CBSE Class 12 Physics Chapter-5 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 5 – Magnetism And Matter 3 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-5 Important Questions
CBSE Class 12 Physics Important Questions Chapter 5 – Magnetism And Matter
3 Marks Questions
1. A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at 60o to a uniform magnetic field. It experiences a torque of 0.063 Nm. (i) calculate the strength of the magnetic field and (ii) what orientation of the bar magnet corresponds to the equilibrium position in the magnetic field?
Ans. (i) Since 
Here 
(ii) The magnet will be in stable equilibrium in the magnetic field if 
i.e When magnet aligns itself parallel to the field
2. A beam of electrons is moving with a velocity of 
and carries a current of 1
A.
(a) How many electrons per second pass a given point?
(b) How many electrons are in 1m of the beam?
(c) What is the total force on all the electrons in 1m of the beam if it passes through the field of 
?
Ans. 
(a) 
(b) Electrons traverse a distance of 
in 1 s
No. of electrons in 1 meter of the beam
(c) Force on 1 meter of the beam of electrons
3. What is the main function of soft iron core used in a moving coil galvanometer? A galvanometer gives full deflection for Ig. Can it be converted into an ammeter of range I < Ig?
Ans. Soft iron core is used the moving coil galvanometer because it increases the strength of the magnetic field thus increases the sensitivity of the galvanometer.
We know S = 
For I < Ig, S becomes negative
Hence it cannot be converted into an ammeter of range I < Ig.
4. Two wires loops PQRSP formed by joining two semicircular wires of radii 
carries a current I as shown in the figure. What is the direction of the magnetic induction at the centre C.?
Ans. Magnetic field due to semicircle QR at C. is
Magnetic field due to semicircle is at C is
Net field 
5. A circular coil is placed in uniform magnetic field of strength 0.10T normal to the plane of coil. If current in the coil is 5.0A. Find.
(a) Total torque on the coil
(b) Total force on the coil
(c) Average force on each electron due to magnetic field
(The coil is made of copper wire of cross- sectional area 
and free electron density in copper is
)
Ans. (a) B = 0.10T
(Normal to plane of the coil)
I = 5.0 A, Area = n =
(b) Total force on the coil = 0 Newton
(c) Fav = q 
)
(I = neAVd)
Fav = 
Fav = 
6. Using Ampere’s circuital law, derive an expression for magnetic field along the axis of a Toroidal solenoid?
Ans. If n be the no, of turns per unit length I be the current flowing through the Toroid
Then from Ampere’s circuital law
7. A short bar magnet of magnetic moment m = 
is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Ans. Moment of the bar magnet, M= 
External magnetic field, B= 0.15 T
(a) The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle
, between the bar magnet and the magnetic field is
.
Potential energy of the system
(b) The bar magnet is oriented 180°to the magnetic field. Hence, it is in unstable equilibrium.
Potential energy = – MB 
8. A closely wound solenoid of 2000 turns and area of cross-section
, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 
is set up at an angle of 
with the axis of the solenoid?
Ans. Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A= 
Current in the solenoid, I= 4 A
(a) The magnetic moment along the axis of the solenoid is calculated as:
M= nAI
= 
= 1.28 A
(b) Magnetic field, B =
Angle between the magnetic field and the axis of the solenoid, 
Torque, 
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 
s
9. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude
. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 
. What is the moment of inertia of the coil about its axis of rotation?
Ans.Number of turns in the circular coil, N= 16
Radius of the coil, r= 10 cm = 0.1 m
Cross-section of the coil, A=
Current in the coil, I= 0.75 A
Magnetic field strength, B=
Frequency of oscillations of the coil, v=
∴Magnetic moment, M= NIA
= 
= 
Frequency is given by the relation:
Where,
I= Moment of inertia of the coil
Hence, the moment of inertia of the coil about its axis of rotation is 
10. A short bar magnet has a magnetic moment of
. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Ans. Magnetic moment of the bar magnet, M= 
(a) Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:
ss
Where,
= Permeability of free space = 
The magnetic field is along the S – N direction.
(b) The magnetic field at a distance of 10 cm (i.e., d= 0.1 m) on the equatorial line of the magnet is given as:
The magnetic field is along the N – S direction.
11. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Ans. Earth’s magnetic field at the given place, H= 0.36 G
The magnetic field at a distance d, on the axis of the magnet is given as:
……(i)
Where,
= Permeability of free space
M= Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
[Using equation (i)]
Total magnetic field,
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
12. A long straight horizontal cable carries a current of 2.5 A in the direction 
south of west to 10°north of east. The magnetic meridian of the place happens to be west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Ans. Current in the wire, I= 2.5 A
Angle of dip at the given location on earth, 
Earth’s magnetic field, H= 0.33 G =
The horizontal component of earth’s magnetic field is given as:
= H cos 
The magnetic field at the neutral point at a distance Rfrom the cable is given by the relation:
Where,
= Permeability of free space = 
Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.
13. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 
with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Ans. Number of turns in the circular coil, N= 30
Radius of the circular coil, r= 12 cm = 0.12 m
Current in the coil, I= 0.35 A
Angle of dip, 
(a) The magnetic field due to current I, at a distance r, is given as:
s
Where,
= Permeability of free space = 
s
= 
The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as:
=
(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of
, the needle will reverse its original direction. In this case, the needle will point from East to West.
14. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is
, and one of the fields has a magnitude of 
If the dipole comes to stable equilibrium at an angle of 
with this field, what is the magnitude of the other field?
Ans. Magnitude of one of the magnetic fields
Magnitude of the other magnetic field =
Angle between the two fields, 
At stable equilibrium, the angle between the dipole and field 
Angle between the dipole and field 
= 
At rotational equilibrium, the torques between both the fields must balance each other.
∴ Torque due to field 
Where,
M= Magnetic moment of the dipole
s
Hence, the magnitude of the other magnetic field is 
15. The magnetic moment vectors 
and 
associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:
, 
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Ans. The magnetic moment associated with the orbital angular momentum is valid with the classical mechanics.
The magnetic moment associated with the orbital angular (l) momentum is given as
For current i and area of cross-section A, we have the relation:
Magnetic moment
Where,
e = Charge of the electron
r = Radius of the circular orbit
T = Time taken to complete one rotation around the circular orbit of radius r
Angular momentum,
Where,
M = Mass of the electron
V = Velocity of the electorn
Dividing equation (1) by equation (2),we get:
Therefore of the two relations,
is in accordance with class physics.