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:
Answer & explanation
Correct answer: option 2
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: $IabB(\hat{j})$