Question

The magnetic field existing in a region is given by $\vec{B} = 0.2(1 + 2x)\hat{k} \, \text{T}.$A square loop of edge $50 \, \text{cm}$ carrying $0.5 \, \text{A}$ current is placed in the $x$-$y$ plane with its edges parallel to the $x$- and $y$-axes, as shown in the figure. The magnitude of the net magnetic force experienced by the loop is ______ $\text{mN}$.

Answer 1

Magnetic Force on a Current-Carrying Wire in a Magnetic Field:
The magnetic force $F$ on a segment of current-carrying wire in a magnetic field is given by:
$F = ILB \sin \theta$ where:
$I$ is the current,
$L$ is the length of the segment,
$B$ is the magnetic field,
$\theta$ is the angle between the magnetic field and the current direction.

In this case, the loop lies in the $x-y$ plane with its edges parallel to the $x-$ and $y-$ axes. Since $\vec{B}$ varies with $x$, the magnetic force on each side of the loop depends on its position in the $x-y$ plane.

Calculate the Force on Each Side of the Loop:
For the left side at $x = 0$:
$B_{\text{left}} = 0.2(1 + 2 \times 0) = 0.2 \, \text{T}$
Force on the left side:
$F_{\text{left}} = ILB_{\text{left}} = 0.5 \times 0.5 \times 0.2 = 0.05 \, \text{N}$

For the right side at $x = 0.5 \, \text{m}$:
$B_{\text{right}} = 0.2(1 + 2 \times 0.5) = 0.2 \times 2 = 0.4 \, \text{T}$
Force on the right side:
$F_{\text{right}} = ILB_{\text{right}} = 0.5 \times 0.5 \times 0.4 = 0.1 \, \text{N}$

Net Force on the Loop:
The forces on the top and bottom sides (parallel to the $x$-axis) will cancel each other out due to symmetry, as the magnetic field along these sides is the same.
Therefore, the net force is due to the difference in the forces on the left and right sides:
$F_{\text{net}} = F_{\text{right}} - F_{\text{left}} = 0.1 - 0.05 = 0.05 \, \text{N}$

Convert to mN:
$F_{\text{net}} = 0.05 \, \text{N} = 50 \, \text{mN}$

Conclusion:
The magnitude of the net magnetic force experienced by the loop is $50 \, \text{mN}$.