NCERT Solutions for Moving Charges and Magnetism Class 12 Class 12 Physics typically covers the topic of moving charges and magnetism, which is an essential part of electromagnetism. For more information, visit-www.vavaclasses.com

1. MOTION OF CHARGES AND MAGNETISM
Reading Materials
1. Write a brief description about the invention of magnetic effects by a current carrying
conductor.
In 1820, the Danish physicist Hans Christian Oersted noticed that a current in a straight
wire caused a noticeable deflection in a nearby magnetic compass needle. It is noticeable
when the current is large and the needle sufficiently close to the wire so that the earth’s
magnetic field may be ignored. Reversing the direction of the current reverses the orien-
tation of the needle. The deflection increases on increasing the current or bringing the
needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in
concentric circles with the wire as the centre. Oersted concluded that moving charges or
currents produced a magnetic field in the surrounding space.
Figure 1:
2. State and explain Lorentz force.
Lorentz force is the force acting on a charge in a combined electric and magnetic field.
The force on an electric charge q moving with a velocity v can be written as
F = q
h
~
E + ~
v × ~
B
i
= Felectric + Fmagnetic
The Lorentz force depends on the following factors
(a) It depends on q, v and B (charge of the particle, the velocity and the magnetic field).
Force on a negative charge is opposite to that on a positive charge.
(b) The magnetic force q [ v × B ] includes a vector product of velocity
(c) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving
charge feels the magnetic force.
3. Show diagrammatically, the direction of F,v and B in the case of Lorentz force.
Figure 2:
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2. 4. Derive an equation for the magnetic force experienced by a current carrying conductor
placed in a magnetic field B?
Consider a rod of a uniform cross-sectional area A and length l. Let the number density
of these charge carriers in it be n. Then the total number of charge carriers in it is nAl.
For a steady current I in this conducting rod, let vd drift velocity . In the presence of an
external magnetic field B, the force on these carriers is:
F = (charge)vd × B
F = (nAl)qvd × B
where q is the value of the charge on a carrier. Now current
I = nAqvd
Thus,
F = [(nqvd)Al] × B = I1 × B
where l is a vector of magnitude l, the length of the rod, and with a direction identical
to the current I
5. What are permittivity and permeability?
Electric permittivity 0 is a physical quantity that describes how an electric field affects
and is affected by a medium. It is determined by the ability of a material to polarise in
response to an applied field, and thereby to cancel, partially, the field inside the material.
Similarly, magnetic permeability µ0 is the ability of a substance to acquire magnetisation
in magnetic fields. It is a measure of the extent to which magnetic field can penetrate
matter.
6. Describe the motion of a charged particle in uniform and non-uniform magnetic field?
In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular
to the velocity of the particle. So no work is done and no change in the magnitude of the
velocity is produced(though the direction of momentum may be changed).
Consider motion of a charged particle in a uniform magnetic field. First consider the case
of v perpendicular to B. The perpendicular force, q(v × B) , acts as a centripetal force
and produces a circular motion perpendicular to the magnetic field. The particle will
describe a circle if v and B perpendicular to each other.
Figure 3:
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3. If velocity has a component along B(ie, velocity is not purpendiculat to B), this component
remains unchanged as the motion along the magnetic field will not be affected by the
magnetic field, since F in that direction is zero There fore the particle will execute a
helical motion
7. Derive an expression for the radius of the circular path followed by a charged particle
crossing a uniform magnetic field normally.
Consider a charged particle q is projected into a uniform magnetic field B with an initial
velocity v perpendicular to the magnetic field. It gets trapped in a circular path. The
force exerted by the field provides the necessary centripetal force.
The magnetic force is,
Fm = qvB
The direction of this force will be perpendicular to both B and v. This force provide the
centripetal force for the charge to move in circular motion. Ler r be radius of circular
path.
