A bar magnet A of magnetic moment M_A is foun | vedclass

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Question

(C) The frequency of oscillation $
u$ of a bar magnet in a vibration magnetometer is given by the formula: $
u = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}

Vedclass · Accepted Answer

$M_A = 4M_B$

Vedclass · Answer

(C) The frequency of oscillation $
u$ of a bar magnet in a vibration magnetometer is given by the formula: $
u = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$.Assuming the moment of inertia $I$ is the same for both magnets,we have $
u \propto \sqrt{M}$.Given that the frequency of magnet $A$ is twice that of magnet $B$,we have $
u_A = 2
u_B$.Therefore,$\frac{
u_A}{
u_B} = \sqrt{\frac{M_A}{M_B}}$.Substituting the values: $2 = \sqrt{\frac{M_A}{M_B}}$.Squaring both sides: $4 = \frac{M_A}{M_B}$,which implies $M_A = 4M_B$.

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

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