First slide

Magnetic dipole in a uniform magnetic field

Question

A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 2^5/4 second. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in second is

Difficult

A

2^1/4
B

2^1/2
C

2
D

2^3/4

Solution

The arrangement of magnets is shown in fig. (6).It is clear from figure that distance between free north and south pole is $\sqrt{2}$ l. Therefore, the dipole moment of the combination = $\sqrt{2}$ M.
The total moment of inertia of the combination = I + I = 2I
The time period of vibration magnetometer in this combination is given by
$T^{'} = 2 π \sqrt{(\frac{I^{'}}{M^{'} B_{H}})} or T^{'} \propto \sqrt{(\frac{I^{'}}{M^{'}})} ∴ T^{'} \propto \sqrt{\frac{2 I}{(\sqrt{2} M)}} . . . . . . (1)$
when one magnet is removed, then
$T \propto \sqrt{(\frac{I}{M})} . . . . . . (2)$
From eqs. (1) and (2), we get
$\frac{T}{T^{'}} = \frac{1}{2^{1 / 4}} or T = \frac{T^{'}}{2^{1 / 4}} ∴ T = \frac{2^{5 / 4}}{2^{1 / 4}} = 2 \sec .$

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