A uniform magnetic field of 2×10−3 T acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is:
(A)2×10−4 N-m along positive Z-direction
(B)2×10−4 N-m along negative Z-direction✓ Correct
(C)2×10−4 N-m along positive X-direction
(D)2×10−4 N-m along positive Y-direction
Answer: (B)
Solution
The magnetic moment of the loop is M=iA. With the 5 A current anticlockwise as viewed from the negative x-axis, the right-hand rule places the area vector along the negative x-direction:
M=5×(0.2×0.1)(−i^)=0.1(−i^)A m2
The torque on the loop is
τ=M×B=0.1(−i^)×(2×10−3j^)=−2×10−4(i^×j^)=−2×10−4k^N m
The torque has magnitude 2×10−4 N·m, directed along the negative Z-axis.
JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option
