Question

A bar magnet of mass 120 g in the form of a rectangular parallelepiped, has dimensions \( l = 40 \, \text{mm}, b = 100 \, \text{mm} \) and \( h = 80 \, \text{mm} \), with its dimension ‘h’ vertical, the magnet performs angular oscillations in the plane of the magnetic field with period \( \pi \) seconds. If the magnetic moment is 3.4 Am\(^2\), determine the influencing magnetic field.

Hint:

The period of oscillation for a bar magnet in a magnetic field depends on its moment of inertia and the strength of the magnetic field. A stronger magnetic field reduces the period.

Answer 1

Step 1: Formula for the period of oscillation.
For a bar magnet performing angular oscillations in a magnetic field, the period \( T \) is related to the magnetic moment \( M \), the mass \( m \), and the magnetic field \( B \) by the formula: \[ T = 2 \pi \sqrt{\frac{I}{M B}} \] where \( I \) is the moment of inertia of the bar magnet.
Step 2: Moment of inertia.
The moment of inertia \( I \) of the bar magnet about its center of mass is given by: \[ I = \frac{1}{12} m (l^2 + b^2) \] Substitute the given values: \( m = 120 \, \text{g} = 0.12 \, \text{kg} \), \( l = 40 \, \text{mm} = 0.04 \, \text{m} \), \( b = 100 \, \text{mm} = 0.1 \, \text{m} \). \[ I = \frac{1}{12} \times 0.12 \times (0.04^2 + 0.1^2) \] \[ I = \frac{1}{12} \times 0.12 \times (0.0016 + 0.01) = \frac{1}{12} \times 0.12 \times 0.0116 = 1.16 \times 10^{-4} \, \text{kg} \cdot \text{m}^2 \]
Step 3: Solve for the magnetic field \( B \).
Given that the period \( T = \pi \), we can substitute into the period formula: \[ \pi = 2 \pi \sqrt{\frac{1.16 \times 10^{-4}}{3.4 \cdot B}} \] Squaring both sides and solving for \( B \): \[ \pi^2 = 4 \pi^2 \times \frac{1.16 \times 10^{-4}}{3.4 B} \] \[ B = \frac{4 \pi^2 \times 1.16 \times 10^{-4}}{3.4 \pi^2} \] \[ B = \frac{1.16 \times 10^{-4}}{3.4} = 5.2 \, \text{T} \]
Step 4: Conclusion.
The influencing magnetic field is \( 5.2 \, \text{T} \).