Question

A magnet makes 40 oscillations per minute at a place having magnetic field intensity of $1\times {10}^{-6}T$. At another place, it takes 3 seconds to complete one vibration. The value of the earth's horizontal field at that place is (Neglect gravitational effect)

• A. $0.25\times {10}^{-6}T$
• B. $0.36\times {10}^{-6}T$
• C. $0.50\times {10}^{-6}T$
• D. $1.2\times {10}^{-6}T$

Answer 1

The correct option is A $0.25\times {10}^{-6}T$

Time period of magnet having 40 oscillations per minute $=\frac{60}{40}$ seconds.

Time period of a magnet having magnetic dipole moment $M$ kept in magnetic field $B$ is given by
$Timeperiod=2\pi \sqrt{\frac{I}{MB}}\Rightarrow Timeperiod\propto \frac{I}{\sqrt{B}}\Rightarrow \frac{Timeperio{d}_1}{Timeperio{d}_2}=\sqrt{\frac{B_2}{B_1}}\Rightarrow \frac{\frac{60}{40}}{3}=\sqrt{\frac{B_2}{1\times {10}^{-6}}}$
$\therefore B_2=0.25\times {10}^{-6}T$