In a certain region, the magnetic field is specified by B = B_0 (1...

Consider the square loop placed in the $xy$-plane with its vertices at $(x, y)$, $(x+l, y)$, $(x+l, y+l)$, and $(x, y+l)$.

The magnetic force on a current element is $d\vec{F} = i(d\vec{l} \times \vec{B})$.

For the two edges parallel to the $x$-axis, the current flows in opposite directions. Since the magnetic field $\vec{B} = B_0 \left(1 + \dfrac{x}{l}\right) \hat{k}$ depends only on $x$, the forces on these two edges are equal in magnitude and opposite in direction, cancelling each other out.

For the edge at $x + l$ parallel to the $y$-axis, the force is:

$\vec{F}_1 = i(l\hat{j}) \times \left[B_0\left(1 + \dfrac{x+l}{l}\right)\hat{k}\right] = i l B_0\left(1 + \dfrac{x+l}{l}\right)\hat{i}$

For the edge at $x$ parallel to the $y$-axis, the current flows in the opposite direction, so the force is:

$\vec{F}_2 = i(-l\hat{j}) \times \left[B_0\left(1 + \dfrac{x}{l}\right)\hat{k}\right] = -i l B_0\left(1 + \dfrac{x}{l}\right)\hat{i}$

The net magnetic force on the loop is the sum of the forces on all edges:

$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = i l B_0 \left(1 + \dfrac{x+l}{l} - 1 - \dfrac{x}{l}\right)\hat{i}$

$\vec{F}_{net} = i l B_0 \left(\dfrac{l}{l}\right)\hat{i} = i B_0 l \hat{i}$

The magnitude of the net magnetic force experienced by the loop is $i B_0 l$.

Answer: $i B_0 l$