First slide

Magnetic dipole in a uniform magnetic field

Question

The time period of a freely suspended magnet is 4 sec. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period in seconds will be

Moderate

A

4
B

2
C

0.5
D

0.25

Solution

$T = 2 π \sqrt{(\frac{I}{{MB}_{H}})}$
After cutting / reduces to l/2. Hence
M' = m(l/2) = (M/2)
$and I = \frac{M}{12} [l^{2} + b^{2}] ≃ \frac{{ml}^{2}}{12}$ (thin bar)
$\begin{array}{l} I^{'} = \frac{1}{12} (\frac{M}{2}) (\frac{l^{2}}{4}) = \frac{I}{8} \\ ∴ T^{'} = 2 π \sqrt{(\frac{(I / 8)}{(M / 2) B_{H}})} \\ = \frac{T}{2} = \frac{4}{2} = 2 \sec . \end{array}$

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