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

qq

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

qq

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

------------