Physics Smart Booklet

4.MOVING CHARGES AND

MAGNETISM

Physics Smart Booklet

Theory + NCERT MCQs + Topic Wise
Practice MCQs + NEET PYQs

1
Physics Smart Booklet

2
Physics Smart Booklet

MOVING CHARGES AND MAGNETISM

Introduction
A gravitational field is associated with a mass. An electrostatic field is associated with a charge. A magnetic field is
in a region surrounding a current carrying conductor.



Magnetic force on a charged particle moving in a magnetic ⃗
F
field
The force exerted by a magnetic field on a moving electric charge or a ⃗
B
current carrying conductor is called magnetic force. 
⃗
v
⃗, Force on a moving charge
v , in a magnetic field B
A charge q moving with a velocity ⃗
⃗ . It is given by
experiences a force F
F = q⃗v × ⃗
⃗ B.
 The magnitude of the magnetic force is F = q vB sin , where  is the angle between ⃗v and ⃗B .
 F is that of ⃗v × ⃗
The direction of ⃗ B.
 v is parallel or anti parallel to ⃗
F is zero, when ⃗ B ( = 0 or 180).
 F is maximum when a charged particle moves in a direction perpendicular to the direction of
⃗
B ( = 90). Fmax = q VB sin 90 = q vB.
 The work by the magnetic force on a charged particle is zero since ⃗
F is perpendicular to ⃗
v . Thus, a magnetic field
cannot change the speed and kinetic energy of a charged particle.
Fleming’s left hand rule

Force
The direction of the force on a charged particle moving perpendicular to a magnetic
ld
field is given by Fleming’s left hand rule. Fie
Stretch the first three fingers of the left hand such that they are mutually
perpendicular. If the forefinger is in the direction of the field, the middle finger in the Ve
loc
it y
direction of velocity of the positively charged particle then the thumb gives the
direction of the mechanical force. Flemin g’s left h an d rule

Motion of a charged particle with ⃗

v perpendicular to B
Consider a positively charged particle moving in a uniform magnetic field. When the initial velocity of the
particle is perpendicular to the field, (in the figure, the magnetic field is perpendicular to the plane of the paper
and inwards) the particle moves in a circular path whose plane is perpendicular to the magnetic field.
mv 2 Bin
 ⃗
v  
Thus, the centripetal force r is provided by the magnetic force qvB, ⃗
v
where r = radius of the circular path. r
mv ⃗
v FB FB +q
+q F
 r = qB .
⃗
v
v qB +q  ⃗
v
=  
The angular speed of the particle,  = r m .
2 πr 2 π 2 πm
= =
The period of circular motion, T = v ω qB and frequency

3
Physics Smart Booklet

Thus, the angular speed of the particle, period of the circular motion and frequencies of rotation do not depend on
the translational speed of the particle or the radius of the orbit, for a given charged particle in a given uniform
magnetic field. This principle is used in the design of a particle accelerator called cyclotron.
Cyclotron
Cyclotron is a device used to accelerate charged particles to very large kinetic energies by applying electric and
magnetic fields.

(a) (b)
Schematic diagram of cyclotron
 Expression for kinetic energy
The maximum kinetic energy, of the ion as it emerges from the cyclotron will then be


 Cyclotron frequency
The frequency f, of the oscillator required to keep the ion in phase is the reciprocal of the time in which the particle

makes one revolution. This is called the cyclotron frequency given by

It can be shown that kinetic energy = 2m2 f2R2

Helical path of a charged particle moving in a magnetic field (0 <  <90°)

 If a charged particle moves in a uniform magnetic field with its velocity at some arbitrary angle 
(0 <  < 90) with respect to a magnetic field ⃗ B , the path is a helix.
 The axis of the helix is along the direction of B.
 The perpendicular component of velocity (v sin ) determines the radius (r) of the helix.
mv sinθ
r=
Then, qB

 The pitch of the helix where T is the period of the circular motion and is given by
The Pitch of the helical path (p) is the distance travelled by the particle along the direction of the field in one
period of revolution of the circular motion.

4
Physics Smart Booklet


Motion of a charge in combined electric and magnetic fields
Lorentz force
⃗ and a magnetic field B
v in the presence of an electric field E
A charge q moving with a velocity ⃗ ⃗ , experiences a
⃗ ⃗ ⃗
force given by F =q E +q( ⃗v × B )] . This force is called the Lorentz force.
This expression for force was deduced by H.A. Lorentz and is based on experimental observations.
Velocity selector
Consider a positively charged particle q, moving with a velocity subjected to uniform electric field and
magnetic field acting at right angles to each other, as shown in the figure.
Fmag = qvB acting upwards … (1)
… (2)
Let us make

… (3)
For a given magnitude E and B, v is fixed.
Thus, a particle whose velocity v = E/B alone travels undeflected. Particles whose velocities differ from this value
get deflected. Hence, if we inject a stream of particles (each of same charge) with varying velocities into a region of
combined fields, we get a fine pencil of particles with a single value of v. In other words this arrangement works as
a velocity selector.
Magnetic force on a current carrying wire
The phenomenon in which a current carrying conductor experiences a force in a magnetic field is called the
mechanical effect of electric current. This force on the conductor is a manifestation of the force acting on the free
electrons in the conductor placed in a magnetic field.
A current carrying conductor placed in a magnetic field experiences a mechanical force. The magnitude of this
force given by F = B I l sin 
P
where B is the magnetic field, I is the current, l is the length of the conductor and 
is the angle between the direction of the current and the magnetic field.  ⃗
B PQ =
 Vector form : ⃗
⃗
F =I l ×B l
I
Q
 The force is maximum when  = 90
i.e., when a current carrying conductor is placed at right angles to the direction of the magnetic field, F max = B I l
 The force exerted is zero when  = 0, or 180° i.e., when a current carrying
Motion

Fo
conductor is placed parallel to the direction of the magnetic field or antiparallel rc
The direction of the force is given by Fleming’s left hand rule. ld
Fie
The rule is stated as follows.
Stretch the fore finger, the middle finger and the thumb of the left hand such that Cu
rre
they are mutually at right angles. If the fore finger shows the direction of the nt

magnetic field and the middle finger the direction of the current, the thumb shows Flem ing’s left hand rule
the direction of the mechanical force on the conductor.

Biot Savart's Law or Laplace's Rule

Y
It states that the magnetic field (dB ) at a point P at a distance r from a conductor of length
dl carrying a current I is, I


dl 5
P
r
dBp
→
Physics Smart Booklet

(i) directly proportional to I .

(ii) directly proportional to dl
(iii) directly proportional to the sine of the angle between the current element and the line joining the mid point of the
conductor to the given point.
2
(iv) inversely proportional to r .
I dl sin θ
dB ∝
i.e., r2
μ 0 I dl sinθ
dB=
∴ 4π r2 .
The direction of dB is perpendicular to the plane containing both the conductor and the given point.
The direction of the magnetic field around a current carrying conductor can be obtained by any one of the
following rules.
1. Maxwell's right handed corkscrew rule
If we imagine a right handed corkscrew to be rotated such that its tip advances in the direction of the current, then
the direction of rotation of the screw head gives the direction of magnetic field lines.
2. Right hand clasp rule
If we imagine a current carrying conductor to be clasped with the right hand such that the thumb indicates the
direction of electric current, then the direction in which the fingers encircle the conductor gives the direction of the
magnetic field lines.
Magnetic Field Lines
Magnetic field can be better understood by visualising it using the concept of magnetic field lines.
Magnetic field lines are imaginary lines with the following characteristics:
1. They are continuous, closed loops.
2. The tangent to a magnetic field line at a point gives the direction of the magnetic field at that point.
3. Two magnetic field lines never intersect.
4. The number of field lines normal to a given surface is called magnetic flux. Magnetic flux is measured in weber
(Wb).
The magnetic flux per unit area is the strength of the magnetic field or simply the magnetic field. It is also called
magnetic flux density (B). It is measured in Wb m2 = T (tesla).
Magnetic field at a point on the axis of a circular coil carrying current
Consider a coil of n turns and radius r carrying a current I. The magnetic field (B) at a
μ0 2 π n I r 2 r BAx
B=
point P on the axis at a distance x from the centre is given by 4 π ( x 2 +r 2 )3/2 . I
O x P

The direction of B is along the axis. When the current is counterclockwise, the field
along the axis is towards the observer facing the coil.

Magnetic field at the centre of a circular coil carrying current

If the current through the coil of n turn of radius r is I, then, the field at the centre of the coil is

B=
4π [
μ0 2 π n I μ0 nI
r
= ]
2r
⃗ at the centre is perpendicular to the plane of the coil.
The field B
If the coil is in the plane of the paper and the current is
(a) clockwise, then ⃗
B is into the plane of the paper.
⃗ is out of the plane of the paper.
(b) anti-clockwise, then B

6
Physics Smart Booklet

Ampere’s circuital law

Ampere’s circuital law for calculation of magnetic field due to distributed currents is analogous to Gauss’ law in
electrostatics used to calculate electric field due to distributed charges.
Statement
The line integral of magnetic field along a closed curve enclosing an arbitrary surface is equal to 0 times the total
current passing through the surface.
Explanation
Figure shows the number of currents I1 I2 I3 ..... producing a magnetic field .

where
and 0 = permeability of free space.
is the magnetic field at a point P on the boundary of the surface making an angle  with the length element

on the boundary. is the sum of all the products over the complete boundary or path. This closed
boundary is called the amperean loop.
Procedure to apply Ampere’s law
Step 1:
If etc., are the currents causing a magnetic field at a point P, mark a surface through which all these
currents pass and indicate the boundary of the surface with P on it.
Step 2:
Mark in any arbitrary direction.
Step 3:
Mark a length element at P and let the angle between and be marked .
Step 4:
Let the sense of carrying out the integration be marked arbitrarily. In the given case, it is marked anticlockwise.
Step 5:
Keep your right hand with the fingers half curled so that the fingers held along the boundary (amperean loop)
indicate the direction of integration. Mark all currents indicated by the direction of the thumb as +ve and those
opposite the thumb as ve.
Step 6:
Evaluate B. If B works out to be +ve, the marked direction of is correct. Otherwise, reverse the direction of .

Magnetic field at a point due to a long straight conductor carrying a current

7
Physics Smart Booklet

Consider an infinitely long conducting wire, carrying a

current I. Let P be a point close to it at a distance r
from it. (r is much less than the length of the
conductor).

