Question

A magnet makes 40 oscillations per minute at a place having magnetic field intensity of 0.1 X ${10}^{-5}$ T. At another place, it takes 2.5 seconds to complete one vibration. The value of earth's horizontal field at that place is

• A. 0.25 X ${10}^{-6}$ T
• B. 0.36 X ${10}^{-6}$ T
• C. 0.66 X ${10}^{-8}$ T
• D. 1.2 X ${10}^{-6}$ T

Answer 1

The correct option is B 0.36 X ${10}^{-6}$ T

period of oscillation $T=2\pi \sqrt{\frac{I}{m{B}_H}}$

from this $T\alpha \frac{1}{\sqrt{B_H}}$

$\frac{T_1}{T_2}=\sqrt{\frac{{B_H}_2}{{B_H}_1}}$

$T_1=\frac{60}{40}$ , $T_2=2.5$ & ${B_H}_1=0.1\times 10-5$

$\therefore \frac{\frac{60}{40}}{2.5}=\sqrt{\frac{{B_H}_2}{0.1\times {10}^{-5}}}$

$=0.36\times {10}^{-6}$T