of hair springs of phosphor-bronze. The springs provide the restoring
torque. A light aluminium pointer attached to the coil measures its deflection on a
suitable scale.
The coil is placed symmetrically between the concave poles of a permanent horse-shoe
magnet. There is a cylindrical soft iron core which not only makes the field radial but
also increases the strength of the magnetic field.
Theory and working: As the field is radial, the plane of the coil always remains parallel
to the field B
. When a current flows through the coil, a torque acts on it. It is
= Force x perpendicular distance = NIbB x a sin 90 = NIB(ab) = NIBA
Here 𝜃=90 , because the normal to the plane of coil remains perpendicular to the field B
in all positions.
The torque deflects the coil through an angle 𝛼. A restoring torque is set up in the coil
due to the elasticity of the springs such that
krestoring
Where k is the torsion constant of the springs i.e., torque required to produce unit angular
twist.
In equilibrium position,
Restoring torque = Deflecting torque
k𝛼 = NIBA
k
NBAI
Thus the deflection produced in the galvanometer coil is proportional to the current
flowing through it.
Functions:
(i) A uniform magnetic field provides a linear current scale.
(ii) A soft iron core makes the field radial. It also increases the strength of the magnetic field
and hence increases the sensitivity of the galvanometer.
One mark questions
1. What is the direction of the force acting on a charge particle q, moving with a velocity v
in a uniform magnetic field B?
Ans: Force, F= q v B sinθ
Obviously, the force on charged particle is perpendicular to both velocity v and magnetic
field B.
2. Magnetic field lines can be entirely confined within the core of a toroid, but not within a
straight solenoid. Why?
Ans: Magnetic field lines can be entirely confined within the core of a toroid because toroid
has no ends. A solenoid is open ended and the field lines inside it which is parallel to the
length of the solenoid, cannot form closed curved inside the solenoid.
3. An electron does not suffer any deflection while passing through a region of uniform
magnetic field. What is the direction of the magnetic field?
Ans: Magnetic field is parallel or antiparallel to velocity of electron i.e., angle between v and
B is 0° or 180°.
4. A beam of a particles projected along +x-axis, experiences a force due to a magnetic field
along the +y-axis. What is the direction of the magnetic field?
Ans: By Fleming’s left hand rule magnetic field must be along negative Z-axis
5. What is the characteristic property of a diamagnetic material?
Ans: These are the substances in which feeble magnetism is produced in a direction opposite to
the applied magnetic field. These substances are repelled by a strong magnet. These
substances have small negative values of susceptibility and positive low value of relative
permeability.
6. An electron and a proton moving with the same speed enter the same magnetic field
region at right angles to the direction of the field. Show the trajectory followed by the
two particles in the magnetic field. Find the ratio of the radii of the circular paths which
the particles may describe.
Ans: Trajectories are shown in fig.
7. The permeability of a magnetic material is 0.9983. Name the type of magnetic materials
it represents.
Ans: As permeability < 1, so magnetic material is diamagnetic.
8. Where on the surface of Earth is the angle of dip zero?
Ans: Angle of dip is zero at equator of earth’s surface.
9. A narrow beam of protons and deuterons, each having the same momentum, enters a
region of uniform magnetic field directed perpendicular to their direction of momentum.
What would be the ratio of the circular paths described by them?
Ans: As q
rqB
mvr
1
So, 1:1: dp rr
10. Mention the two characteristic properties of the material suitable for making core of a
transformer.
Ans: Two characteristic properties: (i) Low hysteresis loss (ii) Low coercivity
11. An electron is moving along positive x axis in the presence of uniform magnetic field
along positive y axis. What is the direction of the force acting on it?
Ans: negative z direction.
12. Why should the spring or suspension wire in a moving coil galvanometer have low
torsional constant?
Ans: Sensitivity of a moving coil galvanometer is inversely proportional to the torsional
constant.
13. Steel is preferred for making permanent magnets whereas soft iron is preferred for
making electromagnets .Give one reason.
Ans: steel-- high retentivity, high coercivity
Soft iron-- high permeability and low retentivity.
14. Where on the surface of the earth is the vertical component of earth’s magnetic field
zero?
Ans: At equator.
Two marks questions
1. Define magnetic susceptibility of a material. Name two elements, one having positive
susceptibility
and the other having negative susceptibility. What does negative susceptibility signify ?
Ans: Magnetic susceptibility: It is defined as the intensity of magnetisation per unit
magnetising field,
It has no unit.
Iron has positive susceptibility while copper has negative susceptibility.
Negative susceptibility of a substance signifies that the substance will be repelled by a
strong magnet or opposite feeble magnetism induced in the substance.
2. Define the term magnetic dipole moment of a current loop. Write the expression for the
magnetic moment when an electron revolves at a speed ‘v’, around an orbit of radius ‘ r’
in hydrogen atom.
Ans: Magnetic moment of a current loop:
M = NIA
i.e., magnetic moment of a current loop is the product of number of turns, current flowing
in the loop and area of loop. Its direction is perpendicular to the plane of the loop.
Magnetic moment of Revolving Electron, M = evr/2
3. Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the
current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer.
Justify.
Ans: Current sensitivity : It is defined as the deflection of coil per unit current flowing in it.
Current Sensitivity, S=NAB/C
Voltage sensitivity : It is defined on the deflection of coil per unit potential difference
across its ends.
Voltage Sensitivty, SV=NAB/GC
where G is resistance of galvanometer.
Justification: When number of turns N is doubled, then the current sensitivity (µN) is
doubled; but at the same time, the resistance of galvanometer coil (G) will also be
doubled, so voltage sensitivity Swill remain unchanged; hence inreasing current
sensitivity does not necessarily increase the voltage sensitivity.
4. A wire of length L is bent round in the form of a coil having N turns of same radius. If a
steady current I flows through it in a clockwise direction, find the magnitude and
direction of the magnetic field produced at its centre.
Ans:
L
IN
r
INB
N
LrrNL
2
00
2
22
5. A point charge is moving with a constant velocity perpendicular to a uniform magnetic
field as shown in the figure. What should be the magnitude and direction of the electric
field so that the particle moves undeviated along the same path?
Ans: Magnitude of electric field is vB and its direction is along positive Y-axis.
6. (i) Write two characteristics of a material used for making permanent magnets.
(ii) Why is core of an electromagnet made of ferromagnetic materials?
Ans: (i) For permanent magnet the material must have high retentivity and high coercivity
(e.g.,steel).
(ii) Ferromagnetic material has high retentivity, so when current is passed in
ferromagnetic material it gains sufficient magnesium immediately on passing a current
through it.
7. Draw magnetic field lines when a (i) diamagnetic, (ii) paramagnetic substance is placed
in an external magnetic field. Which magnetic property distinguishes this behaviour of
the field lines due to the two substances?
Ans:
The magnetic susceptibility of diamagnetic substance is small and negative but that of
paramagnetic substance is small and positive.
8. Deduce the expression for the magnetic dipole moment of an electron orbiting around the
central nucleus.
Ans: Consider an electron revolving around a nucleus (N) in circular path of radius r with
speed v. The revolving electron is equivalent to electric current
r
ve
vr
e
T
eI
2/2
Area of current loop (electron orbit), A = p r2
Magnetic moment due to orbital motion, M= IA=evr/2
------------