Question

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

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

Answer 1

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

$T_1=\frac{60}{40}=1.5$sec and $T_2=2.5$ sec

$T\propto \sqrt{\frac{1}{B_H}}$

$\therefore \frac{T_2}{T_1}=\sqrt{\frac{B_{H1}}{B_{H2}}}$

$\therefore B_{H2}=B_{H1}.{\left(\frac{T_1}{T_2}\right)}^2=0.1\times {10}^{-5}\times {\left(\frac{1.5}{2.5}\right)}^2=0.36\times {10}^{-6}T$