Question

The magnetic field existing in a region is given by $\overrightarrow{B}=B_0\left(1+\frac{x}{l}\right)\overrightarrow{k}$. A square loop of edge l and carrying a current i, is placed with its edges parallel to the x−y axes. Find the magnitude of the net magnetic force experienced by the loop.

Answer 1

Given:
Magnetic field, $\overrightarrow{B}=B_0\left(1+\frac{x}{l}\right)\overrightarrow{k}$
Length of the edge of a square loop = l
Electric current flowing through it = i
As per the question, the loop is placed with its edges parallel to the X−Y axes.

In the figure, arrow denotes the direction of force on different sides of the square.
The net magnetic force experienced by the loop,
$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$
Force on AB:
Consider a small element of length dx at a distance x from the origin on line AB.
Force on this small element,
$dF=i{B}_0(1+\frac{x}{l})dx$
Force on the full length of AB,
$$\begin{gathered} F_{\mathrm{AB}}={\int }_{x=0}^{x=l}i{B}_0(1+\frac{x}{l})dx \\ =i{B}_0{\int }_{x=0}^{x=l}(dx+\frac{1}{l}xdx) \\ =i{B}_0(l+\frac{l}{2}) \\ =\frac{3i{B}_{0}l}{2} \end{gathered}$$
Force on AB will be acting downwards.
Similarly, force on CD,
$$\begin{gathered} F_2=i{B}_0\left(l+\frac{l}{2}\right) \\ =\frac{3i{B}_{0}l}{2} \end{gathered}$$
So, the net vertical force = F₁ − F₂ = 0
Force on AD,
$$\begin{gathered} F_3=i{B}_{0}l\left(1+\frac{0}{l}\right) \\ =i{B}_{0}l \end{gathered}$$
Force on BC
$$\begin{gathered} F_4=i{B}_{0}l\left(1+\frac{1}{l}\right) \\ =2i{B}_{0}l \end{gathered}$$
So, the net horizontal force = F₄−F₃ = iB₀l