Question

A square loop of wire of side length \(\ell\) lies in the \(x y\) -plane, with its center at the origin and its sides parallel to the \(x\) - and \(y\) -axes. It carries a current, \(i\), flowing in the counterclockwise direction, as viewed looking down the \(z\) -axis from the positive direction. The loop is in a magnetic field given by \(\vec{B}=\left(B_{0} / a\right)(z \hat{x}+x \hat{z}),\) where \(B_{0}\) is a constant field strength, \(a\) is a constant with the dimension of length, and \(\hat{x}\) and \(\hat{z}\) are unit vectors in the positive \(x\) -direction and positive \(z\) -direction. Calculate the net force on the loop.

Short answer

Answer: The net force on the square loop of wire is zero.

Step-by-step solution

Calculate forces on each side separately

We will break the loop into 4 segments (top, bottom, left and right) and find the force on each segment one by one.

Determine \(\vec{L}\) for each side and compute \(\vec{F}\) for each side

Define \(\vec{L}_{top}\), \(\vec{L}_{bottom}\), \(\vec{L}_{left}\), and \(\vec{L}_{right}\) as the vectors for the top, bottom, left, and right sides of the loop, respectively. We can represent the vectors as: $$ \vec{L}_{top} = -\ell \hat{y}, \, \vec{L}_{bottom} = \ell \hat{y}, \, \vec{L}_{left} = -\ell \hat{x}, \, \vec{L}_{right} = \ell \hat{x} $$ Now, we will compute the force on each side by using the expression \(\vec{F} = I\vec{L} \times \vec{B}\). Remember that \(\vec{B}=\left(B_{0} / a\right)(z \hat{x}+x \hat{z})\)

Compute Force on the top side

$$ \vec{F}_{top} = i\left(-\ell\hat{y}\right)\times\left[\frac{B_0}{a}\left(z\hat{x}+x\hat{z}\right)\right] = 0 $$ Since the vectors are parallel, there is no force on this side.

Compute Force on the bottom side

$$ \vec{F}_{bottom} = i(\ell\hat{y})\times\left[\frac{B_0}{a}\left(z\hat{x}+x\hat{z}\right)\right]=0 $$ Again, the vectors are parallel, hence there is no force on this side as well.

Compute Force on the left side

$$ \vec{F}_{left} = i(-\ell\hat{x})\times\left[\frac{B_0}{a}\left(z\hat{x}+x\hat{z}\right)\right] = -\frac{B_0 i\ell}{a}\hat{y} $$

Compute Force on the right side

$$ \vec{F}_{right} = i(\ell\hat{x})\times\left[\frac{B_0}{a}\left(z\hat{x}+x\hat{z}\right)\right] = \frac{B_0 i\ell}{a}\hat{y} $$

Compute the net force on the loop

To find the net force on the loop, we will add the force vectors for all the sides: $$ \vec{F}_{net} = \vec{F}_{top} + \vec{F}_{bottom} + \vec{F}_{left} + \vec{F}_{right} = -\frac{B_0 i\ell}{a}\hat{y} + \frac{B_0 i\ell}{a}\hat{y} = \vec{0} $$ So the net force on the square loop of wire is zero.