A rectangular wire-loop of width a is suspended from the insulated...

Let $W$ be the weight of the rectangular wire-loop.

The magnetic field $B$ is present only in the lower region and is directed into the plane of the paper. The magnetic forces on the vertical segments of the loop are equal and opposite in the horizontal direction, cancelling each other out.

The net magnetic force acts on the bottom horizontal segment of width $a$.

When the current $i$ flows in the anticlockwise direction, the current in the bottom segment is directed towards the right. Using the magnetic force formula $\vec{F} = i(\vec{l} \times \vec{B})$, the force is directed upwards.

The magnitude of this upward magnetic force is $F_B = i a B$.

The initial tension in the spring balance is:

$T_1 = W - i a B$

When the current is reversed, it flows in the clockwise direction, and the current in the bottom segment is directed towards the left. The magnetic force is now directed downwards with the same magnitude.

The new tension in the spring balance becomes:

$T_2 = W + i a B$

The change in the tension of the spring is:

$\Delta T = T_2 - T_1$

$\Delta T = (W + i a B) - (W - i a B) = 2 i a B$