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(B) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the m

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$9 	imes 10^{-6} ~T$

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(B) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the horizontal component of the earth's magnetic field.Since the frequency of oscillation $f = \frac{1}{T}$,we have $f = \frac{1}{2\pi} \sqrt{\frac{MH}{I}}$.This implies $f \propto \sqrt{H}$,or $H \propto f^2$.Given that $f_A = 40 	ext{ oscillations/min}$ and $f_B = 20 	ext{ oscillations/min}$.We have $\frac{H_B}{H_A} = \left( \frac{f_B}{f_A} ight)^2$.Substituting the values: $\frac{H_B}{36 	imes 10^{-6}} = \left( \frac{20}{40} ight)^2 = \left( \frac{1}{2} ight)^2 = \frac{1}{4}$.Therefore,$H_B = \frac{36 	imes 10^{-6}}{4} = 9 	imes 10^{-6} ~T$.

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} ~T$,then its value at $B$ is

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