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

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A

$0.25\times 10^{– 6} \, T$

B

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

C

$0.66\times 10^{- 8} \, T$

D

$1.2\times 10^{– 6} \, T$

Solution:

$T=2\pi \sqrt{\frac{I}{M \times B_{H}}}\Rightarrow T \propto \frac{1}{\sqrt{B_{H}}}$
$\Rightarrow \frac{T_{1}}{T_{2}}=\sqrt{\frac{\left(B_{H}\right)_{2}}{\left(B_{H}\right)_{1}}} \Rightarrow \frac{60 / 40}> {2.5}=\sqrt{\frac{\left(B_{H}\right)_{2}}{0.1 \times 10^{-5}}}$
$\Rightarrow\left(B_{H}\right)_{2}=0.36 \times 10^{-6} T$

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