A uniform magnetic field $\vec{B}$ is established along the positive z-direction. A rectangular loop of sides 'a' and 'b' carries a current of I as shown in figure. The torque in the loop is:

IabB \((\hat{j})\)

The correct answer is Option (2) → IabB \((\hat{j})\)

The loop lies in the $YZ$-plane.

Current direction is clockwise when viewed from the positive $X$-axis, so by the right-hand rule, the magnetic moment is along negative $X$-direction:

$\vec{m} = Iab(-\hat{i})$

Magnetic field is along positive $Z$-direction:

$\vec{B} = B\hat{k}$

Torque on the loop:

$\vec{\tau} = \vec{m} \times \vec{B}$

Using cross product:

$\vec{\tau} = (-Iab\hat{i}) \times (B\hat{k})$

$= -IabB(\hat{i} \times \hat{k})$

Since

$\hat{i} \times \hat{k} = -\hat{j}$

therefore,

$\vec{\tau} = -IabB(-\hat{j})$

$\vec{\tau} = IabB\hat{j}$

So the correct answer is:  $IabB(\hat{j})$