Question

prove that under certain condition a magnet vibrating in uniform magnetic field perform angular SHM

Answer 1

Explanation:

$\tau=m \times B$

The magnitude of this torque is given by sine. Here is the restoring torque, and $\Theta$ is the angle between the direction of the magnetic moment (m) and the direction of the magnetic field (B).

At equilibrium, we can say that,

$I \frac{d^{2} \theta}{d t} \cong-m B \sin \theta$

The negative sign in the above expression sine brings us to the conclusion that the restoring torque acting here acts in the opposite direction to the deflecting torque. Also, as the value of is very small in radians, we can approximate $\sin \theta=\theta .$ Thus, using this approximation, we can write

$I \frac{d^{2} \theta}{d t} \cong-m B \theta$

$\mathrm{Or}$

$\frac{d^{2} \theta}{d t} \cong \frac{-m B \theta}{I}$

The above equation is a representation of a simple harmonic motion and the angular frequency can be given as,

$\omega=\frac{m B}{I}$

and thus, the time period can be stated as,

$T=2 \pi \sqrt{\frac{I}{m B}}$

Answer 2

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