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

Magnetic field $$B = 2 \times 10^{-3}$$ T along +Y. Rectangular loop 20 cm × 10 cm in Y-Z plane, current 5 A anticlockwise w.r.t. negative X-axis.

The area vector of the loop in Y-Z plane points along $$\pm\hat{x}$$. Since current is anticlockwise when viewed from negative X-axis, by the right-hand rule, the area vector points along $$-\hat{x}$$.

$$\vec{m} = I\vec{A} = 5 \times (20 \times 10 \times 10^{-4})(-\hat{x}) = 5 \times 0.02(-\hat{x}) = -0.1\hat{x}$$ A m$$^2$$.

Torque: $$\vec{\tau} = \vec{m} \times \vec{B} = (-0.1\hat{x}) \times (2 \times 10^{-3}\hat{y})$$

$$= -0.1 \times 2 \times 10^{-3}(\hat{x} \times \hat{y}) = -2 \times 10^{-4}\hat{z}$$

Magnitude: $$2 \times 10^{-4}$$ N m, direction: along negative Z-direction.

The correct answer is Option B: $$2 \times 10^{-4}$$ N m along negative Z-direction.