Chapter 32, Problem 32.60AP

Review. Consider a capacitor with vacuum between its large, closely spaced, oppositely charged parallel plates. (a) Show that the force on one plate can be accounted for by thinking of the electric field between the plates as exerting a “negative pressure” equal to the energy density of the electric field. (b) Consider two infinite plane sheets carrying electric currents in opposite directions with equal linear current densities J_s. Calculate the force per area acting on one sheet due to the magnetic field, of magnitude μ₀J_s/2, created by the other sheet. (c) Calculate the net magnetic field between the sheets and the field outside of the volume between them. (d) Calculate the energy density in the magnetic field between the sheets. (e) Show that the force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density. This result for magnetic pressure applies to all current configurations, not only to sheets of current.

To determine

To show: The force on one plate can be accounted for by thinking of the electrical field between the plates as exerting a negative pressure equal to the energy density of the electric field.

Answer to Problem 32.60AP

The force on one plate can be accounted for by thinking of the electrical field between the plates as exerting a negative pressure equal to the energy density of the electric field.

Explanation of Solution

The formula to force between the plates of capacitor is,

$F=\frac{Q^2}{2{\in }_{0}A}$

Here,

$Q$ is the charge on capacitor plate.

$A$ is the area of capacitor plate.

${\in }_0$ is the permittivity of free space.

The formula to calculate electric field produced by negative capacitor plate is,

$E_{\text{n}}=\frac{Q}{2{\in }_{0}A}$

The formula to calculate electric field produced by positive capacitor plate is,

$E_{\text{p}}=-\frac{Q}{2{\in }_{0}A}$

The formula to calculate net electric field produced between the plates is,

$E=E_{\text{n}}-E_{\text{p}}$

Substitute $\frac{Q}{2{\in }_{0}A}$ for $E_{\text{n}}$ and $-\frac{Q}{2{\in }_{0}A}$ for $E_{\text{p}}$ in the above equation.

$\begin{array}{c}E=\frac{Q}{2{\in }_{0}A}-\left(-\frac{Q}{2{\in }_{0}A}\right)\\ =\frac{Q}{{\in }_{0}A}\end{array}$

The energy density of capacitor is,

$u_E=\frac{1}{2}{\in }_0{E}^2$

Substitute $\frac{Q}{2{\in }_{0}A}$ for $E$ in the above equation.

$\begin{array}{c}{u}_E=\frac{1}{2}{\in }_0{\left(\frac{Q}{{\in }_{0}A}\right)}^2\\ =\frac{1}{2}{\in }_0\frac{Q^2}{{\in }_0^2{A}^2}\\ =\frac{1}{2}\frac{Q^2}{{\in }_0{A}^2}\end{array}$

Rearrange above equation.

$\begin{array}{c}F=\frac{Q^2}{2{\in }_{0}A}\\ =\frac{1}{2}\frac{Q^2}{{\in }_0{A}^2}\left(A\right)\\ =PA\end{array}$

Here,

$P$ is the pressure.

$A$ is the area.

Hence, the force on one plate can be accounted for by thinking of the electrical field between the plates as exerting a negative pressure equal to the energy density of the electric field.

Conclusion:

Therefore, the force on one plate can be accounted for by thinking of the electrical field between the plates as exerting a negative pressure equal to the energy density of the electric field.

To determine

The force per area acting on one sheet due to the magnetic field.

Answer to Problem 32.60AP

The force per area acting on one sheet due to the magnetic field is $\frac{\mu J_{{}_{\text{x}}}^2}{2}$.

Explanation of Solution

Given info: The current density of capacitor plates is $J_{\text{x}}$ and the magnetic field produced by one plate on another is $\frac{{\mu }_0{J}_{\text{x}}}{2}$.

The formula to calculate the force on one capacitor plate is,

$\overrightarrow{\text{F}}=I\left(\overrightarrow{\text{l}}\times \overrightarrow{\text{B}}\right)$

Here,

$I$ is the current of capacitor plate.

$\overrightarrow{\text{l}}$ is the length vector.

