Question

A uniform magnetic field of \( 2 \times 10^{-3} \, \text{T} \) acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is in the Y-Z plane. The current is in anticlockwise sense with reference to the negative X-axis. Magnitude and direction of the torque is:

• A. \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive Z-direction
• B. \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along negative Z-direction
• C. \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive X-direction
• D. \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive Y-direction

Answer 1

The Correct Option is B

The magnetic moment \( \vec{M} \) is given by:

\[ \vec{M} = I \vec{A}. \]

With \( I = 5 \, \text{A} \), \( A = 0.2 \times 0.1 = 0.02 \, \text{m}^2 \), and \( \vec{A} = 0.02 \, \hat{i} \), we get:

\[ \vec{M} = 5 \times 0.02 \, \hat{i} = 0.1 \, \hat{i}. \]

The torque \( \vec{\tau} \) is given by:

\[ \vec{\tau} = \vec{M} \times \vec{B}. \]

Substituting \( \vec{B} = 2 \times 10^{-3} \, \hat{j} \):

\[ \vec{\tau} = 0.1 \, \hat{i} \times 2 \times 10^{-3} \, \hat{j} = 2 \times 10^{-4} \, (-\hat{k}) = 2 \times 10^{-4} \, \text{Nm}. \]

Therefore, the answer is:

\[ 2 \times 10^{-4} \, \text{Nm} \text{ along the negative Z-direction.} \]