Question

A bar magnet of moment of inertia $9\times {10}^{-5}kg{m}^2$ placed in a vibration magnetometer and oscillating in a uniform magnetic field $16{\pi }^2\times {10}^{-5}T$ makes $20$ oscillations in $15s$. The magnetic moment of the bar magnet is

• A. $3A-m^2$
• B. $2A-m^2$
• C. $5A-m^2$
• D. $6A-m^2$
• E. $4A-m^2$

Answer 1

The correct option is E $4A-m^2$

Time period is given by
$T=2\pi \sqrt{\frac{I}{MB}}$
$T^2=4{\pi }^2\times \frac{I}{MB}$
$M=4{\pi }^2\times \frac{I}{T^{2}B}$
Time period $T=\frac{15}{20}=0.75s$
$\therefore M=4{\pi }^2\times \frac{9\times {10}^{-5}}{16{\pi }^2\times {10}^{-5}\times (0.75{)}^2}$
$M=\frac{9}{4\times 0.5625}$
$M=\frac{9}{2.25}=4A-m^2$