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 2 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
2 Marks Questions
1. A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. What is the work done to turn the magnet, so as the align its magnetic moment?
(i) Opposite to the field direction
(ii) Normal to the field direction?
Ans. Since work done W = MB 
(i) 
W = MB [1- (-1)]
W = 2MB
(ii) 
W = MB 
W = MB
2. An electron in the ground state of hydrogen atom is revolving in anti – clock wise direction in a circular orbit. The atom is placed normal to the electron orbit makes an angle of 30o in the magnetic field. Find the torque experienced by the orbiting electron?
Ans. Magnetic moment associated with electron M = 
3. Define angle of dip. Deduce the relation connecting angle of dip and horizontal component of earth’s total magnetic field with the horizontal direction.
Ans. 
4. A point change +q is moving with speed 
perpendicular to the magnetic field B as shown in the figure. What should be the magnitude and direction of the applied electric field so that the net force acting on the charge is zero?
Ans. Force on the charge due to magnetic field = qVB sin
Since 
to the plane of paper and in words
F = qVB 
F = qVB (along OY)
Force on the charge due to electric field
F = qE
Net force on change is zero if 
qE = qVB
E = VB
(along YO)
5. The energy of a charged particle moving in a uniform magnetic field does not change. Why?
Ans. The force on a charged particle in a uniform magnetic field always acts in a direction perpendicular to the motion of the charge. Since work done by the magnetic field on the charge is zero, hence energy of the charged particle will not change.
6. In the figure, straight wire AB is fixed; white the loop is free to move under the influence of the electric currents flowing in them. In which direction does the loop begin to move? Justify.
Ans. Since current in AB and arm PQ are in same direction therefore wire will attract the arm PQ with a force (say
)
But repels the arm RS with a force (say
)
Sine arm PQ is closer to the wire AB
i.e. the loop will move towards the wire.
7. State two factors by which voltage sensitivity of a moving coil galvanometer can be increased?
Ans. Voltage sensitivity = 
It can be increased by
(1) increasing B using powerful magnets
(2) decreasing k by using phosphor borne strip
8. What is the magnetic moment associated with a coil of 1 turns, area of cross- section 
carrying a current of 2A?
Ans. m = NIA
9. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Ans. Mean radius of a Rowland ring, r= 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N= 3500
Relative permeability of the core material, 
Magnetising current, I= 1.2 A
The magnetic field is given by the relation:
B 
Where,
= Permeability of free space = 
Therefore, the magnetic field is n the core is 4.48 T.
10. At a certain location in Africa, a compass points 
west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 
above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Ans. Angle of declination, 
Angle of dip, 
Horizontal component of earth’s magnetic field, BH= 0.16 G
Earth’s magnetic field at the given location = B
We can relate B and BH as:
Earth’s magnetic field lies in the vertical plane, West of the geographic meridian, making an angle of 60°(upward) with the horizontal direction. Its magnitude is 0.32ss G.
11. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 
with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Ans. Horizontal component of earth’s magnetic field, BH= 0.35 G
Angle made by the needle with the horizontal plane = Angle of dip = 
Earth’s magnetic field strength = B
We can relate B and BH as:
Hence, the strength of earth’s magnetic field at the given location is 0.377 G.
12. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 
with the direction of applied field?
Ans. Magnetic field strength, B = 0.25 T
Magnetic moment, M=
The angle, between the axis of the solenoid and the direction of the applied field is 
Therefore, the torque acting on the solenoid is given as:
13. A closely wound solenoid of 800 turns and area of cross section
carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Ans. Number of turns in the solenoid, n = 800
Area of cross-section, A= 
Current in the solenoid, I= 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M= n I A
= 
= 
14. A short bar magnet placed with its axis at 
with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to
. What is the magnitude of magnetic moment of the magnet?
Ans. Magnetic field strength, B= 0.25 T
Torque on the bar magnet, T=
Angle between the bar magnet and the external magnetic field, 
Torque is related to magnetic moment (M) as:
T= MB sin θ
Hence, the magnetic moment of the magnet is
.
