Question

A magnetic needle of magnetic moment$6.7\times {10}^{-2}A{m}^2$ and moment of inertia$7.5\times {10}^{-6}kg{m}^2$ is performing simple harmonic oscillations in a magnetic field of$0.01T$. Time taken for $10$ complete oscillations is:

• A. $6.65s$
• B. $8.89s$
• C. $6.98s$
• D. $8.76s$

Answer 1

The correct option is A

$6.65s$

Step 1: Given data

Magnetic moment,$M$ $=6.7\times {10}^{-2}A{m}^2$

Moment of inertia $=7.5\times {10}^{-6}kg{m}^2$

Magnetic field, $B$ $=0.01T$

Angular frequency $=\omega$

Moment of inertia $=I$

Step 2: To find the time taken for $10$ complete oscillations

The time period of oscillation

$t=2\pi \sqrt{\frac{I}{BM}}$

We know, Angular frequency

$$\begin{gathered} \omega =\sqrt{\frac{MB}{I}} \\ t=\frac{2\pi }{\omega } \\ =2n\pi \sqrt{\frac{I}{MB}}(for10oscillations,n=10) \\ =10\times 2\pi \sqrt{\frac{I}{MB}} \\ =10\times 2\times 3.14\times \sqrt{\frac{7.6\times {10}^{-6}}{6.7\times {10}^{-2}\times 0.01}} \\ t=6.65s \end{gathered}$$

Hence the time taken for complete $10$ oscillation $t=6.65s$.

Therefore the correct answer is Option A.