Question

A short bar magnet of magnetic moment 2.5 Am\(^2\) is kept in a uniform magnetic field of \(4 \times 10^{-5}\) T. The work done in moving the magnet from its most stable position to most unstable position is

• A. \(40 \times 10^{-5}\) J
• B. \(25 \times 10^{-5}\) J
• C. \(10 \times 10^{-5}\) J
• D. \(20 \times 10^{-5}\) J

Hint:

Remember the key positions for a dipole in a field: - Most Stable: \(\theta=0^\circ\), aligned with the field, minimum potential energy (-MB). - Most Unstable: \(\theta=180^\circ\), anti-aligned with the field, maximum potential energy (+MB). - Zero Potential Energy (by convention): \(\theta=90^\circ\), perpendicular to the field.

Answer 1

The Correct Option is D

The potential energy (U) of a magnetic dipole with moment M in a uniform magnetic field B is given by \( U = -M \cdot B = -MB\cos\theta \), where \(\theta\) is the angle between the magnetic moment and the magnetic field.
The most stable position occurs when the potential energy is minimum. This happens when \( \cos\theta \) is maximum, i.e., \( \cos\theta = 1 \), which corresponds to \( \theta_1 = 0^\circ \). In this position, the magnetic moment is aligned with the field.
The potential energy in the most stable position is \( U_{stable} = -MB\cos(0^\circ) = -MB \).
The most unstable position occurs when the potential energy is maximum. This happens when \( \cos\theta \) is minimum, i.e., \( \cos\theta = -1 \), which corresponds to \( \theta_2 = 180^\circ \). In this position, the magnetic moment is anti-aligned with the field.
The potential energy in the most unstable position is \( U_{unstable} = -MB\cos(180^\circ) = -MB(-1) = +MB \).
The work done in moving the magnet from the stable to the unstable position is the change in its potential energy.
\( W = \Delta U = U_{unstable} - U_{stable} \).
\( W = (+MB) - (-MB) = 2MB \).
We are given the values:
Magnetic moment, \( M = 2.5 \) Am\(^2\).
Magnetic field, \( B = 4 \times 10^{-5} \) T.
Substitute these values to find the work done.
\( W = 2 \times (2.5) \times (4 \times 10^{-5}) \).
\( W = 5 \times 4 \times 10^{-5} = 20 \times 10^{-5} \) J.