Question:
A magnet makes 40 oscillations per minute at a place with \(B_H = 0.1\times10^{-5}T\). At another place, time period is 2.5 s. Find \(B_H\) there.
A magnet makes 40 oscillations per minute at a place with \(B_H = 0.1\times10^{-5}T\). At another place, time period is 2.5 s. Find \(B_H\) there.
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Time period of magnet \(\propto 1/\sqrt{B_H}\).
Updated On: Apr 23, 2026
- \(0.25\times10^{-6}T\)
- \(0.36\times10^{-6}T\)
- \(0.66\times10^{-8}T\)
- \(1.2\times10^{-6}T\)
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The Correct Option is B
Solution and Explanation
Concept:
\[
T = 2\pi\sqrt{\frac{I}{MB_H}} \Rightarrow T \propto \frac{1}{\sqrt{B_H}}
\]
Step 1: Initial time period
\[ 40 \text{ oscillations/min} = \frac{40}{60} = \frac{2}{3}\,Hz \Rightarrow T_1 = \frac{3}{2} = 1.5s \]
Step 2: Use relation
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \] \[ \frac{1.5}{2.5} = \sqrt{\frac{B_2}{0.1\times10^{-5}}} \] \[ \frac{3}{5} = \sqrt{\frac{B_2}{10^{-6}}} \] \[ \frac{9}{25} = \frac{B_2}{10^{-6}} \Rightarrow B_2 = \frac{9}{25}\times10^{-6} = 0.36\times10^{-6} \] Conclusion: \[ 0.36\times10^{-6}T \]
Step 1: Initial time period
\[ 40 \text{ oscillations/min} = \frac{40}{60} = \frac{2}{3}\,Hz \Rightarrow T_1 = \frac{3}{2} = 1.5s \]
Step 2: Use relation
\[ \frac{T_1}{T_2} = \sqrt{\frac{B_2}{B_1}} \] \[ \frac{1.5}{2.5} = \sqrt{\frac{B_2}{0.1\times10^{-5}}} \] \[ \frac{3}{5} = \sqrt{\frac{B_2}{10^{-6}}} \] \[ \frac{9}{25} = \frac{B_2}{10^{-6}} \Rightarrow B_2 = \frac{9}{25}\times10^{-6} = 0.36\times10^{-6} \] Conclusion: \[ 0.36\times10^{-6}T \]
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