Question: A stationary magnetic dipole,$\mathit{m}\mathbf{=}\mathit{m}\hat{\mathbf{z}}$ , is situated above an infinite uniform surface current$K=K\hat{x}$, (Fig. 12.44).

(a) Find the torque on the dipole, using Eq. 6.1.

(b) Suppose that the surface current consists of a uniform surface charge , moving at velocity$v=v\hat{x}$ , so that $\mathit{K}\mathbf{=}\mathit{\sigma }\mathit{v}$, and the magnetic dipole consists of a uniform line charge , circulating at speed (same ) around a square loop of side I , as shown, so that$\mathit{m}\mathbf{=}\mathit{\lambda }\mathit{v}{\mathit{l}}^{\mathbf{2}}$ .Examine the same configuration from the point of view of system, moving $\overline{\mathbf{S}}$in the direction at speed . In $\overline{\mathbf{S}}$, the surface charge is at rest, so it generates no magnetic field. Show that in this frame the current loop carries an electric dipole moment, and calculate the resulting torque, using Eq. 4.4.

The magnetic dipole moment,$m=m\hat{z}$

The uniform surface current,$K=K\hat{x}$

The velocity,$v=v\hat{x}$

The uniform surface current in terms of surface charge,$K=K\hat{x}$

Uniform surface charge is $\sigma$.

Uniform line charge is $\lambda$.

The expression to calculate the torque on the dipole moment is given as follows.

$\mathit{N}\mathbf{=}\mathit{m}\mathbf{\times }\mathit{B}$ …… (1)

The expression to calculate the torque on the dipole moment in terms of electric field is given as follows.

$\mathit{N}\mathbf{=}\mathit{p}\mathbf{\times }\mathit{E}$ …… (2)

(a)

Consider the plates as shown below.

The magnetic field due to uniform surface current above it is given by,

$B=\frac{-{\mu }_0}{2}K\hat{y}$

Calculate the torque on the dipole.

Substitute$\frac{-{\mu }_0}{2}K\hat{y}$for B and $m\hat{z}$for m into equation (1).

$$\begin{gathered} N=m\hat{z}\times \frac{-{\mu }_0}{2}K\hat{y} \\ N=\frac{{\mu }_{0}mK}{2}\hat{x} \end{gathered}$$

Substitute $\lambda v{l}^2$for m and $\sigma v$for K into above equation.

$$\begin{gathered} N=\frac{{\mu }_0\left(\lambda v{l}^2\right)\left(\sigma v\right)}{2}\hat{x} \\ N=\frac{{\mu }_0}{2}\lambda \sigma v^2{l}^2\hat{x} \end{gathered}$$

Hence the torque on the magnetic dipole is$\frac{{\mu }_0}{2}\lambda \sigma v^2{l}^2\hat{x}$ .

(b)

The charge density on the front side of the dipole is and back side is$\overline{\lambda }=\overline{\gamma }{\lambda }_0$ .

Here, the expression for the$\gamma$ is given by,

$\overline{\gamma }=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ ……. (3)

Here, $\overline{v}=\frac{2v}{1+\frac{v^2}{c^2}}$

Substitute$\frac{2v}{1+\frac{v^2}{c^2}}$for into equation (3).

$$\begin{gathered} \overline{\gamma }=\frac{1}{\sqrt{1-\frac{{\left(\frac{2v}{1+\frac{v^2}{c^2}}\right)}^2}{c^2}}} \\ \overline{\gamma }=\frac{1}{\sqrt{1-\frac{\left(\frac{4v^2}{c^2}\right)}{{\left(1+\frac{v^2}{c^2}\right)}^2}}} \\ \overline{\gamma }=\frac{1}{\sqrt{1-\frac{4v^2}{c^2{\left(1+\frac{v^2}{c^2}\right)}^2}}} \\ \overline{\gamma }=\frac{1}{\sqrt{\frac{c^2{\left(1+\frac{v^2}{c^2}\right)}^2-4v^2}{c^2{\left(1+\frac{v^2}{c^2}\right)}^2}}} \end{gathered}$$

Solve further as,

$$\begin{gathered} \overline{\gamma }=\frac{1}{\sqrt{\frac{c^2\left(1+\frac{v^4}{c^4}+2\frac{v^2}{c^2}\right)-4v^2}{c^2{\left(1+\frac{v^2}{c^2}\right)}^2}}} \\ \overline{\gamma }=\frac{c\left(1+\frac{v^2}{c^2}\right)}{\sqrt{c^2+\frac{v^4}{c^2}+2v^2-4v^2}} \\ \overline{\gamma }=\frac{c\left(1+\frac{v^2}{c^2}\right)}{\sqrt{c^2+\frac{v^4}{c^2}-2v^2}} \\ \overline{\gamma }=\frac{c\left(1+\frac{v^2}{c^2}\right)}{c\sqrt{\left(1+\frac{v^4}{c^4}-2\frac{v^2}{c^2}\right)}} \end{gathered}$$

Solve further as,

$$\begin{gathered} \overline{\gamma }=\frac{1+\frac{v^2}{c^2}}{\sqrt{{\left(1-\frac{v^2}{c^2}\right)}^2}} \\ \overline{\gamma }=\frac{1+\frac{v^2}{c^2}}{\left(1-\frac{v^2}{c^2}\right)} \end{gathered}$$

Substitute${\gamma }^2$ for$\frac{1}{{\displaystyle \left[1-{\left(\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)}^2\right]}}$ into above equation.

