CBSE Class 12 Physics Chapter-4 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 4 – Moving Charges and Magnetism 5 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-4 Important Questions
CBSE Class 12 Physics Important Questions Chapter 4 – Moving Charges and Magnetism
5 Marks Questions
1. (a) What is cyclotron? Explain its working principle?
(b) A cyclotron’s oscillator frequency is 10MHz what should be the operating magnetic field for accelerating protons? If radius of its dees is 20cm, what is the K.E. of the proton beam produced by the accelerator? (
, 
,
)?
Ans. (a) It is a device used to accelerate charged particles like protons, deuterons, 
– particle etc.
It is based on the principle that a charged particle can be accelerated to very high energies by making it pass through a moderate electric field a number of times and applying a strong magnetic field at the same time.
(b) v = 10MHz = 10
106 Hz
r = 20cm = 
KE = 
Since 
2. (a) Draw a labelled diagram of a moving coil galvanometer. Prove that in a radial magnetic field, the deflection of the coil is directly proportional to the current flowing in the coil.
(b) A galvanometer can be converted into a voltmeter to measure upto
(i) V volt by connecting a resistance 
series with the coil
(ii) 
volt by connecting a resistance 
in series with coil Find R in terms of 
required to convert – it into a voltmeter that can read upto ‘2v’ volt.
Ans. (a) When a current I is passed through a coil two equal and opposite forces acts on the arms of a coil to form a couple which exerts a Torque on the coil.
=> 
If 
= 
is the angle made by the normal to the plane of coil with B
= NIAB —-(1)
This is called as deflecting torque
As the coil deflected the spring is twisted and a restoring torque per unit twist then the restoring torque for the deflecting & is given by
= k 
—-(2)
In equilibrium
Deflecting Torgue=Restoring Torgue
NIAB = K
I = 
I = G where G = 
(galvanometer constant)
=> 
Thus deflection of the coil is directly proportional to the current flowing in the coil.
(b) We know Ig = 
=> Ig = 
—–(1)
And Ig = 
Equating (1) & (2)
For conversion Ig = 
=> Ig 
3. Two moving coil meters, M1 and M2 have the following particulars:
, 
, 
, 
, 
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of 
Ans. For moving coil meter M1:
Resistance,
Number of turns,
Area of cross-section, 
Magnetic field strength,
Spring constant 
For moving coil meter M2:
Resistance,
Number of turns,
Area of cross-section, 
Magnetic field strength,
Spring constant, 
(a) Current sensitivity of
is given as:
And, current sensitivity of
is given as:
Ratio
Hence, the ratio of current sensitivity of 
is 1.4.
(b) Voltage sensitivity for is given as:
And, voltage sensitivity for is given as:
4. In a chamber, a uniform magnetic field of 6.5 G 
is maintained. An electron is shot into the field with a speed of 
normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e =
, me=
)
Ans.Magnetic field strength, B = 6.5 G = 
Speed of the electron, v = 
Charge on the electron, e =
Mass of the electron, 
Angle between the shot electron and magnetic field, 
Magnetic force exerted on the electron in the magnetic field is given as:
F = evB 
This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.
Hence, centripetal force exerted on the electron,
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,
= 4.2 cm
Hence, the radius of the circular orbit of the electron is 4.2 cm.
5. In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Ans. Magnetic field strength, 
Charge of the electron, 
Mass of the electron, 
Velocity of the electron, 
Radius of the orbit, r = 4.2 cm = 0.042 m
Frequency of revolution of the electron = v
Angular frequency of the electron = 
= 2nv
Velocity of the electron is related to the angular frequency as:
v = r
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
This expression for frequency is independent of the speed of the electron.
On substituting the known values in this expression, we get the frequency as:
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.
6. Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.
Ans. Radius of coil X, 
= 0.16 m
Radius of coil Y, 
= 0.1 m
Number of turns of on coil X, 
Number of turns of on coil Y, 
Current in coil X, 
Current in coil Y, 
Magnetic field due to coil X at their centre is given by the relation,
Where,
= Permeability of free space = 
(towards East)
Magnetic field due to coil Y at their centre is given by the relation, 
(towards East)
Hence, net magnetic field can be obtained as:
7. For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, 
, approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Ans. Radius of circular coil = R
Number of turns on the coil = N
Current in the coil = I
Magnetic field at a point on its axis at distance x is given by the relation,
Where,
= Permeability of free space
(a) If the magnetic field at the centre of the coil is considered, then x = 0.