The variation of magnetic field B with distance r from the conductor is shown in the figure.
The direction of ⃗
B is given by the right hand clasp rule.
For a short conductor the magnetic field at a point P near it can be shown to be given by,

or

Magnetic field due to a solenoid

A nearly uniform magnetic field can be produced using
solenoids. A solenoid is a helical winding of a conducting
wire with neighbouring turns closely spaced (Fig). wound on
a cylinder such that the plane of each turn is parallel to the
next and perpendicular to the axis of the cylinder.
Expression for magnetic field produced by a long
solenoid
From Ampere’s circuital law,
or B = 0nI
Direction of the field is along ab. We see that the magnetic field does
not depend on the position of point P inside the solenoid. Hence the
magnetic field inside a long solenoid (except at the ends) is uniform in magnitude and direction.

  For a small solenoid of length L compared to its radius R, (Fig) the magnetic field at the centre is

.
 For a long solenoid (L >> R)
1  0 and 2  180

B
 Magnetic field at one end of a long solenoid (1 = 0, 2 = 90) is
B0

 The magnetic field inside a solenoid can be greatly increased by B0 / 2

introducing a ferromagnetic material like iron, which has a large
relative permeability r. Then the magnetic field in the solenoid
end centre end
will be .
 A current carrying solenoid when suspended freely comes to rest just like a suspended bar magnet along the
north and south direction.
When looking at one end of a solenoid, (i) if the current is anticlockwise as shown in figure (1), the end face of the

Figure 1. anti-clockwise current Figure 2. clockwise current 8

Physics Smart Booklet

solenoid behaves like a north pole of a bar magnet (ii) if current is clockwise as in figure (2), the end face
behaves like south pole of a bar magnet

Magnetic field due to a toroid

A toroid is a hollow ring on which a large number of closely spaced turns of a
conducting wire are wound. It may be treated as a long solenoid bent into the shape
of a doughnut. If a current I is passed through it, the magnetic field lines inside the
toroid will be concentric circles.
Let B be the field at a point P at distance r from the centre of the toroid O.

From Ampere’s theorem, 2rB = 0  NI 

If is the number turns per unit length we once again have B = 0nI.

Force between two parallel current carrying conductors

If I2 is the current in the second conductor, parallel to I 1. Then the a segment of the second conductor of length L
experiences a force given by .
This force is directed towards the first conductor and has a
magnitude

or

The force per unit length of the second conductor towards the

first conductor is

Since the first conductor also experiences a force in the field produced by the second conductor, the force per unit

length of current I1 should also be equal to and directed towards I2.

μ0 I 1 I 2
Thus, the force per unit length of the conductor = 2 πd I1 I2
 If two straight, parallel conductors carry current in the same direction, they attract each other,
with the same force. F21
F12
 If two straight, parallel conductors carry current in the opposite direction, they repel each
other, with the same force.
I1 I2
Current loop as a magnetic dipole
F21
At large distances as the axis of the current loop, we have F12

R2 = Area A of the loop

This equation is analogous with the equation for the electric field on the axis of an electric dipole,

9
Physics Smart Booklet

Hence, the term IA is the magnetic analogue of the electric dipole moment p hence IA is called the magnetic
moment,

for x >>> R
The magnetic field produced by a current loop is identical with the electric field produced by an electric dipole.
Hence a current loop is equivalent to a magnetic dipole.
We observe that the magnetic moment depends only on the area of the loop but not on its shape.

Magnetic dipole moment of a revolving electron

Consider an electron revolving in a circular path, typically like an electron revolving around a nucleus. Let the
magnitude of the charge on the electron be e and T be its period of revolution.
The current equivalent of the revolving electron is

… (1)

If v is the velocity of the electron,

… (2)

From Eqs., (1) and (2)

… (3)

The magnetic moment produced by the revolving electron is given by

… (4)

The direction of this magnetic moment is into the plane of the paper, as if it is due to a clockwise current I.

Expression for gyromagnetic ratio

Magnetic moment of an electron given by

In vector form
The negative sign indicates that the angular momentum of the revolving electron is opposite to its magnetic
moment.

Considering only the magnitude of the ratio

This ratio is called gyromagnetic ratio of the electron. Its value is 8.8  1010 C kg1.
This has been verified experimentally.

Bohr magneton

We know that, … (1)

10
Physics Smart Booklet

Let us consider the Bohr’s postulates for quantization of orbits, namely,

, where n = 1, 2, 3, … … (2)

The minimum value of magnetic moment is obtained when n = 1

This minimum value of l is called Bohr magneton.

The value of Bohr magneton is = 9.27  1024 Am2

Torque on a magnetic dipole (Current loop)

⃗
A current loop of area A carrying current I has a magnetic moment, M=I ⃗A
The direction of this magnetic moment vector is perpendicular to the plane of the current loop, given by right hand
corkscrew rule.
A current loop suspended in a magnetic field is analogous to an electric dipole in an electric field. Hence, it
⃗ ⃗ ⃗ ⃗
experiences a torque, τ⃗ = M × B  τ⃗ =I A×B .
The magnitude of the torque is τ =IAB sin θ where  is the angle between the direction of magnetic field and the
magnetic moment. The direction of ⃗ ⃗ and ⃗
τ is perpendicular to the plane containing M B . The tendency of the
torque
is to align the magnetic moment along the field. The coil tends to rotate until its plane is perpendicular to the
magnetic
field.
If the coil has n turns,  = n A I B sin . When  = 90, max = nAIB.
Moving coil galvanometer (pointer galvanometer) 0
It is a device used for accurate measurement of very small currents of the order 20 20
of microampere. It works on the principle that a current carrying coil when
suspended in a magnetic field experiences a torque and then deflects N S

For small deflection of the coil, the deflection is proportional to the current.

( )
T T
C
I= θ=Kθ
n BA , hence I   Table galvanometer

where, C = couple per unit twist of the suspension wire, n = number of turns in the coil, A = area of the coil, B =
magnetic field,  = deflection of the coil in radian.

Current sensitivity

Current sensitivity
( θI )
of a galvanometer is numerically equal to the deflection produced in the galvanometer
when unit current flows through it.
θ I nBA
= =
From the equation I=K θ , we get I K C . The galvanometer is sensitive, if it can produce a large
deflection for a small current.
The sensitivity can be increased by
1. increasing the number of turns (n) in the coil

11
Physics Smart Booklet

2. increasing the magnetic field (B)

3. increasing the area of the coil (A)
4. by decreasing the couple/unit twist of the suspension wire (C). This can be achieved by using a thin and long
suspension wire.

Conversion of a pointer galvanometer into an ammeter

A galvanometer can be converted into an ammeter by connecting a low resistance, Amm eter
called shunt, in parallel to the galvanometer coil.
I S I IIgg
G
 Full scale deflection current, I g = G+ S , where I is the maximum current (range) to ISs

be measured, G is the resistance of the galvanometer and S is the shunt resistance  S

low resistance
Ig  I
Ig G C on version of
S= galvanom eter in to am meter

(I−I g )

Conversion of a pointer galvanometer into a voltmeter

Voltmeter
A galvanometer can be converted into a voltmeter by connecting a high resistance
in series with the galvanometer coil. high
resistance
 P.D, V = Ig (G + R) Ig
G
R
where V is the range of voltmeter, I g is the full scale deflection current through the
galvanometer, G is the resistance of the galvanometer and R is the high resistance
Fig.Conversion
15 .1 1 Conversion
of a of a
in series  Ig  V galvanometer into an voltm eter
galvanometer
V into an voltmeter
R= −G

Ig
Illustrations
1. Two identical coils A and B are kept coaxially with a separation equal to their diameter. Coil A carries a current I
in anticlockwise direction and B carries same current I but in clockwise direction. The magnetic field at the
midpoint on the axis between the two coils is

(A) zero (B) (C) (D)

Ans (A)
With respect to the observer at O, the magnetic field at the axial point P due to the
coil A is directed towards O whereas that due to B is directed away from O as
shown in figure. Since P is the mid point of A and B, the magnetic fields are equal
and opposite. Therefore the net magnetic field at the point P is zero.

2. In the figure P and Q are two concentric circular coils having equal number of turns but of radii r1
and r2 carrying currents I1 and I2 respectively. If the magnetic field at the common centre O is zero,
then

(A) (B) (C) (D)

Ans (B)

12
Physics Smart Booklet

The magnetic fields due to the current loops P and Q are in opposite directions.
Since the resultant magnetic field is zero at the common centre,

3. In the loops shown, all curved sections are either semicircles or quarter circles. All the loops carry the same current.
The magnetic fields at the centers have magnitudes B1, B2, B3 and B4. Then,

(i) (ii) (iii) (iv)

(A) B4 is maximum (B) B3 is minimum

(C) B4 > B1 > B2 > B3
(D) B2 cannot be found unless the dimensions of the section B are known.
Ans (C)
For B1 and B4, the contributions due to the different sections add up. For B2 and B3, the contributions due to the
outer sections oppose the contributions due to the inner sections. Thus, B1 and B4 are greater than
B2 and B3.
For B4, there is a section with radius < b and hence it contributes more than the semicircular section of radius b
does for B1. Thus B4 > B1.
For B3, there is a section with radius > b and hence it contributes less than the semicircular section of radius b does
for B2. Thus B3 < B2.
4. Two coils X and Y having the same number of turns, carrying the same current and in
the same sense are arranged coaxially so that they subtend the same solid angle at
point O. If the smaller coil is midway between the larger coil and the point O, then the
ratio of the magnetic fields at O due to the two coils is x/2

(A) 1 (B) 4 (C) 2 (D) 8 x

Ans (C)
It is obvious from the diagram that

The magnetic field at O due to coil X is

The magnetic field at O due to coil Y is

5. A battery is connected between two points A and B on a uniform conducting ring of radius r and resistance R. One
of the arcs AB of the ring subtends an angle θ at the centre. The value of the magnetic field at the centre due to the

13
Physics Smart Booklet

current in the ring is

(A) proportional to 2(2π – θ) (B) inversely proportional to R
(C) zero, only if θ = 2π radian (D) zero, for all values of θ.
Ans (D)

The situation is shown in figure.

The magnetic fields at the centre due to the arcs ACB and ADC are given by

 Resultant magnetic field is zero for all values of θ.

6. The magnitude of magnetic moment of the current loop shown in the figure is
(A) Ia2 (B) Ia2

(C) zero (D)

Ans (B)
The loop can be divided into two square loops.
The magnetic moments of loops are perpendicular to each other.
For each square loop, M = I a2,
we have
The net magnetic moment is
=
7. A conducting wire of uniform cross-section is bent to form an equilateral triangle ABC of side L. A current I enters
at A and leaves at C. The magnetic field at the centroid O of the triangle is

(A) (B) (C) (D) zero.

