Question

The magnetic moment of a bar magnet is \( 0.5 \, \text{Am}^2 \). It is suspended in a uniform magnetic field of \( 8 \times 10^{-2} \, \text{T} \). The work done in rotating it from its most stable to most unstable position is:

• A. \( 16 \times 10^{-2} \, \text{J} \)
• B. \( 8 \times 10^{-2} \, \text{J} \)
• C. \( 4 \times 10^{-2} \, \text{J} \)
• D. \( \text{Zero} \)

Answer 1

The Correct Option is B

To solve this problem, we need to calculate the work done in rotating a bar magnet from its most stable position (parallel to the magnetic field) to its most unstable position (antiparallel to the magnetic field). The formula for the work done \( W \) in rotating a magnetic dipole in a magnetic field is given by:

\[ W = \text{PE}_{\text{unstable}} - \text{PE}_{\text{stable}} \]

where PE (Potential Energy) is calculated using:

\[ \text{PE} = -MB \cos \theta \]

Here, \( M \) is the magnetic moment, \( B \) is the magnetic field, and \( \theta \) is the angle with the magnetic field.

1. **Most Stable Position**: The magnetic dipole is aligned with the magnetic field. Therefore, \( \theta = 0^\circ \).

\[ \text{PE}_{\text{stable}} = -MB \cos 0^\circ = -MB \]

2. **Most Unstable Position**: The magnetic dipole is antiparallel to the magnetic field. Thus, \( \theta = 180^\circ \).

\[ \text{PE}_{\text{unstable}} = -MB \cos 180^\circ = MB \]

Substituting these into the formula for work done:

\[ W = MB - (-MB) = 2MB \]

3. **Substitute Given Values**: \( M = 0.5 \, \text{Am}^2 \) and \( B = 8 \times 10^{-2} \, \text{T} \).

\[ W = 2 \times 0.5 \times 8 \times 10^{-2} = 8 \times 10^{-2} \, \text{J} \]

Therefore, the work done in rotating the magnet is \(\mathbf{8 \times 10^{-2} \, \text{J}}\).

The correct option is: \( \mathbf{8 \times 10^{-2} \, \text{J}} \).

Answer 2

Magnetic Potential Energy:
The potential energy U of a magnetic moment m in a uniform magnetic field B is given by:
U = -mB cos θ.

Stable and Unstable Equilibrium Positions:
1. At stable equilibrium (θ = 0°):
U_stable = -mB cos 0° = -mB.

2. At unstable equilibrium (θ = 180°):
U_unstable = -mB cos 180° = +mB.

Work Done:
The work done in rotating the magnet from the stable to the unstable position is the change in potential energy:
W = ΔU = U_unstable - U_stable = mB - (-mB) = 2mB.

Substitute Values:
Given:
m = 0.5 Am², B = 8 × 10^-2 T,
W = 2 × 0.5 × 8 × 10^-2 = 8 × 10^-2 J.

Answer: 8 × 10^-2 J