Question

2. (a) A magnetic dipole is oscillating in a mag-
netic field obeying the following expression.

What is the time period of oscillation and
mention the nature of oscillation?

​

Answer 1

Answer:

The time period of oscillation is

T =2\pi\sqrt{ \frac{I}{MB}}T=2π

MB

I

And this is angular SHM

Explanation:

As we know that the torque on magnetic dipole due to external uniform magnetic field is given as

\tau = M\times Bτ=M×B

\tau = MB sin\thetaτ=MBsinθ

If the dipole is displaced by small angle then we have

I\alpha = - MB\thetaIα=−MBθ

\alpha = - \frac{MB}{I}\thetaα=−

I

MB

θ

so it is an angular SHM

here angular frequency is given as

\omega = \sqrt{\frac{MB}{I}}ω=

I

MB

so the time period of the motion is given as

T =2\pi\sqrt{ \frac{I}{MB}}T=2π

MB

I

Answer 2

Answer:

here is your answer

Explanation:

τ=

M

×

B

=MBsinθ

For small angle θ

τ=MBθ

α=(

I

MB

)θ

Comparing with angular SHM

w

2

=

I

mB

T=

w

2π

=2π

mB

I

where,

m= magnetic moment

I=moment of inertia

B= magnetic field intensity