Ans (D)
The field at a point due to a straight conductor of finite length is

both directed normally it to the plane of loop.

Similarly, directed normally out of the plane of loop. Resultant

magnetic field at the centroid O is

14
Physics Smart Booklet

B = BAB + BBC  BAC = 0

8. A uniform metallic wire forms the edges of a cube of length l. A current I enters at one edge of the cube and leaves
from the diagonally opposite edge. The magnitude of the magnetic field at the centre of the cube is

(A) zero (B) (C) (D)

Ans (A)
The given system exhibits symmetry with respect to the diagonally opposite corners A
and G. The current distribution is similar in the two halves of the cube with AG as
reference. These current distributions produce fields of same magnitude and opposite
directions at the centre of the cube.
Hence, the resultant magnetic field at the centre of the cube is zero.

9. An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR.
The magnetic field due to this current at the point M is B1. Now, another infinitely long straight conductor QS is
connected at Q so that the current is in QR as well as QS, the current in PQ remaining unchanged. The

magnetic field at M is now B2. The ratio is given by

(A) (B) 1 (C) (D) 2

Ans (C)

The magnetic field at M due to QR is zero in both cases. When QS carries current the magnetic field at M due

to QS is

Magnetic field due to PQ carrying current I

Total field at M is

or

10. Figure shows two long straight wires carrying currents i each and placed perpendicular to each other. The resultant
magnetic field is zero
(A) in quadrant 1 and 2
(B) in quadrant 2 and 4
(C) in quadrant 3 and 4
(D) at the point of intersection.
Ans (B)
Let and represent the magnetic field at any point due to currents in the vertical and
the horizontal wires respectively.
Application of the Fleming’s right hand rule gives the direction of magnetic field at

15
Physics Smart Booklet

different points.
Here B1 = B2 (in magnitude)
From figure, it follows that the fields cancel each other in quadrants (2) and (4). Hence, in the given context choice
(B) is correct.

11. A current of 25 A flows through an overhead power cable from the North to South direction. The magnitude and
direction of magnetic field at a point 5 m below the cable is
(A) 10−6 T towards east (B) 10−6 T towards west
(C) 10−5 T towards north (D) 10−5 T towards south
Ans (A)
Magnitude of the magnetic field is given by

From the right hand clasp rule, it follows that direction of B is towards east.

12. Three conductors are arranged as shown in the diagram. If the magnitude of the forces on the
conductors A, B and C are (assume x to be small compared to the length of the conductors) FA,
FB and FC respectively, then
(A) (B)
(C) (D)
Ans (D)

13. A solenoid has a length 1 m and inner diameter 4 cm and it carries a current of 5 A. It consists of five close packed
layers, each with 800 turns along its length. The magnitude of the magnetic field at the centre is nearly equal to
(A) zero (B) 25 mT (C) 5 mT (D) 1 mT
Ans (B)

Number of turns per unit length, turns per meter.

Length, . The solenoid is long and can be treated as an ideal solenoid.

The magnetic field at the centre is given by B = µ0nI = (4π × 10−7) × = 25 mT.
14. *When a proton, a deuteron and an α-particle are projected into a uniform magnetic field at right angles to the field,
the radii of their respective paths are found to be in the ratio The respective accelerating potentials for
providing the required velocities are in the ratio
(A) (B) (C) (D) 2 : 2 : 1
Ans (D)
Work done by the accelerating potential is equal to the kinetic energy gained by the charged particle in the electric
field. Given:

16
Physics Smart Booklet

We have, . For a given B, ∴

= 2:2:1

15. A particle with charge to mass ratio, q/m = α is shot with a speed v towards a wall at a distance d perpendicular to
the wall. The minimum value of that must exist in this region for the particle not to hit the wall is

(A) (B) (C) (D) .

Ans (A)
The situation is shown in the figure.
Let the particle projected at the point O, undergo deflection due to the applied magnetic field in a direction
normally inwards and let it just miss hitting the wall at A.

Then,

For the particle not to hit the wall i.e., to just miss hitting the wall,

… (1)

16. A neutral atom which is at rest at the origin of the co-ordinate system emits an electron in the
z- direction. The product ion is P. A uniform magnetic field exists in the positive x-direction.
(A) The electron and the ion P will move along circular paths of equal radii.
(B) The electron has same time period as that of the ion P
(C) The electron has same kinetic energy as that of the ion P
(D) The electron starts moving in a circle around the ion P.
Ans (A)
The total momentum of the system is conserved. Hence, the electron and the ion P move in opposite directions with
equal momenta.

Radius of the circular path,

In the given situation, p, e and B are same for both the electron and the ion.
Hence, both the particles describe circles of same radius.
17. Two particles of masses m and 2m carrying charges q and −2q respectively are projected towards each other with
the same speed v into a region of uniform magnetic field directed normally inwards into the plane of the
diagram. If d is the initial separation between the particles, then the maximum value of the speed v so that the
particles do not collide, is (consider only the magnetic force of interaction on each).

(A) (B) (C) (D)

Ans (B)
Under the influence of the external uniform magnetic field, the two particles describe circular paths of radii r1 and

r2 given by and

Hence, r1 = r2

17
Physics Smart Booklet

The particles do not collide if or

or . Hence, the maximum speed of projection must be .

18. An electron and a proton enter a region of uniform magnetic field at right angles to it with the same linear
momentum. Then
(A) the radius of the circular path of electron is less than that of the proton
(B) the radius of the circular path of electron is greater than that of the proton
(C) the radius of the circular path of electron is same as that of the proton
(D) both the proton and the electron move in opposite directions along a straight line.
Ans (C)
For a charged particle describing a circular path in a region of a uniform magnetic field, the centripetal force is
provided by the force due to the magnetic field.

i.e., = Bqv ⇒ r = .

The radius of the proton path .

The radius of the electron path ⇒ = .

But it is given that the linear momentum (mv) of both the particles is same. Therefore = 1. Thus, both the proton

and the electron describe circular paths of same radii.

19. The figure shows four possible directions for the velocity of a proton moving through a
region of uniform electric field and magnetic field . The direction of so that the net
force on the proton is zero is
(A) N (B) S (C) E (D) W
Ans (B)
= q[ + ( × )] = 0 
Since is normally outwards, is normally inwards. i.e., magnetic force is directed normally inwards.
The application of Fleming’s left hand rule indicates that the velocity is along south direction (S).
20. A charged particle with charge q enters a region of uniform and mutually orthogonal fields with a
velocity perpendicular to both . The particle comes out without any change in magnitude or direction
of its velocity. Then

(A) (B) (C) (D)

Ans (B)
As are mutually perpendicular and does not change.

… (1)

… (2)

18
Physics Smart Booklet

21. A conductor AB carries a current i in a magnetic field If and the force on the conductor is , then
(A)
(B)
(C)
(D) is the resultant effect of the electric forces on the electrons in the conductor.
Ans (A)
Force on a current carrying conductor is .
22. A long straight wire carrying current lies along the axis perpendicular to the plane of a metal ring. The conductor
will
(A) exert a force on the ring if the ring carries a current.
(B) exert a force on the ring if the ring has a static charge distributed uniformly on the rim of the ring.
(C) exert a force on the ring if the ring has a static charge distributed non-uniformly on the rim of the ring.
(D) not exert any force on the ring.
Ans (D)
The straight wire carrying current produces a magnetic field at every point on the rim of the ring. But this

field B
is along the current element at every point on the ring, where i is the current in the ring.
Mechanical force on each element of the ring is zero.
as
Even if the ring is charged, the force is zero as the magnetic field does not interact with static charges irrespective
of
the nature of their distributions.
23. A conductor PQ of length L carrying a current I is placed perpendicular to a long
straight conductor xy carrying a current I, as shown in the figure. The force on PQ will
be
(A) upwards (B) downwards
(C) to the right (D) to the left.
Ans (A)
The magnetic field at any point on the wire PQ due to the current i in the wire XY is directed normally into the
plane of the diagram.
Now applying Fleming’s left hand rule, we notice that the mechanical force on PQ is directed upwards.
24. A bent wire AB carrying a current I is placed in a region of uniform magnetic field . The force on the wire AB is

(A) BIL (B) BIL

(C) zero (D) .

Ans (D)
. Here is directed from end A to end B of the wire.

Here and θ = 90°. ∴

25. A copper rod of length L and mass m is sliding down a smooth inclined plane of inclination θ
with a constant speed v. A current I is flowing in the conductor perpendicular to the plane of
diagram inwards. A vertically upward magnetic field exists in this region. The magnitude

19
Physics Smart Booklet

of the required magnetic field is

(A) (B)

(C) (D)

Ans (C)
Magnetic force on the rod, Fm = BIL. It acts in the direction as shown in figure. The rod will move with a constant
speed if the net force on the rod is zero.

BIL cos θ = mg sin θ or B = tan θ.

26. Two very long straight parallel wires carry currents I and 2I in opposite directions. The distance between the wires
is r. At a certain instant of time a point charge q is located at a point equidistant from the two wires in the plane of
the wires. Its instantaneous velocity is perpendicular to this plane. The force due to the magnetic field acting on
the charge at this instant is

(A) zero (B) (C) (D)

Ans (A)

The field acts parallel to the direction of motion of the charged particle. Hence, the force on it is
zero.
Force on charged particle is
F = qvB sin θ = qvB × 0 = 0.
27. An electron moves parallel to a current carrying conductor with a velocity 10 7 ms−1. If the conductor carries a
current of 10 A, then the magnitude of the force experienced by the electron is
(A) N (B) N (C) N (D) N.
Ans (D)
Force experienced by the electron F = Bqv sinθ.

The magnetic field at a distance r from a straight conductor carrying current I is,

The magnetic field exerts a force on the electron given by

Here θ = 90°,

F=

= 8 × 10–17 N.
28. Three long straight wires X, Y and Z are connected parallel to each other across a battery of negligible internal
resistance. The resistances of the three wires are in the ratio 1 : 2 : 3. If the net force experienced by the middle
wire Y is zero, then the ratio of the distances of the middle wire from the other two wires is
(A) 3 : 1 (B) 1 : 2 (C) 2 : 3 (D) 3 : 4

20
Physics Smart Booklet

Ans (A)
The wires X, Y and Z are in parallel.


