A uniform magnetic field of $2\times10^{-3}\ \text{T}$ acts along the positive $Y$-direction. A rectangular loop of sides $20\ \text{cm}$ and $10\ \text{cm}$ with a current of $5\ \text{A}$ is in the $Y$-$Z$ plane. The current is in the anticlockwise sense with reference to the negative $X$ axis. The magnitude and direction of the torque is:
Answer: (B) $2\times10^{-4}\ \text{N m}$ along negative $Z$-direction
Anticlockwise as seen with reference to the $-X$ axis means the magnetic moment points along $-\hat i$:
$$\vec m = 5\times(0.2\times0.1)(-\hat i) = -0.1\,\hat i\ \text{A m}^2$$
$$\vec\tau = \vec m\times\vec B = (-0.1\,\hat i)\times(2\times10^{-3}\,\hat j) = -2\times10^{-4}\,\hat k\ \text{N m}$$
Magnitude $2\times10^{-4}\ \text{N m}$ along the negative $Z$-direction.
Solution by Sreeraj P, M.Sc Physics