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Q.

With a standard rectangular bar magnet of length (l) breadth (b < < l) and magnetic moment M, the time period of magnet in a vibration magnetometer is 4 s. If the magnet is cut normal to its length into 4 equal pieces, the time period (in seconds) with one of the pieces is

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a

2

b

16

c

1

d

$\frac{1}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{I}{{MB}_{H}}}; \\ I_{1} = \frac{m}{12} (l^{2} + b^{2}) = \frac{m l^{2}}{12}; I_{2} = \frac{m / 4}{12} (\frac{l^{2}}{16} + b^{2}) = \frac{I_{1}}{64}; Also, M_{2} = \frac{M_{1}}{4} \\ \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1}}{I_{2}} \times \frac{M_{2}}{M_{1}}} \Rightarrow T_{2} = 1 \sec \end{array}$

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With a standard rectangular bar magnet of length (l) breadth (b < < l) and magnetic moment M, the time period of magnet in a vibration magnetometer is 4 s. If the magnet is cut normal to its length into 4 equal pieces, the time period (in seconds) with one of the pieces is