⇒ … (in magnitude)

29. A circular loop of area 0.02 m 2 carrying a current of 10 A is held with its plane perpendicular to a uniform
magnetic field, 0.2 T. The torque acting on the loop is
(A) 0.01 N m (B) 0.001 N m (C) zero (D) 0.8 N m
Ans (C)
τ = nIAB sin θ
τ = nIAB sin (0) = 0
30. A wire of length l is bent in the form of a circular loop and is suspended in a region of a uniform magnetic field B.
When a steady current I is passed through the loop, the maximum torque experienced by it is

(A) (B) (C) 4π2Il 2B (D)

Ans (B)
The torque experienced by the loop τ = BIA sin θ. The maximum torque experienced τmax = BIA.

Length of the loop l = 2πR. ⇒ R = .

Area of the loop A = =

Torque, τmax =

31. The coil of a suspended coil galvanometer has very high resistance. When a momentary current is passed through
the coil, it
(A) gets deflected and comes to rest slowly
(B) shows a steady deflection
(C) oscillates with the same amplitude
(D) oscillates with the decreasing amplitude.
Ans (D)
When a momentary current is passed, the coil is subjected to a sudden, unbalanced deflecting couple. Gradually
under the action of the restoring couple, the coil comes to rest after executing several oscillations with decreasing
amplitude.
32. A metal wire is bent to form a square loop of side L. It carries a current i and is placed in a region of a uniform
magnetic field normal to the field . If the shape of the loop is slowly changed to a circle without changing its
length, the amount of work done is

21
Physics Smart Booklet

(A) (B) (C) (D) zero

Ans (A)
Work done,

33. The sensitivity of a suspended coil galvanometer depends on

(A) moment of inertia of the coil
(B) the deflection θ
(C) the horizontal component of earth’s magnetic field
(D) the torsional constant of the suspension wire and the spring.
Ans (D)

The expression for current flowing through a suspended coil galvanometer is I = θ or current sensitivity

= . For a given coil, depends on C, the torsional constant or the couple per unit twist of the suspension

wire.
34. Two moving coil galvanometers 1 and 2 are with identical field magnets and suspension torque constants, but with
coils of different number of turns N1 and N2, area per turn A1 and A2 and resistances R1 and R2. When they are
connected in series in the same circuit, they show deflection θ1 and θ2.

Then is

(A) (B) (C) (D)

Ans (A)

Given that I1 = I2

So

35. If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter
(A) a low resistance in parallel (B) a high resistance in parallel
(C) a high resistance in series (D) a low resistance in series
Ans (C)
An ammeter can be converted into a voltmeter by connecting a high resistance in series with it.
36. A voltmeter of range (0V  30 V) is to be connected to a voltage line of 150 V. The maximum current that the
voltmeter can withstand is 5 mA. In order to connect the voltmeter safely to the voltage line, the series resistance

22
Physics Smart Booklet

required is
(A) 240 kΩ (B) 24 kΩ (C) 2.4 kΩ (D) 240 Ω
Ans (B)

The resistance to be connected in series is given by Rs = – R. and I = 5 mA. The resistance of the voltmeter, R

= = = 6 × 103Ω. V′ is the line voltage to be measured (new range of the voltmeter).

Rs = – 6 × 103 = 24 × 103 Ω.

37. The resistance of a galvanometer is 50 Ω and it requires 2 µA per two division deflection. The value of the shunt
required in order to convert a galvanometer into ammeter of range 5 A is (The number of divisions on the
galvanometer scale on one side is 30)
(A) 0.2 Ω (B) 0.002 Ω (C) Ω (D) Ω
Ans (C)
The shunt S to be connected is given by

S=

Ig = ×N= × 30

 Ig = 30 × 10–6 A.

S= ≅ 3 × 10–4 Ω

38. Of the following graphs, the one which shows the variation of the series resistance to be connected with a moving
coil galvanometer so as to convert it into a multi range voltmeter is

(A) (B) (C) (D)

Ans (C)

is the series resistance required for conversion into voltmeter of higher range.

(Compare with y = mx + C)
39. An α particle and a proton having equal velocity are moving inside a uniform magnetic field. The field is
perpendicular to the direction of the velocity for both particles. Their radii rα and rp are in the ratio (assume mα = 4
mp).

(A) (B) (C) (D)

Ans (C)

We have the radius of the circular orbit given by

and 

23
Physics Smart Booklet

40. A uniform magnetic field is acting at an angle θ with respect to the direction of velocity (v) of a particle of charge q
and mass m. The pitch of the helical path is

(A) (B) (C) (D)

Ans (B)

For the motion of the particle,

By the time, a charged particle completes one circular motion, due to the component of velocity along the
field, the charged particle moves through a distance namely the pitch.
∴ Pitch, ∆x = T . v cos θ

Each of the following questions consists of a Statement−I and a Statement−II. Examine both of them and select one
of the options using the following codes:
(A) Statement-I and Statement-II are true and
Statement-II is the correct explanation of Statement-I.
(B) Statement-I and Statement-II are true, but Statement-II is not the correct explanation of Statement -I
(C) Statement-I is true, but Statement -II is false
(D) Statement-I is false, but Statement -II is true
41. Statement-I: The deflection in the galvanometer is directly proportional to the current passing through it.
Statement-II: The magnetic field in which the coil of galvanometer is suspended is radial.
(A) A (B) B (C) C (D) D
Ans (A)
Both assertion and reason are correct and reason is correct explanation of assertion as radial magnetic field helps to
make the scale of galvanometer linear.
42. Statement-I: We can increase the range of an ammeter but cannot decrease the range.
Statement-II: Minimum range of an ammeter is fixed.
(A) A (B) B (C) C (D) D
Ans (A)
A suitable shunt resistance can be used in parallel with the ammeter to increase its range. But its range cannot be
decreased as shunt cannot be made negative. The reason is true and explains the assertion correctly.
43. Statement-I: An ammeter connected in parallel with a resistance may get damaged.
Statement-II: An ammeter has low resistance.
(A) A (B) B (C) C (D) D
Ans (A)
An ammeter has low resistance. When an ammeter is connected in parallel with a resistance, the effective
resistance of the circuit decreases. As a result, current in the circuit increases, which may damage the ammeter.
44. Statement-I: A charged particle is accelerated by a potential difference V. It then enters perpendicularly to a
uniform magnetic field. It moves in a circle. Its angular momentum about centre is say L. Now if V is doubled, L is
also doubled.
Statement-II: If V is doubled, kinetic energy is doubled and therefore, L is also doubled.
Ans (B)

24
Physics Smart Booklet

NCERT LINE BY LINE

1. A current element (where dx = 1 cm) is placed at the origin and carries a large current
of 10 A. The magnetic field on y-axis at distance of 50 cm from it is [NCERT Pg. 148)

(a) (b) (c) (d)

2. Consider a tightly wound 100 turn coil of radius 12 cm carrying current of 10 A. What is
magnetic field at centre of this coil. [NCERT Pg. 146]

(a) (b) (c) (d)

3. A straight wire carrying current of 15 A is bent into a semicircular arc of radius 2.5 cm. The
magnetic field at the centre of semicircular arc is [NCERT Pg. 150]

(a) (b) (c) (d)

4. Consider a tightly wound 200 turns coil of radius 10 cm carrying current of 10 A. The
magnitude of magnetic field at the centre of the coil is [NCERT Pg. 151]

(a) (b) (c) (d)

25
Physics Smart Booklet

5. A long straight wire of circular cross-section of radius 5 cm is carrying a steady current of 20 A,

uniformly distributed over its cross-section. The magnetic field induction at 2 cm from the axis
of the wire is
[NCERT Pg. 149]

(a) (b) (c) (d)

6. A long straight cylindrical wire carries current I and current is uniformly distributed across
cross-section of conductor. Figures below shows a plot of magnitude of magnetic field with
distance from centre of the wire. The correct graph is (NCERT Pg. 150]

(a) (b)

(c) (d)
7. A closely wound solenoid 80 cm long has 5 layers of winding of 400 turns each. The diameter
of solenoid is 1.8 cm. If it carries current of 8 A then magnitude of magnetic field intensity
inside solenoid near its centre is (NCERT Pg. 173]

(a) (b) (c) (d)

8. A circular coil of 30 turns and radius 8 cm carries a current of 6 A. It is suspended in a uniform
horizontal magnetic field of 1.0 T. The field lines make an angle of 60° with the normal of the
coil. The magnitude of counter torque that must be applied to prevent the coil from turning is
[NCERT Pg. 169]
(a) 3.133 Nm (b) 0.236 N m (c) 30.8 N m (d) 35 N m
9. In a chamber, a uniform magnetic field of 1.2 T is maintained. An electron is shot into the field
with a speed of 3.2 X 10 6 m s–1 normal to the field. The radius of circular orbit in which it starts
circular path is (m0 = 9.1 X 10–31 kg) [NCERT Pg. 169]

(a) 15.16 m (b) 627 m (c) 12.42 m (d) 22.4 m

10. Two moving coil galvanometers M1 and M2 have the following particulars. N1 = 30, B1= 0.25 T,
A1 = 7.2 X 10–3 m2, G1 = 10 and N2 = 60, B2 = 0.50 T, A2 = 1.8 X 10–3 m2, G2 = 5 respectively.

