Question

12) Derive an expression for angular velocity of magnetic needle oscillating in the magnetic field.​

Answer 1

Explanation:

(a)This is done by placing a small compass needle of known magnetic moment m and moment of inertia I and allowing it to oscillate in the magnetic field.

The torque on the needle is,

τ=m×B

In magnitude, τ=mBsinθ

Here, τ= Restoring torque, θ= angle between m and B

Therefore, in equilibrium,

I

dt

2

d

2

θ

=−mBsinθ

Negative sign indicates that the restoring torque is in opposition to deflecting torque.

This represents a simple harmonic motion.

The square of angular frequency is ω

2

=mB/I and the time period is

τ=2π(I/mB)

−1/2

(b) Since the compass needle is oriented vertically,

(i) Horizontal component of earth's magnetic field will be zero.

(ii)Angle of dip will be 90 .........