Question

The frequency of oscillation of a small magnet in a magnetic field of induction 'B' is '\(n\)'. If the frequency of oscillations of the same magnet in a field of induction 'X' falls to \((\frac{n}{3})\), the value of 'X' is

• A. \(3B\)
• B. \(9B\)
• C. \(\frac{B}{3}\)
• D. \(\frac{B}{9}\)

Hint:

Frequency of a magnet oscillating in a field varies as the square root of B.

Answer 1

The Correct Option is D

Step 1: Formula:
For a small magnet in a field, \(T=2\pi\sqrt{\dfrac I{MB}}\), so \(f=\dfrac1{2\pi}\sqrt{\dfrac{MB}I}\). Thus \(f\propto\sqrt B\).

Step 2: Ratio:
\[ \frac{f_X}{f_B}=\sqrt{\frac XB}=\frac{n/3}{n}=\frac13 \]

Step 3: Solve:
\(\dfrac XB=\dfrac19\Rightarrow X=\dfrac B9\).

Step 4: Check the Options:
\(3B\) and \(9B\) would raise the frequency. \(B/3\) forgets the square root. Only \(B/9\) makes the frequency fall to \(n/3\).

Final Answer:
The field is \(B/9\), option (D). \[ \boxed{\text{(D) } \frac{B}{9}} \]