NCERT solution Class 12 magnetism and matter

NCERT Solutions

Class 12 Physics

Magnetism and Matter

NCERT Questions

Q.1
Q.2
Q.3
Q.4
Q.5
Q.6
Q.7
Q.8
Q.9
Q.10
Q.11
Q.12
Q.13
Q.14
Q.15

Q.7

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)?

Given:

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:

W = -MB (cos θ₂ – cos θ₁)

= - 1.5 x 0.22 (cos 90⁰ – cos 0⁰)

=-0.33(0 -1)

=0.33J

(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:

W = -MB (cos θ₂ – cos θ₁)

= - 1.5 x 0.22 (cos 180⁰ – cos 0⁰)

=-0.33(-1 -1)

=0.66J

(I) For case (i) : θ₂ = θ₁= 90⁰

Therefore torque τ = MB sin θ

= 1.5 x 0.22 sin 90⁰

=0.33 J

(II) For case (ii) : θ₂ = θ₁= 180⁰

Therefore torque τ = MB sin θ

= MB sin 180⁰

=0J

Video solution:

Magnetism and Matter Class 12 Physics NCERT Chapter 5 NCERT Solutions

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