Q. A magnet makes 40 oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{- 5}T$. At another place, it takes $2.5\ sec$ to complete one vibration. The value of earth’s horizontal field at that place is+
A) $0.25 \times 10^{- 6}T$
B) $0.36 \times 10^{- 6}T$
C) $0.66 \times 10^{- 8}T$
D) $1.2 \times 10^{- 6}T$
Explanation
(b) $T = 2\pi\sqrt{\frac{I}{M \times B_{H}}} \Rightarrow T \propto \frac{1}{\sqrt{B_{H}}}$
$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{\left( B_{H} \right)_{2}}{\left( B_{H} \right)_{1}}} \Rightarrow \frac{60/40}{2.5} = \sqrt{\frac{\left( B_{H} \right)_{2}}{0.1 \times 10^{- 5}}}$
$\Rightarrow \left( B_{H} \right)_{2} = 0.36 \times 10^{- 6}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.