1. Given data

• Magnetic field: $\vec B = 2\times 10^{-3}\,\hat j\,\text{T}$
• Rectangular loop sides: $20\text{ cm} = 0.20\text{ m}$ and $10\text{ cm} = 0.10\text{ m}$
• Current: $I = 5\text{ A}$
• Plane of loop: $YZ$-plane

So, area of the loop is $A = 0.20\times 0.10 = 0.02\,\text{m}^2$

2. Magnetic moment of the loop

Magnitude of magnetic moment: $\mu = IA = 5\times 0.02 = 0.1\,\text{A m}^2$

Direction of $\vec \mu$ is given by right-hand rule.

The current is anticlockwise when viewed from the negative $X$-axis. That means, as seen from the $-X$ side, the magnetic moment points towards the observer, i.e. along negative $X$-direction.

Hence, $\vec \mu = -0.1\,\hat i\,\text{A m}^2$

3. Torque on a current loop

Torque is given by $\vec \tau = \vec \mu \times \vec B$

Substitute: $\vec \tau = (-0.1\hat i) \times (2\times 10^{-3}\hat j)$

$\vec \tau = -0.1\times 2\times 10^{-3}(\hat i\times \hat j)$

Since $\hat i \times \hat j = \hat k$ we get $\vec \tau = -2\times 10^{-4}\hat k\,\text{N m}$

Thus,

• Magnitude of torque $= 2\times 10^{-4}\,\text{N m}$
• Direction $= -\hat k$ = negative $Z$-direction

4. Matching with options

This corresponds to:

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

5. Comparison with stored answer

Stored correct answer: A

This matches the derived answer.