Question

A magnetic dipole of magnetic moment $M$ is suspended by a string in a uniform horizontal magnetic field as shown in figure. If in horizontal plane this dipole is slightly tilted and released, it will execute simple harmonic motion, find its period of oscillation. Consider the dipole as a uniform rod of mass $m$ and length $l$.

Given,
$M\to$ Dipole moment

• A. $T=2\pi \sqrt{\frac{m{l}^2}{MB}}$
• B. $T=2\pi \sqrt{\frac{2m{l}^2}{7MB}}$
• C. $T=2\pi \sqrt{\frac{m{l}^2}{12MB}}$
• D. $T=2\pi \sqrt{\frac{m{l}^2}{2MB}}$

Answer 1

The correct option is C $T=2\pi \sqrt{\frac{m{l}^2}{12MB}}$

If the dipole is tilted from its equilibrium position by a small angle $\theta$, in horizontal plane, restoring torque on magnetic dipole is given by,

$\tau =MB\sin \theta \simeq MB\theta (\because \theta$ is very small)

The angular acceleration of dipole is given by,

$\alpha =-\frac{\tau }{I}$

$\Rightarrow \alpha =-\frac{MB}{\left(m{l}^2/12\right)}\theta$

$\Rightarrow \alpha =-\frac{12MB}{\left(m{l}^2\right)}\theta$

The above equation states that resoring angular acceleration of the dipole is directly proportional to the angular displacement. Hence, it will execute SHM. Comparing this with standard angular SHM,

$\alpha =-{\omega }^2\theta$, we get

$\omega =\sqrt{\frac{12MB}{m{l}^2}}$

$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m{l}^2}{12MB}}$

Hence, option $\left(C\right)$ is the correct answer.