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A bar magnet freely suspended in a vibration magnetometer has time period T at a place A and T/2 at a place B. If horizantal Component of earth's magnetic field at A is 36 × 10^–6 T, then its value at ‘B’ is

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$144 \times 10^{- 6} T$

$100 \times 10^{- 6} T$

$72 \times 10^{- 6} T$

None

answer is C.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{I}{{MB}_{H}}} \\ ∴ \frac{B_{2}}{B_{1}} = \frac{T_{1}^{2}}{T_{2}^{2}} \Rightarrow B_{2} = 144 \times 10^{- 6} T \end{array}$

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