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

Concept: The potential energy ($ U $) of a magnetic dipole (current loop) in a uniform magnetic field $ B $ is given by the scalar product: $$ U = -\vec{M} \cdot \vec{B} = -MB \cos \theta $$ where $ M $ is the magnetic moment and $ \theta $ is the angle between the magnetic moment and the magnetic field.

Step 1: Calculating Initial Potential Energy. Initially, the angle $ \theta_1 = 0^\circ $. $$ U_{\text{initial}} = -MB \cos(0^\circ) $$ Since $ \cos(0^\circ) = 1 $: $$ U_{\text{initial}} = -MB $$ Substituting values: $ U_{\text{initial}} = -(0.5)(0.4) = -0.2 \, \text{J} $.

Step 2: Calculating Final Potential Energy. Finally, the loop is rotated to $ \theta_2 = 180^\circ $. $$ U_{\text{final}} = -MB \cos(180^\circ) $$ Since $ \cos(180^\circ) = -1 $: $$ U_{\text{final}} = -MB(-1) = +MB $$ Substituting values: $ U_{\text{final}} = +(0.5)(0.4) = +0.2 \, \text{J} $.

Step 3: Calculating the Change in Potential Energy. The change in potential energy ($ \Delta U $) is the difference between the final and initial states: $$ \Delta U = U_{\text{final}} - U_{\text{initial}} $$ $$ \Delta U = 0.2 \, \text{J} - (-0.2 \, \text{J}) $$ $$ \Delta U = 0.2 + 0.2 = 0.4 \, \text{J} $$