26
Physics Smart Booklet

The spring constants are identical to both galvanometers. The ratio of their current sensitivity
is
[NCERT Pg. 173]
(a) 1:1 (b) 2:1 (c) 4 : 1 (d)1:4
11. A toroid ring has inner radius 21 cm and outer radius 23 cm in which 4400 turns of wire are
wound. If the current in the wire is 10 A, then magnetic field inside the core of the toroid will
be
[NCERT Pg. 170]

(a) (b) (c) (d)

12. Two concentric circular coils X and Y of radius 20 cm and 25 cm respectively lie in the same
vertical plane. Coil X has 40 turns and coil Y has 100 turns. If coil X and Y carries currents of 18
A each but in opposite
sense, the net magnetic field due to the coils at their centre is [NCERT Pg. 170]

(a) (b) (c) (d)

13. A galvanometer has resistance of 60 . It is converted in to an ammeter by connecting a shunt
resistance of 1.2 . Its range becomes [NCERT Pg. 172]
(a) 68 (b) 50 (c) 51 (d) 60
14. To convert a galvanometer into a voltmeter of large range, we connect a resistance with
galvanometer. The resistance [NCERT Pg. 165]
(a) Is connected in parallel and of higher value
(b) Is connected in series and of lower value
(c) Is connected in parallel and of lower value
(d) Is connected in series and of higher value
15. Magnetic moment associated with an electron moving at speed v in a circular orbit of radius r
is (in magnitudes) [NCERT Pg. 162]

(a) evr (b) (c) (d)

16. The horizontal component of earth’s magnetic field at a certain place is 3.2 x 10 –5 T and field is
directed from south to North. A long straight conductor is carrying a current of 3 A. What is
force per unit length experienced by it when it is placed on horizontal table and current in wire
is from west to east? [NCERT Pg. 156]
(a) 9.6 X 10 Nm upwards
–6 –1
(b) 9.6 X 10 Nm . downwards
–5 –1

(c) 3.6 X 10–5 Nm–1, upwards (d) 9.6 X 10–5 Nm–1, horizontal
17. Two long straight parallel wires A and B carrying current of 20 A and 10 A is same direction
are separated by a distance of 5 cm. The force of 15 cm section of wire B is [NCERT Pg. 173]

(a) , attractive (b) , repulsive

(c) , attractive (d) , attractive

18. A cyclotron’s oscillatory frequency is 10 MHz. What should be the operating magnetic field for
accelerating deuterons? [NCERT Pg. 146]

27
Physics Smart Booklet

(a) 0.96 T (b) 1.52 T (c) 0.46 T (d) 1.32 T

19. A charge C moving with speed of v m s–1 crosses electric field Vm–1 and

magnetic field . The electric field and magnetic fields are crossed and velocity v is also
perpendicular to both. If the charge particle crosses both fields undeflected, the value of v is
[NCERT Pg. 140]

(a) (b) (c) (d)

20. A proton is moving with speed of 2 x 10 m s enters a uniform magnetic field B = 1.5 T. At the
5 –1

entry velocity vector makes an angle of 30° to the direction of the magnetic field. The pitch of
helical path it
describes is nearly [NCERT Pg. 138]
(a) 6.25 mm (b) 4.37 mm (c) 7.25 mm (d) 1.67 mm

NCERT BASED PRACTICE QUESTIONS

1. A magnetic field line is used to find the direction of
(a) south – north (b) a bar magnet
(c) a compass needle (d) magnetic field
2. An electric current passes through a straight wire in the direction of south to north.
Magnetic compasses are placed at points A and B as shown in the figure.

What is your observation?

(a) the needle will not deflect
(b) only one of the needles will deflect
(c) both the needles will deflect in the same direction
(d) the needles will deflect in the opposite directions
3. The magnetic field lines due to a straight wire carrying a current are
(a) straight (b) circular
(c) parabolic (d) elliptical
4. Magnetic field produced at the centre of a current carrying circular wire is
(a) directly proportional to the square of the radius of the circular wire
(b) directly proportional to the radius of the circular wire
(c) inversely proportional to the square of the radius of the circular wire
(d) inversely proportional to the radius of the circular wire
5. The magnetic field lines inside a long, current carrying solenoid are nearly
(a) straight (b) circular
(c) parabolic (d) elliptical
6.‘ A soft iron bar is introduced inside a current carrying solenoid. The magnetic field
inside the solenoid

28
Physics Smart Booklet

(a) will become zero (b) will decrease

(c) will increase (d) will remain unaffected
7. The direction of the force on a current-carrying wire placed in a magnetic field
depends on
(a) the direction of the current
(b) the direction of the field
(c) the direction of current as well as field
(d) neither the direction of current nor the direction of field
8. The direction of induced curent is obtained by
(a) Fleming’s left-hand rule
(b) Maxwell’s right-hand thumb rule
(c) Ampere’s rule
(d) Fleming’s right-hand rule
9. Who first discovered the relationship between electricity and magnetism?
(a) Faraday (b) Newton
(c) Maxwell (d) Oersted
10.: In an electric motor, the energy transformation is from
(a) electrical chemical (b) chemical to light
(c) mechanical to electrical (d) electrical to mechanical
11. A commentator changes the direction of current in the coil of
(a) a DC motor
(b) a DC motor and an AC generator
(c) a DC motor and a DC generator
(d) an AC generator
12. ___ Which of the following devices works on the principle of electromagnetic
induction?
(a) Ammeter (b) Voltmeter
(c) Generator (d) Galvanometer

13. A device used for measuring small currents due to changing magnetic field is
known as
(a) galvanometer (b) ammeter
(c) voltmeter (d) potentiometer
14.: An electric generator actually acts as
(a) source of electric charge (b) source of heat energy
(c) an electromagnet (d) a converter of energy
15. Electromagnetic induction is the
(a) charging of a body with a positive charge
(b) production of a current by relative motion between a magnet and a coil
(c) rotation of the coil of an electric motor
(d) generation of magnetic field due to a current carrying solenoid
16. For making a strong electromagnet, the material of the core should be
(a) soft iron (b) steel
(c) brass (d) copper
17. Magnetic field inside a long solenoid carrying current is

29
Physics Smart Booklet

(a) same at all points (uniform)

(b) different at poles and at the centre
(c) zero
(d) different at all points
18. You have a coil and a bar magnet. You can produce an electric current by
1. moving the magnet but not the coil
2. moving the coil but not the magnet
3. moving either the magnet or the coil
4. using another DC source
19.: A fuse in an electric circuit acts as a
1. current multiplication
2. voltage multiplication
3. power multiplier
4. safety device
20. The magnetic lines of force inside a current carrying solenoid are
1. along the axis and parallel to each other
2. perpendicular to the axis and parallel to each other
3. circular and do not intersect each other
4. circular and intersect each other
21. Who was the first person to notice the magnetic effect of electric current?
(a) Faraday (b) Ampere (c) Oersted (d) Volta
22. The magnetic field produced due to the current passing through a conductor is
proportional to the
(a) electric current (b) conducting material
(c) length of conductor (d) diameter of conductor
23. ‘The magnetic field produced at the center of a circular wire is proportional to and
inversely proportional to
(a) radius of loop, current (b) current, radius of loop
(c) length of conductor, current (d) weight of conductor, current
24. The magnetic field of a solenoid is quite similar to that of
(a) a straight conductor (b) a horse-shoe magnet
(c) a bar magnet (d) any magnet
25. The principle of magnetic induction was given by
(a) Michael Faraday (b) Galileo
(c) Oersted (d) Ampere
26. The direction of a magnetic field is taken
(a) north to south and back (b) south to north and back
(c) north to south only (d) south to north only
27. In our domestic electric supply we use following three colours of wire.
(a) red, blue, green (b) red, yellow, blue
(c) red, black, green (d) black, green, yellow
28. The magnetic field due to electric current in a conductor is
(a) in the direction of electric current
(b) in the direction opposite to electric current
(c) circular around the conductor

30
Physics Smart Booklet

(d) in the center of the conductor

29. Which device is used to convert electric energy into mechanical energy ?
(a) electric generator (b) solenoid
(c) electric motor (d) electric iron
30. The principle of electric generator is
(a) conversion of electrical energy into mechanical energy
(b) conversion of electrical energy into thermal energy
(c) conversion of mechanical energy into electrical energy
(d) conversion of electrical energy into light energy
31. Magnetic lines of force inside a solenoid are...
(a) from N to S (b) from S to N
(c) circular (d) Qintersect one another
32. A magnetized wire of magnetic moment M and length L is bent in the form of a
semicircle of radius ‘r’. The new magnetic moment is
(a) M (b) M/2 (c) M/ (d) 2M/
33 In a hydrogen atom the electron is making 6.6×10 revolutions per second around
15

the nucleus in an orbit of radius 0.528 Å. The equivalent magnetic dipole moment is
approximately ( in Am2)
(a) 10-10 (b) 10-15 (c) 10-2 (d) 10-25
34. Two short magnets of dipole moments M and 2M are arranged on the table so that
the axial line of the weaker magnet and the equatorial line of the stronger magnet
are coinciding. If the separation between the magnets is 2d, what is the magnetic
flux density midway between these magnets? Ignore the earth’s magnetic field.
(a) (b) (c) (d)

35. An electron and a proton with the same momentum enter perpendicularly into a
uniform magnetic field
a. Both particles will deflect equally,
b. The proton will deflect more than the electron,
c. The electron will deflect less than the proton
d. None
36. Two parallel beams of electrons moving in the same direction will
a. Repel each other,
b. Attract each other,
c. Neither attract nor repel each other.
d. none
37. When an electron moves in a magnetic field ‘B’ with velocity ‘V’ the force acting on it
is perpendicular to
a .V but not to B, b. both V and B,
c. B but not V d. none
38. If an electron and proton enter into a magnetic field with the same velocity, the
electron shall experience a/an force than the proton.

31
Physics Smart Booklet

a. Greater, b. Lesser,
c. Equal d. none
39. Magnetism derives its name from a region in Asia Minor (Modern Turkey) where it
was found in for form of certain iron core.
a. Magnesia b. Magnesium,
c. Electromagnetism d. None of these
40 . If a magnet is broken into two pieces, then
a. Two magnets are obtained,
b. North pole is obtained,
c. South pole is obtained,
d. One north pole and one south pole is obtained
41. A magnet can be demagnetized by
a. Heating,
b. By dropping it several time,
c. breaking into two pieces,
d. both heating and by dropping it several time

TOPIC WISE PRACTICE

Topic 1: Motion of Charged Particle in Magnetic Field
1. A particle of mass m and charge q enters a magnetic field B perpendicularly with a velocity v. The radius
of the circular path described by it will be
(1) Bq/mv (2) mq/Bv (3) mB/qv (4) mv/Bq
2. The figure shows a thin metalic rod whose one end is pivoted at point O. The rod rotates about the end O
in a plane perpendicular to the uniform magnetic field with angular frequency w in clockwise direction.
Which of the following is correct?

(1) The free electrons of the rod move towards the outer end.

32
Physics Smart Booklet

(2) The free electrons of the rod move towards the pivoted end.
(3) The free electrons of the rod move towards the mid-point of the rod.
(4) The free electrons of the rod do not move towards any end of the rod as rotation of rod has no effect
on motion of free electrons.
3. A charged particle enters into a magnetic field with a velocity vector making an angle of 30º with respect
to the direction of magnetic field. The path of the particle is
(1) circular (2) helical (3) elliptical (4) straight line
4. A particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by the path
described by the particle is proportional to
(1) the velocity (2) the momentum (3) the kinetic energy (4) None of these
5. An electric charge +q moves with velocity in an electromagnetic field given by
and .The y-component of the force experienced by + q is :
(1) 11 q (2) 5 q (3) 3 q (4) 2 q
6. A straight steel wire of length l has a magnetic moment M. When it is bent in the form of a semi-circle its
magnetic moment will be
(1) M (2) M/ (3) 2 M/ (4) M
7. The magnetic force acting on a charged particle of charge – 2 C in a magnetic field of 2T acting in y

direction, when the particle velocity is , is

(1) 4 N in z direction (2) 8 N in y direction (3) 8 N in z direction (4) 8 N in – z direction
8. A charged particle goes undeflected in a region containing electric and magnetic fields. It is possible that
(1)
(2) is not parallel to
(3) but is not parallel to
(4) but is not parallel to
9. A positively charged particle enters in a uniform magnetic field with velocity perpendicular to the
magnetic field. Which of the following figures shows the correct motion of charged particle?

