Question

A freely suspended short bar magnet makes 20 oscillations per minute at a place where earths horizontal field is 40$\mu$ T. Another short bar magnet of moment 1.6 A-m${}^2$ is placed at 20 cm east pointing its north pole to north. The number of oscillations per minute are (nearly)

• A. 14
• B. 80
• C. 32
• D. 50

Answer 1

The correct option is A 14

The time period of oscillation is given by
$T=2\pi \sqrt{\frac{I}{MB}}$
The magnetic field due to the bar magnet on an equatorial line of a short magnet is given by
$B=\frac{{\mu }_o}{4\pi }\frac{M}{d^3}={10}^{-7}\times \frac{1.6}{(20\times {10}^{-2}{)}^3}$
$B=20\mu$ T
$\frac{f_2}{f_1}=\sqrt{\frac{B_2}{B_1}}$
$B_2=40\mu T-20\mu T=20\mu T$
$f_2=\sqrt{200}$ oscillations per minute
$f_2\approx 14$ oscillations per minute