Question

A bar magnet of magnetic moment M = 0.5A m2 is under the influence of a magnetic field 8 T. Find the work done (J) to move the magnet from stable to unstable equilibrium position.

Answer 1

Correct Answer: 8

To find the work done in moving a bar magnet from a stable to an unstable equilibrium position in a magnetic field, we need to understand the energy change involved. The potential energy \(U\) of a magnetic dipole in a magnetic field is given by:

\[ U = -M \cdot B \cdot \cos(\theta) \]

where:

• \(M\) is the magnetic moment of the bar magnet (0.5 A·m²).
• \(B\) is the magnetic field strength (8 T).
• \(\theta\) is the angle between the magnetic moment and the magnetic field direction.

In a stable equilibrium, \(\theta = 0^\circ\) (cos(0) = 1), so the initial potential energy \(U_{\text{stable}}\) is:

\[ U_{\text{stable}} = -M \cdot B \cdot \cos(0^\circ) = -0.5 \cdot 8 \cdot 1 = -4 \, \text{J} \]

In an unstable equilibrium, \(\theta = 180^\circ\) (cos(180) = -1), so the potential energy \(U_{\text{unstable}}\) is:

\[ U_{\text{unstable}} = -M \cdot B \cdot \cos(180^\circ) = -0.5 \cdot 8 \cdot (-1) = 4 \, \text{J} \]

The work done \(W\) in moving the magnet from stable to unstable equilibrium is the change in potential energy:

\[ W = U_{\text{unstable}} - U_{\text{stable}} = 4 - (-4) = 8 \, \text{J} \]

Thus, the work done is 8 J, which falls within the expected range of 8,8. The solution is correct and verified.