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Physics Question on torque

A uniform magnetic field of 2×10−3 T2 \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

2×10−4 N m2 \times 10^{-4} \, \text{N} \, \text{m} along positive Z-direction

B

2×10−4 N m2 \times 10^{-4} \, \text{N} \, \text{m} along negative Z-direction

C

2×10−4 N m2 \times 10^{-4} \, \text{N} \, \text{m} along positive X-direction

D

2×10−4 N m2 \times 10^{-4} \, \text{N} \, \text{m} along positive Y-direction

Answer

2×10−4 N m2 \times 10^{-4} \, \text{N} \, \text{m} along negative Z-direction

Explanation

Solution

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

M⃗=IA⃗.\vec{M} = I \vec{A}.

With I=5 AI = 5 \, \text{A}, A=0.2×0.1=0.02 m2A = 0.2 \times 0.1 = 0.02 \, \text{m}^2, and A⃗=0.02 i^\vec{A} = 0.02 \, \hat{i}, we get:

M⃗=5×0.02 i^=0.1 i^.\vec{M} = 5 \times 0.02 \, \hat{i} = 0.1 \, \hat{i}.

The torque τ⃗\vec{\tau} is given by:

τ⃗=M⃗×B⃗.\vec{\tau} = \vec{M} \times \vec{B}.

Substituting B⃗=2×10−3 j^\vec{B} = 2 \times 10^{-3} \, \hat{j}:

τ⃗=0.1 i^×2×10−3 j^=2×10−4 (−k^)=2×10−4 Nm.\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:

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