Question

Question: (a) Find the force on a square loop placed as shown in Fig. 5.24(a), near an infinite straight wire. Both the loop and the wire carry a steady current I.

(b) Find the force on the triangular loop in Fig. 5.24(b).

Short answer

(a) The force on a square of side a carrying current l at a distance s from an infinite straight wire carrying current l is $\frac{{\mu }_0{l}^2{a}^2}{2\pi s\left(a+s\right)}$.

(b) The force on a triangle of side a carrying current l at a distance s from an infinite straight wire carrying current l is $\frac{{\mu }_0{l}^2}{2\pi }\left[\frac{\mathrm{a}}{\mathrm{s}}-\frac{2}{\sqrt{3}}\mathrm{In}\left(1+\frac{\sqrt{3}\mathrm{a}}{2\mathrm{s}}\right)\right]$.

Step-by-step solution

Given data

(a) A square of side a carrying current l at a distance s from an infinite straight wire carrying current l .

(b) A triangle of side a carrying current l at a distance s from an infinite straight wire carrying current l .

Magnetic field from an infinite wire and force on a current carrying wire in the presence of a magnetic field

Magnetic field at a distance s from an infinite straight wire carrying current l is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\tau \tau s}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, ${\mu }_0$is the permeability of free space.

Force on a wire of length l carrying current l in a magnetic field B is

$\mathbf{F}\mathbf{=}\mathbf{IBI}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Force on the square loop

In the first figure, the force on the left and right hand sides of the square should be exactly equal and opposite. So they cancel out.

From equation (1), magnetic field near the bottom side from the infinite wire is

$B=\frac{{\mu }_{0}l}{2\pi s}$

The field is directed outward.

Thus, from equation (2), force on the bottom wire is

$$\begin{gathered} F=l\times \frac{{\mu }_{0}l}{2\pi s}\times a \\ =\frac{{\mu }_0{l}^{2}a}{2\pi s} \end{gathered}$$

The force is directed upward.

From equation (1), magnetic field near the top side from the infinite wire is

$B=\frac{{\mu }_{0}l}{2\pi \left(a+s\right)}$

The field is directed outward.

Thus, from equation (2), force on the top wire is

$$\begin{gathered} F=l\times \frac{{\mu }_{0}l}{2\pi \left(a+s\right)}\times a \\ =\frac{{\mu }_0{l}^{2}a}{2\pi \left(a+s\right)} \end{gathered}$$

The force is directed downward.

The net force on the square is then

$$\begin{gathered} F=\frac{{\mu }_0{l}^{2}a}{2\pi s}-\frac{{\mu }_0{l}^{2}a}{2\pi \left(a+s\right)} \\ =\frac{{\mu }_0{l}^{2}a}{2\pi s}\left(\frac{1}{s}-\frac{1}{a+s}\right) \\ =\frac{{\mu }_0{l}^2{a}^2}{2\pi s\left(a+s\right)} \end{gathered}$$

Thus, the net force on the square loop is $\frac{{\mu }_0{l}^2{a}^2}{2\pi s\left(a+s\right)}$directed downward.

Force on the triangular loop

The force on the lower side of the triangle is just like before.

$F=\frac{{\mu }_0{l}^{2}a}{2\pi s}$

The force is directed upward.

The forces on the other two sides will each have two components, a vertical and a horizontal. The horizontal components will cancel out. The vertical components will add and point downward.

From equation (1), the field at a distance y from the infinite wire is

$B=\frac{{\mu }_{0}l}{2\pi y}$

From equation (2), the force on a small section of the wire with horizontal component dx with $y=\sqrt{3}x$is

$dF=\frac{{\mu }_0{l}^{2}dx}{2\sqrt{3}\pi x}$

This is the vertical component of the force on the small section. The net force on one ire is thus

$$\begin{gathered} F=\frac{{\mu }_0{l}^2}{2\sqrt{3}\pi }{\int }_{\frac{s}{\sqrt{3}}}^{\frac{s}{\sqrt{3}}+\frac{a}{2}}\frac{dx}{x} \\ =\frac{{\mu }_0{l}^2}{2\sqrt{3}\pi }\left[\mathrm{In}\left(\frac{s}{\sqrt{3}}+\frac{a}{2}\right)-\mathrm{In}\left(\frac{s}{\sqrt{3}}\right)\right] \\ =\frac{{\mu }_0{l}^2}{2\sqrt{3}\pi }\mathrm{In}\left(1+\frac{\sqrt{3}a}{2s}\right) \end{gathered}$$

The net downward vertical force from the two sides is thus

$$\begin{gathered} F=2\times \frac{{\mu }_0{l}^2}{2\sqrt{3}\pi }\mathrm{In}\left(1+\frac{\sqrt{3}a}{2s}\right) \\ =\frac{{\mu }_0{l}^2}{\sqrt{3}\pi }\mathrm{In}\left(1+\frac{\sqrt{3}a}{2s}\right) \end{gathered}$$

The net force on the triangle is then

$$\begin{gathered} F=\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{l}}^{\mathbf{2}}\mathbf{a}}{\mathbf{2}\mathbf{\pi }\mathbf{s}}-\frac{{\mu }_0{l}^2}{\sqrt{3}\pi }\mathrm{In}\left(1+\frac{\sqrt{3}a}{2s}\right) \\ =\frac{{\mu }_0{l}^2}{2\pi }\left[\frac{\mathrm{a}}{\mathrm{s}}-\frac{2}{\sqrt{3}}\mathrm{In}\left(1+\frac{\sqrt{3}\mathrm{a}}{2\mathrm{s}}\right)\right] \end{gathered}$$

Thus, the net force on the triangle is $\frac{{\mu }_0{l}^2}{2\pi }\left[\frac{\mathrm{a}}{\mathrm{s}}-\frac{2}{\sqrt{3}}\mathrm{In}\left(1+\frac{\sqrt{3}\mathrm{a}}{2\mathrm{s}}\right)\right]$pointed upward.