Q147 / 160 · AP EAMCET 2007 · Physics

A bar-magnet of moment of inertia $49 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \mathrm{~T}$. The time period of vibration is $8.8 \mathrm{~s}$. The magnetic moment of the bar magnet is

  • A
    $350 \mathrm{~A}-\mathrm{m}^2$
  • B
    $490 \mathrm{~A}-\mathrm{m}^2$
  • C
    $3300 \mathrm{~A}-\mathrm{m}^2$
  • Correct answer
    D
    5000 A-m ${ }^2$

Solution

Time period of magnet is $\begin{aligned} \quad T & =2 \pi \sqrt{\frac{I}{M H}} \\ \text { or } \quad M & =\frac{4 \pi^2 I}{T^2 H} \\ \text { or } \quad M & =\frac{4 \times(3.14)^2 \times 49 \times 10^{-2}}{(8.8)^2 \times 0.5 \times 10^{-4}} \\ \text { or } \quad M & =5000 \mathrm{~A}-\mathrm{mf}^2\end{aligned}$

Hence, the correct answer is option: 'D'.

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