(1) (2) (3) (4)

10. A thin circular wire carrying a current I has a magnetic moment M. The shape of the wire is changed to a
square and it carries the same current. It will have a magnetic moment

(1) M (2) (3) (4)

11. A charged particle enters in a magnetic field in a direction perpendicular to the magnetic field. Which of
the following graphs show the correct variation of kinetic energy of the particle with time t?

(1) (2) (3) (4)

33
Physics Smart Booklet

12. An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre
of the circle. The radius of the circle is proportional to

(1) (2) (3) (4)

5
13. A proton moving with a velocity 3 × 10 m/s enters a magnetic field of 0.3 tesla at an angle of 30º with
the field. The radius of curvature of its path will be (e/m for proton = 108 C/kg)
(1) 2 cm (2) 0.5 cm (3) 0.02 cm (4) 1.25 cm
14. A beam of electrons is moving with constant velocity in a region having simultaneous perpendicular
electric and magnetic fields of strength 20 Vm–1 and 0.5 T respectively at right angles to the direction of
motion of the electrons. Then the velocity of electrons must be
(1) 8 m/s (2) 20 m/s (3) 40 m/s (4) 1/40m/ s
15. A uniform magnetic field acts at right angles to the direction of motion of electron. As a result, the
electron moves in a circular path of radius 2cm. If the speed of electron is doubled, then the radius of the
circular path will be
(1) 2.0 cm (2) 0.5 cm (3) 4.0 cm (4) 1.0 cm
16. A proton moving with a constant velocity passes through a region of space without any change in its
velocity. If E and B represent the electric and magnetic fields respectively, this region of space may not
have
(1) E = 0, B = 0 (2) E = 0, B 0 (3) E 0, B = 0 (4) E 0, B 0
17. A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated
magnetic moment m is given by
(1) qvR2 (2) qvR2/2 (3) qvR (4) qvR/2
18. A charged particle moves with velocity in a magnetic field . The force acting on the
particle has magnitude F. Then,
(1) F = 0, if aD = dA
(2) F = 0, if aD = –dA
(3) F = 0, i f aA = – dD
(4) F (a2 + b2)1/2 × (A2 + D2)1/2
19. If a particle of charge 10–12 coulomb moving along the - direction with a velocity 105 m/s experiences a
force of 10–10 newton in - direction due to magnetic field, then the minimum magnetic field is
(1) 6.25 × 103 Tesla in - direction
–15
(2) 10 Tesla in - direction
(3) 6.25 × 10–3 Tesla in - direction
(4) 10–3 Tesla in - direction

20. A certain region has an electric field and a uniform magnetic field

The force experienced by a charge 1C moving with velocity ms–1 is

(1) (2) (3) (4)

21. A cathode ray beam is bent in a circle of radius 2 cm by a magnetic induction 4.5 × 10–3 weber/m2. The
velocity of electron is
(1) 3.43 × 107 m/s (2) 5.37 × 107 m/s (3) 1.23 × 107 m/s (4) 1.58 × 107 m/s

34
Physics Smart Booklet

22. A proton and an a-particle enter a uniform magnetic field perpendicularly with the same speed. If proton
takes 25 second to make 5 revolutions, then the time period for the particle would be
(1) 50 sec (2) 25 sec (3) 10 sec (4) 5 sec
23. A wire of length L metre carrying a current I ampere is bent in the form of a circle. Its magnitude of
magnetic moment will be
(1) IL/4 (2) I2L2/4 (3) IL2/4 (4) IL2/8
24. What is cyclotron frequency of an electron with an energy of 100 e V in the magnetic field of 1 × 10–4
weber / m2 if its velocity is perpendicular to magnetic field?
(1) 0.7 MHz (2) 2.8 MHz (3) 1.4 MHz (4) 2.1 MHz
3
25. A charged particle with velocity 2 × 10 m/s passes undeflected through electric and magnetic field.
Magnetic field is 1.5 tesla. The electric field intensity would be
(1) 2 × 103 N/C (2) 1.5 × 103 N/C (3) 3 × 103 N/C (4) 4/3 × 10–3 N/C
26. An electron moving with kinetic energy 6×10–16 joules enters a field of magnetic induction 6 × 10–3
weber/m2 at right angle to its motion. The radius of its path is
(1) 3.42 cm (2) 4.23 cm (3) 5.17 cm (4) 7.7 cm
Topic 2: Magnetic Field Lines, Biot-Savart's Law and Ampere's Circuital Law
27. A current I flows along the length of an infinitely long, straight, thin walled pipe. Then
(1) the magnetic field at all points inside the pipe is the same, but not zero
(2) the magnetic field is zero only on the axis of the pipe
(3) the magnetic field is different at different points inside the pipe
(4) the magnetic field at any point inside the pipe is zero
28. A portion of a conductive wire is bent in the form of a semicircle of radius r as shown below in fig. At the
centre of semicircle, the magnetic induction will be

(1) zero (2) infinite (3) gauss (4) tesla

29. A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm
diameter carrying same current. The strength of magnetic field far away is
(1) twice the earlier value
(2) same as the earlier value
(3) one-half of the earlier value
(4) one-quarter of the earlier value
30. The correct plot of the magnitude of magnetic field vs distance r from centre of the wire is, if the radius
of wire is R

(1) (2) (3) (4)

31. If in a circular coil A of radius R, current I is flowing and in another coil B of radius 2R a current 2I is
flowing, then the ratio of the magnetic fields BA and BB, produced by them will be
(1) 1 (2) 2 (3) 1/2 (4) 4

35
Physics Smart Booklet

32. If the magnetic field at P can be written as K tan , the K is

(1) (2) (3) (4)

33. A long solenoid has 100 turns per cm and carries a current 6A. The magnetic field at its centre is 3.14 ×
10–2 Weber/m2. Another long solenoid has 50 turns per cm and it carries a current 2A. The value of the
magnetic field at its centre is
(1) 5.66 × 10–3 weber/m2 (2) 5.23 × 10–5 weber/m2
(3) 7.23 × 10–5 weber/m2 (4) 6.23 × 10–4 weber/m2
34. Two long straight wires are set parallel to each other. Each carries a current i in the same direction and the
separation between them is 2r. The intensity of the magnetic field midway between them is

(1) i/r (2) 4 i/r (3) zero (4) i/4r

35. A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and
the other in x-z plane. If the current in the loop is i, the resultant magnetic field due to the two
semicircular parts at their common centre is

(1) (2) (3) (4)

36. The figure shows n (n being an even number) wires placed along the surface of a cylinder of radius r.
Each wire carries current i in the same direction. The net magnetic field on the axis of the cylinder is

(1) (2) (3) zero (4)

37. A current of I ampere flows in a wire forming a circular arc of radius r metres subtending an angle at the
centre as shown. The magnetic field at the centre O in tesla is

(1) (2) (3) (4)

36
Physics Smart Booklet

38. The magnetic field at a distance r from a long wire carrying current i is 0.4 tesla. The magnetic field at a
distance 2r is
(1) 0.2 tesla (2) 0.8 tesla (3) 0.1 tesla (4) 1.6 tesla
39. The magnetic induction at a point P which is at a distance of 4 cm from a long current carrying wire is
10–3 T. The field of induction at a distance 12 cm from the current will be
(1) 3.33 × 10–4 T (2) 1.11 × 10–4 T (3) 3 × 10–3 T (4) 9 × 10–3 T
40. A current i ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at
any point inside the tube is

(1) (2) zero (3) tesla (4) tesla

41. If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any
point at a large distance from the centre becomes approximately
(1) double (2) three times (3) four times (4) one-fourth
42. Two long parallel wires P and Q are held perpendicular to the plane of paper with distance of 5 m
between them. If P and Q carry current of 2.5 amp. and 5 amp. respectively in the same direction, then the
magnetic field at a point halfway between the wires is
(1) /17 (2) / 2p (3) / 2p (4) 3 / 2p
43. Two concentric circular coils of ten turns each are situated in the same plane. Their radii are 20 and 40 cm
and they carry respectively 0.2 and 0.4 ampere current in opposite direction. The magnetic field in
weber/m2 at the centre is
(1) /80 (2) 7 /80 (3) (5/4) (4) zero
44. A solenoid of length 1.5 m and 4 cm diameter possesses 10 turns per cm. A current of 5A is flowing
through it, the magnetic induction at axis inside the solenoid is ( = 4 × 10–7 weber amp–1 m–1)
(1) 4 × 10–5 gauss (2) 2 × 10–5 gauss (3) 4 × 10–5 tesla (4) 2 × 10–5 tesla
45. At what distance from a long straight wire carrying a current of 12 A will the magnetic field be equal to 3
× 10–5 Wb/m2?
(1) 8 × 10–2 m (2) 12 × 10–2 m (3) 18 × 10–2 m (4) 24 × 10–2 m
46. A coaxial cable consists of a thin inner conductor fixed along the axis of a hollow outer conductor. The
two conductors carry equal currents in opposites directions. Let B1 and B2 be the magnetic fields in the
region between the conductors and outside the conductor, respectively Then,
(1) B1 0, B2 0 (2) B1 = B2 = 0(3) B1 0, B2 = 0 (4) B1 = 0, B2 0
47. The figure shows a system of infinite concentric circular current loops having radii R1, R2, R3 Rn. The
loops carry net current i alternately in clockwise and anticlockwise direction. The magnitude of net
magnetic field of the centre of the loops is

(1)

37
Physics Smart Booklet

(2)

(3)

(4)
48. Axis of a solid cylinder of infinite length and radius R lies along y-axis, it carries a uniformly distributed

current i along +y direction. Magnetic field at a point is

(1) (2) (3) (4)

Topic 3: Force and torque on current carrying conductor and moving coil Galvanometer
49. A current of 10 A is flowing in a wire of length 1.5 m. A force of 15 N acts on it when it is placed in a
uniform magnetic field of 2 T. The angle between the magnetic field and the direction of the current is
(1) 30° (2) 45° (3) 60° (4) 90°
50. P, Q and R are long straight wires in air, carrying currents as shown. The force on Q is directed

(1) to the left

(2) to the right
(3) to the plane of the diagram
(4) along the current in Q
51. A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane
becomes
(1) inclined at 45º to the magnetic field
(2) inclined at any arbitrary angle to the magnetic field
(3) parallel to the magnetic field
(4) perpendicular to magnetic field
52. Two thin long parallel wires separated by a distance b are carrying a current i amp each. The magnitude
of the force per unit length exerted by one wire on the other is

(1) (2) (3) (4)

53. A current of 5 ampere is flowing in a wire of length 1.5 metres. A force of 7.5 N acts on it when it is
placed in a uniform magnetic field of 2 tesla. The angle between the magnetic field and the direction of
the current is
(1) 30° (2) 45° (3) 60° (4) 90°
54. A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on
segments PS, SR, and P RQ are F1, F2 and F3 respectively and are in the plane of the paper and along
the directions shown, the force on the segment QP is

38
Physics Smart Booklet

(a) (2) (3) (4)

55. The figure shows two long straight current carrying wire separated by a fixed distance d. The magnitude
of current, flowing in each wire varies with time but the magnitude of current in each wire is equal at all
times. Which of the following graphs shows the correct variation of force per unit length f between the
two wires with current i?

