Question

A current loop of magnetic moment $ 0.5 \, \text{Am}^2 $ is placed in a magnetic field of $ 0.4 \, \text{T} $. If the loop is rotated from an angle of $ 0^\circ $ to $ 180^\circ $, what is the change in potential energy?}

• A. $ -0.4 \, \text{J} $
• B. $ 0.4 \, \text{J} $
• C. $ -0.2 \, \text{J} $
• D. $ 0.2 \, \text{J} $

Hint:

The work done in rotating a dipole from $ \theta_1 $ to $ \theta_2 $ is always $ \Delta U = MB(\cos \theta_1 - \cos \theta_2) $. For a $ 180^\circ $ rotation from the stable equilibrium ($ 0^\circ $), the work is always $ 2MB $.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Potential energy of a magnetic dipole in a field depends on its orientation. Work must be done to rotate it against the magnetic torque.
Key Formula or Approach:
\( U = -MB \cos \theta \).
\( \Delta U = U_{final} - U_{initial} \).
Step 2: Detailed Explanation:
Initial angle \( \theta_1 = 0^\circ \).
\( U_1 = -MB \cos 0^\circ = -MB = -(0.5)(0.4) = -0.2 \, \text{J} \).
Final angle \( \theta_2 = 180^\circ \).
\( U_2 = -MB \cos 180^\circ = -MB(-1) = +MB = +(0.5)(0.4) = +0.2 \, \text{J} \).
Change \( \Delta U = U_2 - U_1 = 0.2 - (-0.2) = 0.4 \, \text{J} \).
Step 3: Final Answer:
The change in potential energy is \( 0.4 \, \text{J} \).