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(B) The time period of a magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$, where $I$ is the moment of inertia and $M$

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The time period of a magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$, where $I$ is the moment of inertia and $M$ is the magnetic moment.This implies $T \propto \frac{1}{\sqrt{B}}$, or $T^2 B = 	ext{constant}$.At the first place, $B_1 = 0.1 	imes 10^{-5} \, T = 10^{-6} \, T$. The frequency is $40 \, 	ext{oscillations/min}$, so the time period $T_1 = \frac{60}{40} = 1.5 \, s$.At the second place, the time period $T_2 = 2.5 \, s$. Let the magnetic field be $B_2$.Using the relation $T_1^2 B_1 = T_2^2 B_2$, we get:$(1.5)^2 	imes 10^{-6} = (2.5)^2 	imes B_2$$2.25 	imes 10^{-6} = 6.25 	imes B_2$$B_2 = \frac{2.25}{6.25} 	imes 10^{-6} \, T$$B_2 = 0.36 	imes 10^{-6} \, T$.

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

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