1. Given data

• Square loop side: $a=2\,\text{m}$
• Current: $I=2\,\text{A}$
• Magnetic field: $\vec B = B_0(1+4x)\hat k$
• $B_0=5\,\text{T}$

So, $\vec B = 5(1+4x)\hat k$

The loop lies in the $x$-$y$ plane with edges parallel to the axes.

2. Magnetic force on a current element

For a wire segment, $d\vec F = I\,d\vec l \times \vec B$

Since $\vec B$ is along $\hat k$, forces on opposite sides may cancel partially. Because the field depends on $x$, the vertical sides (parallel to $y$-axis) will experience unequal forces.

3. Forces on horizontal sides

The top and bottom sides are parallel to the $x$-axis.

For these sides, $d\vec l \parallel \hat i$ or $-\hat i$, and $\vec B \parallel \hat k$. Thus forces are along $\mp \hat j$.

But for every value of $x$, the field is the same on top and bottom segments, and currents are opposite. Hence their forces cancel exactly.

So, net force from horizontal sides is: $F_{\text{horizontal}}=0$

4. Forces on vertical sides

Let the left side be at $x=x_1$ and the right side at $x=x_2$, with $x_2-x_1=2$

Force magnitude on a vertical side of length $a$ is: $F=I a B(x)$

because $B$ is constant along that side (depends only on $x$).

So, $F_1 = I a B(x_1), \qquad F_2 = I a B(x_2)$

These two forces are opposite in direction, so net force magnitude is $F_{\text{net}} = I a\,[B(x_2)-B(x_1)]$

Now, $B(x)=B_0(1+4x)$

Hence, $B(x_2)-B(x_1)=B_0[ (1+4x_2)-(1+4x_1)] = 4B_0(x_2-x_1)$

Using $x_2-x_1=2$, $B(x_2)-B(x_1)=4B_0\cdot 2 = 8B_0$

Therefore, $F_{\text{net}} = I a (8B_0)$

Substitute values: $F_{\text{net}} = 2\times 2\times 8\times 5 = 160\,\text{N}$

5. Final answer

The net magnetic force experienced by the loop is $\boxed{160\,\text{N}}$