(1) (2) (3) (4)

56. A moving coil galvanometer has a resistance of 900 . In order to send only 10% of the main current
through this galvanometer, the resistance of the required shunt is
(1) 0.9 (2) 100 (3) 405 (4) 90
57. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field

such that is perpendicular to the plane of the loop. The magnetic force acting on the loop is
(1) (2) (3) zero (4)
58. A current of 3 A is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a
magnetic field of strength 500 gauss and makes an angle of 30º with the direction of the field. It
experiences a force of magnitude
(1) 3 × 10–4 N (2) 3 × 10–2 N (3) 3 × 102 N (4) 3 × 104 N
59. In figure, an external torque changes the orientation of loop from one of lowest potential energy to one of
highest potential energy. The work done by the external torque is closest to
(1) 0.5 J (2) 0.2 J (3) 0.3 J (4) 0.4 J
60. Through two parallel wires A and B, 10A and 2A of currents are passed respectively in opposite
directions. If the wire A is infinitely long and the length of the wire B is 2m, then force on the conductor
B, which is situated at 10 cm distance from A, will be
(1) 8 × 10–7 N (2) 8 × 10–5 N (3) 4 × 10–7 N (4) 4 × 10–5 N
61. A circular loop of area 0.02 m2 carrying a current of 10A, is held with its plane perpendicular to a
magnetic field induction 0.2 T. The torque acting on the loop is
(1) 0.01 Nm (2) 0.001 Nm (3) zero (4) 0.8 Nm

39
Physics Smart Booklet

62. A current carrying conductor placed in a magnetic field experiences maximum force when angle between
current and magnetic field is
(1) 3 /4 (2) /2 (3) /4 (4) zero
63. To increase the range of voltmeter having resistance G from V to V/n, a shunt of how much resistance
should be connected in parallel to it?
(1) n3G (2) n2G (3) nG (4) G/n
64. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G,
the resistance of ammeter will be :

(1) (2) (3) (4)

65. A very long straight wire carries a current I. At the instant when a charge + Q at point P has velocity , as
shown, the force on the charge is

(1) along oy (2) opposite to oy (3) along ox (4) opposite to ox

NEET PREVIOUS YEARS

1. Current sensitivity of a moving coil galvanometer is 5 div/ mA and its voltage sensitivity (angular
deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is [2018]
(1) 40 (2) 25 (3) 500 (4) 250
2. A metallic rod of mass per unit length 0.5 kg m–1 is lying horizontally on a smooth inclined plane which
makes an angle of 30°with the horizontal. The rod is not allowed to slide down by flowing a current
through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current
flowing in the rod to keep it stationary is [2018]
(1) 7.14 A (2) 5.98 A (3) 11.32 A (4) 14.76 A
3. An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same
current 'I along the same direction is shown in fig. Magnitude of force per unit length on the middle wire
'B' is given by [2017]

40
Physics Smart Booklet

(1) (2) (3) (4)

4. A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 A and subjected
to magnetic field of strength 0.85 T. Work done for rotating the coil by 180º against the torque is[2017]
(1) 4.55 J (2) 2.3 J (3) 1.15 J (4) 9.1 J
5. A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY
carrying a current I, the net force on the loop will be : [2016]

(1) (2) (3) (4)

6. A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its
cross-section. The ratio of the magnetic fields B and B', at radial distances a/2 and 2a respectively, from
the axis of the wire is: [2016]
(1) 1/4 (2) 1/2 (3) 1 (4) 4
7. A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to
field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by
proton is 1 MeV the energy acquired by the alpha particle will be: [2015]
(1) 0.5 MeV (2) 1.5 MeV (3) 1 MeV (4) 4 MeV
8. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field
produced at the centre has magnitude: [2015]

(1) Zero (2) (3) (4)

9. Two identical long conducting wires AOB and COD are placed at right angle to each other, with one
above other such that ‘O’ is their common point for the two. The wires carry I1 and I2 currents
respectively. Point ‘P’ is lying at distance ‘d’ from ‘O’ along a direction perpendicular to the plane
containing the wires. The magnetic field at the point ‘P’ will be: [2014]

(2) (3) (4)

(1)

10. A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the

41
Physics Smart Booklet

magnetic field, B with the distance d, from the centre of the conductor, is correctly represented by the
figure: [NEET – 2019]

1) 2) 3) 4)
11. Ionized hydrogen atoms and a-particles with same momenta enters perpendicular to a constant magnetic

field B. The ratio of their radii of their paths will be [NEET – 2019]
(1) 2 :1 (2) 1 : 2 (3) 4 : 1 (4) 1 : 4
12. Two toroids 1 and 2 have total number of turns 200 and 100 respectively with average radii 40 cm and 20
cm respectively. If they carry same current i, then the ratio of the magnetic fields along the two is :
[NEET – 2019 (ODISSA)]
(1) 1 : 1 (2) 4 : 1 (3) 2 : 1 (4) 1 : 2
13. A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the
circular loop is R. The total magnetic field at the centre P of the loop is : [NEET – 2019 (ODISSA)]

(1) Zero (2) outward

(3) inward (4) inward

14. A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the

centre of the solenoid is [NEET – 2020]

1) 2) 3) 4)
15. An infinitely long straight conductor carries a current of 5A as shown. An electron is moving with a speed
of 10 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is
20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.
[NEET-2021]

42
Physics Smart Booklet

1) 2) 3) 4)
16. A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-
section. The variation of magnetic field BI due to the cable with the distance ‘r’ from the axis of the cable
is represented by [NEET-2021]

1) 2) 3) 4)

17. In the product

For q = 1 and and

What will be the complete expression for ? [NEET-2021]

1) 2) 3) 4)
18. A uniform conducting wire of length 12a and resistance ‘R’ is wound up as a current carrying coil in the
shape of [NEET-2021]
i) an equilateral triangle of side ‘a’ ii) a square of side ‘a’
The magnetic dipole moments of the coil in each case respectively are

1) 3Ia2 and Ia2 2) 3Ia2 and 4Ia2 3) 4Ia2 and 3Ia2 4) Ia2 and 3 Ia3
19. A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic
field strength at the centre of the solenoid is [NEET-2022]

1) 2) 3) 4)
20. A square loop of side 1 m and resistance is placed in a magnetic field of 0.5 T. If the plane of loop is
perpendicular to the direction of magnetic field, the magnetic flux through the loop is [NEET-2022]
1) 2 weber 2) 0.5 weber 3) 1 weber 4) zero weber
21. Given below are two statements: [NEET-2022]
Statements I: Biot-Savert’s law gives us the expression for the magnetic field strength of an infinitesimal
current element (Idl) of a current carrying conductor only
Statement II: Biot-Savert’s law is analogous to Coulomb’s inverse square law of charge q, with the former

43
Physics Smart Booklet

being related to the field produced by a scalar source, Idl while the latter being produced by a vector source q .
In light of above statements choose the most appropriate answer from the options give below:
1) Both Statement I and Statement II are correct
2) Both Statement I and Statement II are incorrect
3) Statement I is correct but Statement II is incorrect
4) Statement I is incorrect and Statement II is correct
22. From Ampere’s circuit law for a long straight wire of circuit cross-section carrying a steady current, the
variation of magnitude field in the inside and outside region of the wire is: [NEET-2022]
(1) Uniform and remains constant for both the regions.
(2) a linearly increasing function of distance upto the boundary of the wire and then linearly decreasing
for the outside region.
(3) a linearly increasing function of distance r upto the boundary of the wire and then decreasing one with
1/r dependence for the outside region.
(4) a linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing
one for the outside for the outside region.

NCERT LINE BY LINE QUESTIONS – ANSWERS

1) c 2) b 3) a 4) b 5) d
6) d 7) b 8) a 9) a 10) a

44
Physics Smart Booklet

11) b 12) d 13) c 14) d 15) b

16) a 17) d 18) d 19) d 20) c
NCERT BASED PRACTICE QUESTIONS
1) D 2) D 3) B 4) D 5) A
6) C 7) C 8) D 9) D 10) D
11) D 12) C 13) B 14) A 15) B
16) A 17) B 18) C 19) D 20) C
21) C 22) A 23) B 24) C 25) A
26) B 27) C 28) C 29) C 30) C
31) B 32) D 33) D 34) E 35) A
36) B 37) B 38) C 39) A 40) A
41) D

TOPIC WISE PRACTICE QUESTIONS - ANSWERS

1) 4 2) 2 3) 2 4) 3 5) 1 6) 3 7) 4 8) 1 9) 3 10) 4
11) 4 12) 4 13) 2 14) 3 15) 3 16) 3 17) 4 18) 1 19) 4 20) 1
21) 4 22) 3 23) 3 24) 2 25) 3 26) 1 27) 4 28) 4 29) 2 30) 2
31) 1 32) 2 33) 1 34) 3 35) 2 36) 3 37) 1 38) 1 39) 1 40) 2
41) 3 42) 3 43) 4 44) 4 45) 1 46) 3 47) 2 48) 1 49) 1 50) 1
51) 4 52) 2 53) 1 54) 2 55) 3 56) 2 57) 3 58) 2 59) 3 60) 2
61) 3 62) 3 63) 3 64) 3 65) 1

NEET PREVIOUS YEARS QUESTIONS-ANSWERS

1) 4 2) 3 3) 3 4) 4 5) 1 6) 3 7) 3 8) 3 9) 4
10) 3 11) 1 12) 1 13) 1 14) 2 15) 3 16) 2 17) 1 18) 4
19) 2 20) 2 21) 3 22) 3

TOPIC WISE PRACTICE QUESTIONS - SOLUTIONS

1. (4) Force,

2. (2) The application of equation on the element dl of the rod gives force on positive charge
towards the outer end. Therefore electrons will move towards pivoted end.

