First slide

Magnetic dipole in a uniform magnetic field

Question

A thin rectangular magnet suspended freely has a period of oscillation equal to T Now, it is broken into equal halves (each having half of original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T', then the ratio T'/T is

Moderate

A

1/4
B

$\frac{1}{(2 \sqrt{2})}$
C

1/2
D

2

Solution

The moment of inertia of original magnet
$I = \frac{{ml}^{2}}{12}$
When it is broken in two parts, each part has a mass half and length half. The new moment of inertia I', of each part becomes
$I^{'} = \frac{(m / 2) (l / 2)^{2}}{12} = \frac{{ml}^{2}}{12} \times \frac{1}{8} = \frac{I}{8}$
Also lf becomes half
We know that, $T = 2 π \sqrt{(\frac{I}{mH})}$
$Now T^{'} = 2 π \sqrt{[\frac{(I / 8)}{(M / 2) H}]} = 2 π \sqrt{[(\frac{I}{mH}) \times \frac{1}{4}]} T^{'} = T \sqrt{(\frac{1}{4})} = \frac{T}{2} or \frac{T^{'}}{T} = \frac{1}{2}$

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