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(A) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_{H}}}$, where $I$ is the moment of inertia, $M$ is the magneti

Vedclass · Accepted Answer

$H_{A}=8 	imes 10^{-6} \,T$

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(A) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_{H}}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B_{H}$ is the horizontal component of the earth's magnetic field.From the formula, we see that $T \propto \frac{1}{\sqrt{B_{H}}}$.Given that $T_{A} = 2T_{B}$, we can write the ratio as $\frac{T_{A}}{T_{B}} = 2$.Using the proportionality, $\frac{T_{A}}{T_{B}} = \sqrt{\frac{B_{HB}}{B_{HA}}}$.Substituting the values, $2 = \sqrt{\frac{32 	imes 10^{-6}}{B_{HA}}}$.Squaring both sides, $4 = \frac{32 	imes 10^{-6}}{B_{HA}}$.Therefore, $B_{HA} = \frac{32 	imes 10^{-6}}{4} = 8 	imes 10^{-6} \,T$.

$A$ vibration magnetometer is used at two different places $A$ and $B$ on the earth. The time period of a magnet suspended freely in the magnetometer at $A$ is twice that at $B$. If the horizontal component of the earth's magnetic field at $B$ is $32 \times 10^{-6} \,T$, then its value at $A$ is

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