The centripetal force is,
Fc =
mv2
r
Since centripetal force = magnetic force,
Fc = Fm
mv2
r
= qvB
r =
mv
qB
Figure 4:
8. What is pitch? Derive an equation for the same?
For the circular path of a particle, r = mv
qB
The larger the momentum, the larger is the radius and bigger the circle described. If ω
is the angular frequency, then v = ωr.
So ω = 2πf = qB
m
, which is independent of the velocity or energy. Here f is the frequency
of rotation. The independence of f from energy has important application in the design
of a cyclotron .The time taken for one revolution is T = 2π
ω
.
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4. If there is a component of the velocity parallel to the magnetic field, it will make the
particle move along the field and the path of the particle would be a helical one. The
distance moved along the magnetic field in one rotation is called pitch p. We have
p = v||T =
2πmv||
qB
The radius of the circular component of motion is called the radius of the helix.
9. Explain the working of velocity selector.
consider the simple case in which electric and magnetic fields are perpendicular to each
other and also perpendicular to the velocity of the particle.
Figure 5:
The force acting on the charge is,
F = Fe + Fm
The electric force is
Fe = qE
The magnetic force is
Fm = qV B
If the electric and magnetic forces are in opposite directions the total force on the charge
is zero and the charge will move in the fields undeflected. Then
Fe = Fm
qE = qvB
v =
E
B
This condition can be used to select charged particles of a particular velocity out of a beam
containing charges moving with different speeds. The crossed E and B fields, therefore,
serve as a velocity selector.
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5. 10. State and explain Biot-Savart Law ?
Consider a XY carrying current I. Consider an infinitesimal element dl of the conductor.
The magnetic field dB due to this element is to be determined at a point P which is
at a distance r from it. Let θ be the angle between dl and the displacement vector r.
According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional
to the current I, the element length |dl|,and inversely proportional to the square of the
distance r
Its direction is perpendicular to the plane containing dl and r
~
dB ∝
I ~
dl × ~
r
r3
~
dB =
µ0
4π
I ~
dl × ~
r
r3
The magnitude of the field is
~
dB =
µ0
4π
Idlsinθ
r2
where
µ0
4π
= 4π × 10−7
TmA−1
and µ0 = permeability of free space(vacuum).
11. Compare the differences and similarities between Biot-Savart law and Coulomb’s law?
(a) Both are long range, since both depend inversely on the square of distance from the
source to the point of interest.
(b) The electrostatic field is produced by a scalar source, namely, the electric charge.
The magnetic field is produced by a vector source I dl
(c) The electrostatic field is along the displacement vector joining the source and the field
point. The magnetic field is perpendicular to the plane containing the displacement
vector r and the current element I dl.
(d) There is an angle dependence in the Biot-Savart law which is not present in the
electrostatic case.
12. Derive an expression for the magnetic field due to a current-carrying coil at a point on
its axis.
Consider a circular coil having radius a and centre O from which current I flows in
anticlockwise direction. The coil is placed at YZ plane so that the centre of the coil
coincide along X-axis. P be the any point at a distance x from the centre of the coil
where we have to calculate the magnetic field. Let dl be the small current carrying
element at any point A at a distance r from the point P where
x =
√
a2 + r2
and the angle between r and dl is 90°.
5
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6. Figure 6: Magntic field due to a circular coil carrying current
Then from Biot-Savart law, the magnetic field due to current carrying element dl is
~
dB =
µ0
4π
Idlsinθ
x2
Here θ = 90o
~
dB =
µ0
4π
Idl
x2
The net magnetic field at P is,
B =
I
dBsinφ
=
I
µ0
4π
Idl
x2
sinφ
=
µ0
4π
I
x2
sinφ
I
dl
But, sinφ = r
√
a2+r2 and
H
dl = 2πr the circumference of the coil.
B =
µ0
4π
I
x2
r
√
a2 + r2
B =
µ0Ir2
2(a2 + r2)3/2
13. State Ampere’s circuital law? Derive expression for magnetic field due to a straight
current carrying conductor.
The line integral of magnetic field around a closed loop is equal to µ0 times the total
current passing through the loop.
