Question

Suppose that the magnetic field in some region has the form

$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{kz}\stackrel{\mathbf{⏜}}{\mathbf{x}}$

(where kis a constant). Find the force on a square loop (side a),lying in the yz

plane and centered at the origin, if it carries a current I,flowing counterclockwise,

when you look down the xaxis.

Short answer

The force on a square loop of side a,lying in the yzplane and centered at the origin, carrying a current I,flowing counterclockwise in the presence of a magnetic field$\overrightarrow{B}=kz\stackrel{⏜}{x}$ is $I{a}^{2}k\stackrel{⏜}{z}$.

Step-by-step solution

Given data

There is a magnetic field of the form$\overrightarrow{B}=kz\stackrel{⏜}{x}$ .

There is a square loop of side alying in the yzplane, centered at the origin and carries a current I,flowing counterclockwise.

Force on a current carrying wire in a magnetic field

The force on a wire carrying current $I$, length $I$in a magnetic field$B$is

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{|}\overrightarrow{\mathbf{|}}\mathbf{\times }\overrightarrow{\mathbf{B}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Force on the current carrying loop

The force on $y=a/2$and $y=-a/2$wires are exactly equal and opposite and so they cancel out.

From equation (1), force on the z=a/2 wire

$$\begin{gathered} {\overrightarrow{F}}_1=IaB{\backslash }_{z-a/2}\stackrel{⏜}{z} \\ =Iak\frac{a}{2}\stackrel{⏜}{z} \\ =\frac{I{a}^{2}k}{2}\stackrel{⏜}{z} \end{gathered}$$

From equation (1), force on the z=-a/2 wire

$$\begin{gathered} {\overrightarrow{F}}_2=IaB{\backslash }_{z-a/2}\left(-\stackrel{⏜}{z}\right) \\ =Iak\left(-\frac{a}{2}\right)\left(-\stackrel{⏜}{z}\right) \\ =\frac{I{a}^{2}k}{2}\stackrel{⏜}{z} \end{gathered}$$

Thus, the net force on the wire is $I{a}^{2}k\stackrel{⏜}{z}$ .