NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter is a study material designed for students pursuing science in their senior secondary education. The chapter explores the principles and applications of magnetism and matter, and the solutions are structured in a concise and simple manner, making it easy for students to understand the concepts and score good marks in their exams.
This chapter covers various topics such as magnetic materials, magnetic field and lines of force, Hysteresis, domain theory of ferromagnetism, and many more. These concepts are crucial for students who plan to pursue further studies in physics, as it provides a strong foundation in the subject. By the end of the course, students will have a thorough understanding of the principles and applications of magnetism and matter, which will be useful in their future studies and career.
Magnetism And Matter NCERT Solutions – Class 12 Physics
Q1 :A short bar magnet placed with its axis at 30º with a uniform externalmagnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10-2J. What is the magnitude of magnetic moment of the magnet?
Answer :
Magnetic field strength, B= 0.25 T
Torque on the bar magnet, T= 4.5 ×10 – 2J
Angle between the bar magnet and the external magnetic field,θ= 30°
Torque is related to magnetic moment (M) as:
T= MB sin θ
Hence, the magnetic moment of the magnet is 0.36 J T – 1.
Q2 :A short bar magnet of magnetic moment m = 0.32 JT-1 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?
Answer :
Moment of the bar magnet, M= 0.32 J T – 1
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 θ, between the bar magnet and the magnetic field is 0°.
Potential energy of the system = – MB cos θ
(b)The bar magnet is oriented 180°to the magnetic field. Hence, it is in unstable equilibrium.
θ = 180°
Potential energy = – MBcos θ
Q3 : A closely wound solenoid of 800 turns and area of cross section
2.5 x 10-4 m2 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?
Answer :
Number of turns in the solenoid, n = 800
Area of cross-section, A= 2.5 x 10-4m2
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
= 800 x 3 x 2.5 x 10-4
= 0.6 J T-1
Q4 :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 30° with the direction of applied field?
Answer :
Magnetic field strength, B = 0.25 T
Magnetic moment, M= 0.6 T – 1
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
Q5 :A bar magnet of magnetic moment 1.5 J T-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
Answer :
(a)Magnetic moment, M= 1.5 J T – 1
Magnetic field strength, B= 0.22 T
(i)Initial angle between the axis and the magnetic field, θ1= 0°
Final angle between the axis and the magnetic field, θ2= 90°
The work required to make the magnetic moment normal to the direction of magnetic field is given as:
(ii)Initial angle between the axis and the magnetic field, θ1= 0°
Final angle between the axis and the magnetic field, θ2= 180°
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
(b)For case (i): θ = θ2 = 90o
∴Torque, τ = MB sinθ
For case (ii): θ = θ2 = 1800
∴Torque, τ = MB sin θ
= MB sin180o= 0 J
Q6 :A closely wound solenoid of 2000 turns and area of cross-section
1.6 x 10-4 m2, 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 7.5 x 10-2T is set up at an angle of 30º with the axis of the solenoid?
Answer :
Number of turns on the solenoid,n = 2000
Area of cross-section of the solenoid, A= 1.6 ×10 – 4 m2
Current in the solenoid, I= 4 A
(a)The magnetic moment along the axis of the solenoid is calculated as:
M= nAI
= 2000 ×1.6 ×10 – 4 ×4
= 1.28 Am2
(b)Magnetic field, B = 7.5 ×10 -2T
Angle between the magnetic field and the axis of the solenoid, θ= 30°
Torque, τ = MB sinθ
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 x 10-2 Nm
Q7 :A short bar magnet has a magnetic moment of 0.48 J T-1. 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.
Answer :
Magnetic moment of the bar magnet, M= 0.48 J T – 1
(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:
Where,
μ0= 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.