I
B.dl = µ0I
Consider a straight conductor carrying a current I. P is a point at a distance r from the
conductor. To find the magnetic field at P imagine a circular loop passing through P
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7. with conductor at the center. Let B be the magnetic file at P.
Then as per Ampere’s circuital law,
I
B.dl = µ0I
I
Bdlsinθ = µ0I
I
Bdlsin90o
= µ0I
I
Bdl = µ0I
B
I
dl = µ0I
I
dl = 2πr, the circumfrence of loop
B2πr = µ0I
B =
µ0I
2πr
14. Suggest a method to find out the direction of magnetic field produced by a current carrying
straight conductor?
Right-hand thumb rule : Hold the wire in your right hand with your extended thumb
pointing in the direction of the current. The direction of the curled fingers give the
direction of magnetic field.
15. What is a solenoid? Derive an equation for the magnetic field produced by a solenoid?
Solenoid consists of a long wire wound in the form of a helix where the neighboring turns
are closely spaced. So each turn can be regarded as a circular loop. The net magnetic
field is the vector sum of the fields due to all the turns.
Figure 7:
Consider a rectangular Amperian loop abcd. Along cd the field is zero . Along transverse
sections bc and ad, the field component is zero. Thus, these two sections make no con-
tribution.
Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h
Let n be the number of turns per unit length, then the total number of turns is nh. The
enclosed current is, I (n h), where I is the current through each turn in the solenoid. From
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8. Ampere’s circuital law
BL = µ0I
Bh = µ0I(nh)
B = µ0I
The direction of the field is given by the right-hand rule. The solenoid is commonly used
to obtain a uniform magnetic field.
16. What is a toroid, Derive an equation for the magnetic field produced by a toroid?
The toroid is a hollow circular ring on which a large number of turns of a wire are closely
wound. It can be viewed as a solenoid which has been bent into a circular shape to close
on itself.
Figure 8:
It can be viewed as a solenoid which has been bent into a circular shape to close on itself.
It is shown in the figure carrying a current I. We shall see that the magnetic field in the
open space inside (point P) and exterior to the toroid (point Q) is zero. The field B inside
the toroid is constant in magnitude for the ideal toroid of closely wound turns.
We shall now consider the magnetic field at S. From Ampere’s law,L = 2πr.
The current enclosed I is (for N turns of toroidal coil) N I.
B(2πr) = µ0NI
B =
µ0NI
2πr
n =
N
2πr
, no of turns per unit length
B = µ0nI
17. Derive an expression for the force acting per unit length between two long straight parallel
current-carrying conductors.
Force per unit length between two long straight parallel conductors: Suppose two long
thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum
(or air) carrying currents I1 and I2 respectively. It has been observed experimentally that
when the currents in the wire are in the same direction, they experience an attractive force
(fig. a) and when they carry currents in opposite directions, they experience a repulsive
force (fig. b). Let the conductors PQ and RS carry currents I1 and I2 in same direction
and placed at separation r.
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9. Figure 9:
Consider a current–element ‘ab’ of length L of wire RS. The magnetic field produced by
current-carrying conductor PQ at the location of other wire RS
B1 =
µ0I1
2πr
According to Maxwell’s right hand rule, the direction of B1 will be perpendicular to the
plane of paper and directed inwards. Due to this magnetic field, each element of other
wire experiences a force. The direction of current element is perpendicular to the mag-
netic field; therefore the magnetic force on length L of the second wire is
F = B1I2Lsin90o
F =
µ0I1
2πr
I2L
The force acting per unit length on the conductor
f =
F
L
=
µ0I1I2
2πr
18. Define one Ampere?
The ampere is the value of that steady current which, when maintained in each of the
two very long, straight, parallel conductors, placed one metre apart in vacuum, would
produce on a force equal to 2 × 10˘7
newtons per metre on each other.
19. Show that a rectangular current carrying coil experiences a torque when it is placed in a
magnetic field.
Consider a rectangular coil of length l and breadth b carrying a current I placed in a uni-
form magnetic field B. θ be the angle between plane of rectangular coil and magnetic field.
