First slide

Magnetic dipole in a uniform magnetic field

Question

A bar magnet A of magnetic moment M_A is found to oscillate at a frequency twice that of an identical magnet B of magnetic moment M_B when placed in a vibrating magnetometer. We may say that

Moderate

A

M_A = 2M_B
B

M_A = 8M_B
C

M_A = 4M_B
D

M_B = 8M_A

Solution

Frequency of bar magnet A is twice the frequency of bar magnet B. Thus if T be the time period of oscillation of B, then time period of 4 will be T/2. Hence
$\begin{array}{l} \frac{T}{2} = 2 π \sqrt{(\frac{I}{M_{A} B_{H}})} \\ and T = 2 π \sqrt{(\frac{I}{M_{B} B_{H}})} \end{array}$
The two expressions show that M_A = 4M_B.

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