Animated Solution for Physics - Magnetic Effects of Current: A circular coil having $N$ turns and radius $r$ carries a current $I$. It is held in the XZ-plane in a magnetic field $B\hat{i}$. The torque on the coil due to the magnetic field (in N-m) is

Imagine a circular coil placed flat in the $XZ$-plane. Let's establish our 3D coordinate system first, and then place the coil right there. We'll assume the current $I$ flows in the anti-clockwise direction when viewed from the positive $Y$-axis. The coil has a radius $r$ and consists of $N$ turns.

Any current-carrying coil acts like a tiny magnet. Its strength is given by its magnetic moment, $\vec{m}$. The formula is very simple: the number of turns, multiplied by the current, multiplied by the area of the coil.

And its direction? We use the right-hand thumb rule! Curl your fingers along the direction of the current, and your thumb points along the positive $Y$-axis. Since the area of a circle is $\pi r^2$, we can write the exact vector form for the magnetic moment:

Now, the question states that this coil is bathed in a uniform magnetic field. The field $\vec{B}$ is directed along the positive $X$-axis, which is represented by $\hat{i}$.

When a magnetic dipole is placed in an external magnetic field, it experiences a twisting force, a torque. The formula for this torque is the cross product of the magnetic moment vector and the magnetic field vector.

Let's plug in our vectors. We need to take the cross product of $NI\pi r^2\hat{j}$ with $B\hat{i}$.

Remember your cross products? $\hat{j}\times \hat{i}$ gives us $-\hat{k}$.

This means the torque acts along the negative $Z$-axis, trying to flip the coil! However, the question asks for the magnitude of this torque. We just drop the direction vector and the negative sign.

And that matches perfectly with option (b)!