Q151 / 160 · AP EAMCET 2022 (06 Jul Shift 1) · Physics
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
- Correct answerA$\mathrm{H}_{\mathrm{A}}=8 \times 10^{-6} \mathrm{~T}$
- B$\mathrm{H}_{\mathrm{A}}=32 \times 10^{-6} \mathrm{~T}$
- C$\mathrm{H}_{\mathrm{A}}^{\mathrm{A}}=4 \times 10^{-6} \mathrm{~T}$
- D$\mathrm{H}_{\mathrm{A}}=16 \times 10^{-6} \mathrm{~T}$
Solution
$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}_{\mathrm{H}}}} \\ & \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{B}_{\mathrm{H}}}}\end{aligned}$
$\begin{aligned} & \mathrm{T}_{\mathrm{A}}=\mathrm{T}_{\mathrm{B}} \sqrt{\frac{\mathrm{B}_{\mathrm{HB}}}{\mathrm{B}_{\mathrm{HA}}}} \Rightarrow \mathrm{B}_{\mathrm{HA}}=\left(\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}\right)^2 \mathrm{~B}_{\mathrm{HB}} \\ & =\left(\frac{1}{2}\right)^2 \times 32 \times 10^{-6} \\ & =8 \times 10^{-6} \mathrm{~T}\end{aligned}$
Hence, the correct answer is option: 'A'.