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10. Figure 10:
The the force acting on side PQ,
F1 = BIlsinθ = BIlsin90o
= BIl
The the force acting on side RS,
F2 = BIlsinθ = BIlsin90o
= BIl
Applying the Flemming Left Hand Rule we find the directions of these forces are opposite
to each other.
As the force acting on the sides PQ and RS are equal and opposite along different lines
of action they constitute a couple. Hence the rectangular coil experiences a torque.
τ = Force × distance
= BIl × bsinθ
= BIAsinθ
where A =lb, area of coil.
τ = BIAsinθ
Let N be the number of turns of the coil, then
τ = NIABsinθ
20. Derive an equation for the magnetic dipole moment of a revolving electron?
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11. The electron performs uniform circular motion around a stationary heavy nucleus of
charge +Ze. This constitutes a current I, where,
I =
e
T
T is the time period of revolution. Let r be the orbital radius of the electron, and v the
orbital speed. Then,
T =
2πr
v
Substituting for T in first equation,
I =
ev
2πr
There will be a magnetic moment, usually denoted by µl,associated with this circulating
current
µl = Iπr2
=
evr
2
=
e
2me
(mevr)
=
e
2me
l
where l = angular momentum of electron. From Bohr’s second postulate l = nh
2π
where n
ia an integer.
µl =
e
2me
nh
2π
µl =
nhe
4πme
For n =1, we get
µl =
eh
4πme
= 9.27 × 10−24
Am2
This value is called the Bohr magneton.
21. Explain the construction and working of a moving coil galvanometer with the help of a
diagram.
Construction
The moving coil galvanometer is made up of a rectangular coil that has many turns and it
is usually made of thinly insulated or fine copper wire that is wounded on a metallic frame.
The coil is free to rotate about a fixed axis. A phosphor-bronze strip that is connected
to a movable torsion head is used to suspend the coil in a uniform radial magnetic field.
Essential properties of the material used for suspension of the coil are conductivity and a
low value of the torsional constant. A cylindrical soft iron core is symmetrically positioned
inside the coil to improve the strength of the magnetic field and to make the field radial.
The lower part of the coil is attached to a phosphor-bronze spring having a small number
of turns. The other end of the spring is connected to binding screws.
11
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12. The spring is used to produce a counter torque which balances the magnetic torque and
hence help in producing a steady angular deflection. A plane mirror which is attached
to the suspension wire, along with a lamp and scale arrangement is used to measure the
deflection of the coil. Zero-point of the scale is at the centre.
Figure 11: Moving coil galvanometer
Working
The torque acting on a coil in a magnetic field is given by,
τ = NIABsinθ
Since the field is radial by design, we have taken sinθ = 1 as θ = 90o
always.The mag-
netic torque NIAB tends to rotate the coil. A spring S provides a counter torque kφ
that balances the magnetic torque NIAB; resulting in a steady angular deflection φ. In
equilibrium
kφ = NIAB
where k is the torsional constant of the spring; i.e. the restoring torque per unit twist.
I =
k
NAB
φ
I = Kφ
where
K =
k
NAB
is called Galvanometer constant.
I ∝ φ
22. How will you convert a galvanometer to ammeter?
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13. Figure 12: Conversion of Galvanometer to Ammeter
A galvanometer is converted to ammeter by connecting a low resistance S called shunt
resistance in parallel to it. Let G be the resistance of galvanometer, S the resistance of
shunt resistor, Ig the maximum safe current that can pass through the galvanometer. Let
I be the current to be measured. Then the current flowing through the shunt resistor is
I − Ig.
The potential difference across galvanometer and the shunt are equal.
(I − Ig)S = IgG
S =
IgG
I − Ig
23. How will you convert a galvanometer to voltmeter?
A galvanometer is converted to voltmeter by connecting a high resistance R in series to
it.
Figure 13: Conversion of Galvanometer to Voltmeter
Let V be the potential difference to be measured. then
V = IgG + IgR
IgR = V − IgG
R =
V
Ig
− G
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