$\overline{\gamma }={\gamma }^2\left(1+\frac{v^2}{c^2}\right)$

The net charge on the back side is $\overline{\lambda }=\overline{\gamma }{\lambda }_0$given by,

$q_{+}=\overline{\lambda }\frac{l}{\gamma }$

Substitute$\overline{\gamma }{\lambda }_0$ for $\overline{\gamma }$into above equation.

$q_{+}=\overline{\gamma }{\lambda }_0\frac{l}{\gamma }$

Substitute${\gamma }^2\left(1+\frac{v^2}{c^2}\right)$ for$\overline{\gamma }$ and$\frac{\lambda }{\gamma }$ for into above equation.

$$\begin{gathered} q_{+}={\gamma }^2\left(1+\frac{v^2}{c^2}\right)\frac{\lambda }{\gamma }\frac{l}{\gamma } \\ {q}_{+}=\lambda l\left(1+\frac{v^2}{c^2}\right) \end{gathered}$$

The net charge on the front side is given by,

$q_{-}={\lambda }_0\frac{l}{\gamma }$

Substitute$\frac{\lambda }{\gamma }$ for ${\lambda }_0$into above equation.

$$\begin{gathered} q_{-}=\frac{\lambda }{\gamma }\frac{l}{\gamma } \\ {q}_{-}=\frac{\lambda l}{{\gamma }^2} \end{gathered}$$

The total dipole moment is given by,

$p=\left(q_{+}\right)\frac{l}{\gamma }\hat{y}-\left(q_{-}\right)\frac{l}{\gamma }\hat{y}$

Substitute$\lambda l\left(1+\frac{v^2}{c^2}\right)$ for $q_{+}$and $\frac{\lambda l}{{\gamma }^2}$for$q_{-}$ into above equation.

$$\begin{gathered} p=\lambda l\left(1+\frac{v^2}{c^2}\right)\frac{l}{2}\hat{y}-\frac{\lambda l}{{\gamma }^2}\frac{l}{2}\hat{y} \\ p=\frac{\lambda l^2}{2}\left[\left(1+\frac{v^2}{c^2}\right)-\frac{1}{{\gamma }^2}\right]\hat{y} \end{gathered}$$

Substitute $\frac{1}{{\displaystyle \sqrt{1-{\left(\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\right)}^2}}}$for into above equation.

$$\begin{gathered} p=\frac{\lambda l^2}{2}\left[\left(1+\frac{v^2}{c^2}\right)-\frac{1}{{\left(\sqrt{1-\frac{v^2}{c^2}}\right)}^2}\right]\hat{y} \\ p=\frac{\lambda l^2}{2}\left[\left(1+\frac{v^2}{c^2}\right)-\frac{1}{1-\frac{v^2}{c^2}}\right]\hat{y} \\ p=\frac{\lambda l^2}{2}\left[1+\frac{v^2}{c^2}-1+\frac{v^2}{c^2}\right]\hat{y} \\ p=\frac{\lambda l^2}{2}\left[\frac{2v^2}{c^2}\right]\hat{y} \end{gathered}$$

Solve further as,

$p=\frac{\lambda l^2{v}^2}{c^2}\hat{y}$

The electric field intensity due to charged plate is given by,

$E=\frac{{\sigma }_0}{2{\epsilon }_0}\hat{z}$

Calculate the resultant torque.

Substitute$\frac{{\sigma }_0}{2{\epsilon }_0}\hat{z}$ for E and$\frac{\lambda l^2{v}^2}{c^2}\hat{y}$ for P into equation (2).

$$\begin{gathered} N=\frac{\lambda l^2{v}^2}{c^2}\hat{y}\times \frac{{\sigma }_0}{2{\epsilon }_0}\hat{z} \\ N=\frac{\lambda l^2{v}^2}{c^2}\left(\frac{{\sigma }_0}{2{\epsilon }_0}\right)\hat{x} \\ N=\frac{\lambda l^2{v}^2}{c^2{\epsilon }_0}\left(\frac{{\sigma }_0}{2}\right)\hat{x} \end{gathered}$$

Substitute$\frac{\sigma }{\gamma }$ for ${\sigma }_0$and$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mu }_0$}\right.$ for$c^2{\epsilon }_0$ into above equation.

$$\begin{gathered} N=\frac{\lambda l^2{v}^2}{\frac{1}{{\mu }_0}}\left(\frac{\sigma }{2\gamma }\right)\hat{x} \\ N={\mu }_0\lambda l^2{v}^2\left(\frac{\sigma }{2\gamma }\right)\hat{x} \\ N=\frac{1}{\gamma }\frac{{\mu }_0}{2}\lambda \sigma l^2{v}^2\hat{x} \end{gathered}$$

Hence the resultant torque is$\frac{1}{\gamma }\frac{{\mu }_0}{2}\lambda \sigma l^2{v}^2\hat{x}$.