Question

A magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magetic moment is quadrupled and the earth's field is doubled, the numbers of oscillations made per minute would be:

• A. $\frac{n}{2\sqrt{2}}$
• B. $\frac{n}{\sqrt{2}}$
• C. $2\sqrt{2}n$
• D. $\sqrt{2n}$

Answer 1

The correct option is C $2\sqrt{2}n$

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{MB}}$

$M=$ Magnetic moment

So, $n=\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$

$n\to$ Frequency

$M^{\prime }=4M$ and $B^{\prime }=2B$

$n^{\prime }=\frac{1}{2\pi }\sqrt{\frac{M^{\prime }{B}^{\prime }}{I}}=\frac{1}{2\pi }\sqrt{\frac{\left(4M\right)\left(2B\right)}{I}}$

$n^{\prime }=2\sqrt{2}\frac{1}{2\pi }\sqrt{\frac{MB}{I}}=2\sqrt{2}n$

So, number of oscillations per minute will increase to $2\sqrt{2}n$

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

Why this Question? Note: The time period of a bar magnet oscillating in an external magnetic field is given by, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{MB}}$