Question

A short bar magnet having magnetic moment $4\, Am ^{2}$, placed in a vibrating magnetometer, vibrates with a time period of $8\, s$. Another short bar magnet having a magnetic moment $8\, Am ^{2}$ vibrates with a time period of $6\, s$. If the moment of inertia of the second magnet is $9 \times 10^{-2}\, kg - m ^{2}$, the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field.)

• A. $9 \times 10^{-2} kg - m ^{2}$
• B. $8 \times 10^{-2} kg - m ^{2}$
• C. $5.33 \times 10^{-2} kg - m ^{2}$
• D. $12.2 \times 10^{-2} kg - m ^{2}$

Answer 1

Answer: (A)

We know that the time period of a vibrating bar magnet
$T=2 \pi \sqrt{\frac{I}{M B_{H}}}$
Given, $T_{1}=8\, s ,\, I_{1}=I,\, M=4\, Am ^{2}$
$8=2 \pi \sqrt{\frac{1}{4 \times B_{H}}}$ ...(i)
Given, $T_{2}=6\, s ,\, I_{2}=9 \times 10^{-2}\, kg - m ^{2},\, M=8\, Am ^{2}$
$6=2 \pi \sqrt{\frac{9 \times 10^{-2}}{8 \times B_{H}}}$ ...(ii)
Dividing Eqs. (i) and (ii)
$\frac{8}{6}=\sqrt{\frac{2I}{9 \times 10^{-2}}}$
Squaring both sides and solving, we have
$I=8 \times 10^{-2} kg - m ^{2}$