A vibration magnetometer is used at two different places $\mathrm{A}$ and $\mathrm{B}$ on the earth. The…

A vibration magnetometer is used at two different places $\mathrm{A}$ and $\mathrm{B}$ on the earth. The time period of a magnet suspended freely in the magnetometer at $\mathrm{A}$ is twice that at $\mathrm{B}$. If the horizontal component of the earth's magnetic field at $\mathrm{B}$ is $32 \times 10^{-6} \mathrm{~T}$, then its value at $\mathrm{A}$ is
  1. $\mathrm{H}_{\mathrm{A}}=8 \times 10^{-6} \mathrm{~T}$
  2. $\mathrm{H}_{\mathrm{A}}=32 \times 10^{-6} \mathrm{~T}$
  3. $\mathrm{H}_{\mathrm{A}}^{\mathrm{A}}=4 \times 10^{-6} \mathrm{~T}$
  4. $\mathrm{H}_{\mathrm{A}}=16 \times 10^{-6} \mathrm{~T}$

Solution

$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}} \\ & \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}\end{aligned}$ $\begin{aligned} & \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \sqrt{\frac{\mathrm{B}_{\mathrm{HB}}}{\mathrm{B}_{\mathrm{HA}}}} \Rightarrow \mathrm{B}_{\mathrm{HA}}=\left(\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}\right)^2 \mathrm{~B}_{\mathrm{HB}} \\ & =\left(\frac{1}{2}\right)^2 \times 32 \times 10^{-6} \\ & =8 \times 10^{-6} \mathrm{~T}\end{aligned}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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