This is the familiar result for magnetic field at the centre of the coil.
(b) Radii of two parallel co-axial circular coils = R
Number of turns on each coil = N
Current in both coils = I
Distance between both the coils = R
Let us consider point Q at distance d from the centre.
Then, one coil is at a distance of 
from point Q.
Magnetic field at point Q is given as:
Also, the other coil is at a distance of 
from point Q.
Magnetic field due to this coil is given as:
Total magnetic field, 
For d << R, neglecting the factor 
, we get:
Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.
8. An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.
Ans. Magnetic field strength, B = 0.15 T
Charge on the electron,
Mass of the electron, 
Potential difference, V = 2.0 
Thus, kinetic energy of the electron = 
……(1)
Where,
v = velocity of the electron
(a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.
Magnetic force on the electron is given by the relation,
B ev
Centripetal force 
……(2)
From equations (1) and (2), we get
= 1 mm
Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
(b) When the field makes an angle 
with initial velocity, the initial velocity will be,
From equation (2), we can write the expression for new radius as: 
= 0.5 mm
Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
9. A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is
, make a simple guess as to what the beam contains. Why is the answer not unique?
Ans. Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 
Electrostatic field, E =
Mass of the electron = m
Charge of the electron = e
Velocity of the electron = v
Kinetic energy of the electron = eV
…… (1)
Since the particle remains unelected by electric and magnetic fields, we can infer that the electric field is balancing the magnetic field.
……..(2)
Putting equation (2) in equation (1), we get
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are
, etc.
10. A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Ans.Magnetic field strength, B = 1.5 T
Radius of the cylindrical region, r = 10 cm = 0.1 m
Current in the wire passing through the cylindrical region, I = 7 A
(a) If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region.
Thus, l = 2r = 0.2 m
Angle between magnetic field and current,
Magnetic force acting on the wire is given by the relation,
F = BIl 
= 
= 2.1 N
Hence, a force of 2.1 N acts on the wire in a vertically downward direction.
(b) New length of the wire after turning it to the Northeast-Northwest direction can be given as:
Angle between magnetic field and current, θ = 45°
Force on the wire,
F = 
= BIl
=2.1 N
Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angle 
because 
is fixed.
(c) The wire is lowered from the axis by distance, d = 6.0 cm
Suppose wire is passing perpendicularly to the axis of cylindrical magnetic field then lowering 6 cm means displacing the wire 6 cm from its initial position towards to end of cross sectional area.
= 8cm
Thus the length of wire in magnetic field will be 16 cm as AB= L =2x =16 cm
Now the force,
F = iLB 
as the wire will be perpendicular to the magnetic field.
The direction will be given by right hand curl rule or screw rule i.e. vertically downwards.
11. A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?
Ans.Magnetic field strength, B = 3000 G = 
= 0.3 T
Length of the rectangular loop, l = 10 cm
Width of the rectangular loop, b = 5 cm
Area of the loop,
= 
= 
Current in the loop, I = 12 A
Now, taking the anti-clockwise direction of the current as positive and vise-versa:
(a) Torque, 
From the given figure, it can be observed that A is normal to the y–z plane and B is directed along the z-axis.
The torque is 
along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.
(b) This case is similar to case (a). Hence, the answer is the same as (a).
(c) Torque 
From the given figure, it can be observed that A is normal to the x–z plane and B is directed along the z-axis.
The torque is along the negative x direction and the force is zero.
(d) Magnitude of torque is given as:
Torque is at an angle of 
with positive x direction. The force is zero.
(e) Torque
= 0
Hence, the torque is zero. The force is also zero.
(f) Torque
= 0
Hence, the torque is zero. The force is also zero.
In case (e), the direction of 
and 
is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.
Whereas, in case (f), the direction of and is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.
12. A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? 
Ans.Length of the solenoid, L = 60 cm = 0.6 m
Radius of the solenoid, r = 4.0 cm = 0.04 m
It is given that there are 3 layers of windings of 300 turns each.
Total number of turns, 
= 900
Length of the wire, l = 2 cm = 0.02 m
Mass of the wire, m = 2.5 g = 
Current flowing through the wire, i = 6 A
Acceleration due to gravity, 
Magnetic field produced inside the solenoid, 
Where,
= Permeability of free space =
I = Current flowing through the windings of the solenoid
Magnetic force is given by the relation, F = Bil
Also, the force on the wire is equal to the weight of the wire.
Hence, the current flowing through the solenoid is 108 A.