3. (2) Here velocity vector have two components

(i) v cos , parallel to magnetic field
(ii) v sin , perpendicular to magnetic field. Due to component v cos , the particle will have a linear
motion but due to v sin , the particle will have simultaneously a circular motion. The resultant of the two
is a helical path.

45
Physics Smart Booklet

4. (3) As

Area
5. (1) Lorentz force acting on the particle

6. (3) Given: Length of iron rod = L

Magnetic moment of rod = M
Solution: As we know,
Magnetic moment is given by,
M=m×L (1)
According to question, when we bent the rod the pole strength of the rod remains unchanged. However, when the
rod is bent in form of a semicircular arc the separation between the two poles become 2r (r is the radius of the
semicircular arc).

∴πr=L ;
Therefore, the new magnetic moment will be,
M′=m×2r

7. (4) The magnetic force acting on the charged particle is given by

Force is of 8N along – z-axis.

46
Physics Smart Booklet

8. (1) a) The charged particle will get accelerated in the direction or opposite to the electric field and will not
deflected since ∥
b) If ∥ , deflection due to magnetic field can be balanced by acceleration due to electric field.
c) ∥ ⇒ mag=0 Since ∥ the particle will get deflected.
d) ∥ ; ∥
⇒ the particle will get deflected.

9. (3) Force on a moving charge in a magnetic field is

Thus if the particle is moving along the magnetic field, = 0.
Hence the particle continues to move along the incident direction, in a straight line.
When the particle is moving perpendicular to the direction of magnetic field, the force is perpendicular to both
direction of velocity and the magnetic field.
Then the force tends to move the charged particle in a plane perpendicular to the direction of magnetic field, in a
circle.
If the direction of velocity has both parallel and perpendicular components to the direction magnetic field, the
perpendicular component tends to move it in a circle and parallel component tends to move it al
10. (4) Initially for circular coil L = 2 r and M = i × r2

-------------(i)

Finally for square coil side and

----------------(ii)

Solving equation (i) and (ii)

11. (4) The change in K.E. is equal to work done by net force which is zero because the magnetic force is
perpendicular to velocity. K.E. remains constant.

12. (4)

13. (2)

14. (3) The electron moves with constant velocity without deflection. Hence, force due to magnetic field is
equal and opposite to force due to electric field.

47
Physics Smart Booklet

15. (3) or
As v is doubled, the radius also becomes double.
Hence, radius = 2 × 2 = 4 cm.
16. (3) (i) When no field is present E=0,B=0, the proton experiences no force. Thus it moves with a constant velocity.
(ii) When E=0 and B=0, then there will be a probability that proton may move parallel to magnetic field. In this
situation, there will be no force acting on proton.
(iii) When both fields are present
E=0,B=0, then let E, B and v may be mutually perpendicular to each other. In this case, the electric and magnetic
forces acting on the proton may be equal and opposite. Thus, there will be no resultant force on the proton.
17. (4) Magnetic moment m = IA

Since T Also,

18. (1)

19. (4)

(when )

Tesla in - direction
20. (1) F=qE+q(v×B)
this is loranze force

21. (4)
22. (3) Time taken by proton to make one revolution

As so

or
23. (3) If r is the radius of the circle,

then or,
Area

48
Physics Smart Booklet

24. (2) Cyclotron frequency

25. (3) E = vB = 2 103 1.5 = 3 103 N/C

26. (1) ;

27. (4) There is no current inside the pipe. Therefore

28. (4) The straight part will not contribute magnetic field at the centre of the semicircle because every
element of the straight part will be 0º or 180º with the line joining the centre and the element

Due to circular portion, the field is

Hence total field at

29. (2) and so it is independent of thickness.

The current is same in both the wires, hence magnetic field induced will be same.
30. (2) The magnetic field from the centre of wire of radius R is given by

and
From the above descriptions, we can say that the graph (2) is a correct representation.

31. (1) In coil A,

Hence,
32. (2)

Let us compute the magnetic field due to any one segment:

49
Physics Smart Booklet

Resultant field will be

33. (1) ;

34. (3)
35. (2) Magnetic fields due to the two parts at their common centre are respectively,

and

Resultant field

36. (3) Since n is an even number, we can assume the wires in pairs such that the two wires forming a pair is
placed diametrically opposite to each other on the surface of cylinder. The fields produced on the axis by
them are equal and opposite and can get cancelled with each other.

37. (1)

38. (1)
When r is doubled, the magnetic field becomes half, i.e., now the magnetic field will be 0.2 T.

39. (1)
As the distance is increased to three times, the magnetic induction reduces to one third. Hence,

40. (2) Magnetic induction inside a thin walled tube is zero. (According to Ampere's Law)

50
Physics Smart Booklet

41. (3) ;
So, when radius is doubled, magnetic field becomes four times.

42. (3)

43. (4)
44. (4)
45. (1) Current (I) = 12 A and magnetic field (2) = 3 × 10–5 Wb/m2. Consider magnetic field at distance r.

Magnetic field,

46. (3) Apply Ampere's circular law to the coaxial circular loops L1 and L2

The magnetic field is B1 at all points on L1 and B2 at all points of L2. for L1 and 0 for L2.
Hence, B1 0 but B2 = 0

47. (2) Field at the centre of a circular current loop is given by . Since the currents are alternately in
opposite directions therefore the correct net field at centre is given by vector sum of field produced by
each loop which are alternately in opposite directions.
48. (1) The magnitude of magnetic field at P

(independent on y-coordinate)
Unit vector in direction of magnetic field is

51
Physics Smart Booklet

(shown by dotted lines)

49. (1) or
50. (1) Parallel current attracts while opposite current repel each other.
51. (4) A current carrying coil behaves as a magnetic dipole. Therefore, in a uniform magnetic field coil will get
aligned such that the dipole moment of the coil is parallel to the magnetic field. And we know that dipole moment
of a coil is perpendicular to its plane.
Therefore, coil will align itself such that its plane is perpendicular the direction of magnetic field.
52. (2) Given

Hence force per unit length is

53. (1)
54. (2)

55. (3) The force per unit length is given by

i.e.,
56. (2) Ig = 0.1I, Is = 0.9 I ; S = Ig Rg / Is
= 0.1 900 / 0.9 = 100 .

52
Physics Smart Booklet

57. (3) Total force on the current carrying closed loop

should be zero, if placed in uniform magnetic field.

Fhorizontal=(F3−F1)
Fvertical=F2

Resultant of and is

where
Since total force = 0, hence force on QP is equal
to F in magnitude but opposite direction.
58. (2) Force on a current carrying conductor is given as F=ILB sin θ where θ is angle between length L and field B.
i.e. 300
Put B=500×10−4Tesla and L=0.4m with I =3A we get
F=3×10−2N so n=3
59. (3) The potential energy of a current carrying loop kept in external magnetic field is
U=−
Hence work done in moving form lowest potential energy to highest potential energy=MB−(−MB)=2MB
=2×0.75×0.2J =0.3J

60. (2)

61. (3) Area (1) = 0.01 m2; Current (I) = 10A;

Angle = 900 and magnetic field (2) =0.1T

Therefore actual angle

And torque acting on the loop

62. (3) F=iBlsinθ. This is maximum when sinθ=1 or θ=π/2.

63. (3) ;

64. (3) As 0.2% of main current passes through the galvanometer hence current through the shunt.

Total resistance of Ammeter

53
Physics Smart Booklet

65. (1) The direction of B is along

The magnetic force

NEET PREVIOUS YEARS QUESTIONS-EXPLANATIONS

1. (4) Current sensitivity of moving coil galvanometer

....(i)
Voltage sensitivity of moving coil galvanometer,

...(ii)
Dividing eqn. (i) by (ii)
Resistance of galvanometer

2. (3) From figure, for equilibrium,

mg sin 30° = I/B cos30°

3. (3) Force per unit length between two parallel current carrying conductors,

Since same current flowing through both the wires i1 = i2 = I, so

Magnitude of force per unit length on the middle wire 'B'

54
Physics Smart Booklet

4. (4) Work done, W = MB(cos –cos )

When it is rotated by angle 180° then
W = MB (cos0° – cos180°) = MB (1 + 1)
W = 2MB = 2 (NIA)B
= 2 × 250 × 85 × 10–6[1.25 × 2.1 × 10–4] × 85 × 10–2 = 9.1 mJ
5. (1) The direction of current in conductor XY and AB is same
(attractive)

FBC opposite to FAD =

Therefore the net force on the loop
Fnet = FAB + FBC + FCD + FAD

6. (3) Consider two amperian loops of radius a/2 and 2a as shown in the diagram.
Applying ampere's circuital law for these loops, we get
∮B.dL=μ0Ienclosed
For the smaller loop,

⇒B×2π =μ0× ×

=μ0I× =

⇒B1
B′×2π(2a)=μ0I

7. (3) As we know, F = qvB =

Since R is same so, KE

Therefore KE of a particle

8. (3) Radius of circular orbit = r

55
Physics Smart Booklet

No. of rotations per second = n

i.e.,
Magnetic field at its centre, Bc =?
As we know, current

equivalent current
Magnetic field at the centre of circular orbit,

9. (4) Net magnetic field,

10.

11.

56
Physics Smart Booklet

12. For a toroid magnetic field, B=μ0ni

Where, n= number of turns per unit length

Now,

13. Magnetic field due to

(Into the plane)

Magnetic field due to

(out of the plane )

For parallel combination

Net magnetic field

14.

15.

16. ;

57
Physics Smart Booklet

17.

Given

Thus, calculating values of RHS,

Comparing L.H.S and R.H.S,

From (2) and (3)

B = –6 and B0 = –8
Hence ,
18. Current in the loop will be V/R = I which is same for both loops.
Now magnetic moment of Triangle loop = NIA

and magnetic moment of square loop = NIA

19.

20.
21. Statement I is correct, Statement II is wrong because Idl is a vector source while in case of
coulomb law, charge is a scalar source.

22. when

when

when

58
Physics Smart Booklet

59