Question

A magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $B_H=0.1\times {10}^{-5}$. At another place, it takes $2.5amp$ to complete one vibration. The value of the earth's horizontal field at the place :

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

Answer 1

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

By using $T=2\pi \sqrt{\frac{1}{M{B}_H}}\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{{\left(B_H\right)}_2}{{\left(B_H\right)}_1}}$

$\Rightarrow \frac{60/40}{2.5}=\sqrt{\frac{{\left(B_H\right)}_2}{0.1\times {10}^{-5}}}$

${\left(B_H\right)}_2\Rightarrow 0.36\times {10}^{-6}T$