$\overrightarrow{\text{B}}$ is the magnetic field vector.

Substitute $J_{\text{x}}\times w$ for $I$ and $\frac{{\mu }_0{J}_{\text{x}}}{2}$ for $\overrightarrow{\text{B}}$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{F}}=J_{\text{x}}\times w\left(l\right)\left(\frac{{\mu }_0{J}_{\text{x}}}{2}\right)\sin 90°\\ =J_{\text{x}}w\left(l\right)\left(\frac{{\mu }_0{J}_{\text{x}}}{2}\right)\\ =w\left(l\right)\left(\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}\right)\end{array}$

The formula to calculate the force per area acting on one sheet is,

$P=\frac{F}{A}$

Substitute $w\left(l\right)\left(\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}\right)$ for $F$ and $wl$ for $A$ in the above equation.

$\begin{array}{c}P=\frac{w\left(l\right)\left(\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}\right)}{wl}\\ =\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}\end{array}$

Conclusion:

Therefore, the force per area acting on one sheet due to the magnetic field is $\frac{\mu J_{{}_{\text{x}}}^2}{2}$.

To determine

The net magnetic field between the sheets and the field outside of the volume between them.

Answer to Problem 32.60AP

The net magnetic field between the sheets is ${\mu }_0{J}_{\text{x}}$ and the field outside of the volume between them is zero.

Explanation of Solution

The formula to calculate the magnetic field due to positive sheet is,

$B_{\text{p}}=\frac{{\mu }_0{J}_{\text{x}}}{2}$

The formula to calculate the magnetic field due to positive sheet is,

$B_{\text{n}}=-\frac{{\mu }_0{J}_{\text{x}}}{2}$

The formula to calculate the net magnetic field between the sheets is,

$B=B_{\text{p}}-B_{\text{n}}$

Substitute $\frac{{\mu }_0{J}_{\text{x}}}{2}$ for $B_{\text{p}}$ and $-\frac{{\mu }_0{J}_{\text{x}}}{2}$ for $B_{\text{n}}$ in the above equation.

$\begin{array}{c}B=\frac{{\mu }_0{J}_{\text{x}}}{2}-\left(-\frac{{\mu }_0{J}_{\text{x}}}{2}\right)\\ ={\mu }_0{J}_{\text{x}}\end{array}$

The formula to calculate the net magnetic field outside the sheets is,

$B=B_{\text{p}}-B_{\text{p}}$

Substitute $\frac{{\mu }_0{J}_{\text{x}}}{2}$ for $B_{\text{p}}$ in the above equation.

$\begin{array}{c}B=\frac{{\mu }_0{J}_{\text{x}}}{2}-\frac{{\mu }_0{J}_{\text{x}}}{2}\\ =0\end{array}$

Conclusion:

Therefore, the net magnetic field between the sheets is ${\mu }_0{J}_{\text{x}}$ and the field outside of the volume between them is zero.

To determine

The energy density in the magnetic field between the sheets.

Answer to Problem 32.60AP

The energy density in the magnetic field between the sheets is $\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}$.

Explanation of Solution

The formula to calculate energy density in the magnetic field between the sheets is,

$u_{\text{B}}=\frac{1}{2\mu }{B}^2$

Substitute ${\mu }_0{J}_{\text{x}}$ for $B$ in the above equation.

$\begin{array}{c}{u}_{\text{B}}=\frac{1}{2{\mu }_0}{\left({\mu }_0{J}_{\text{x}}\right)}^2\\ =\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}\end{array}$

Conclusion:

Therefore, the energy density in the magnetic field between the sheets is $\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}$.

To determine

To show: The force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density.

Answer to Problem 32.60AP

The force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density.

Explanation of Solution

Given info: The current density of capacitor plates is $J_{\text{x}}$ and the magnetic field produced by one plate on another is $\frac{{\mu }_0{J}_{\text{x}}}{2}$.

From part (b) the force per area acting on one sheet due to the magnetic field is $\frac{{\mu }_0{J}_{\text{x}}}{2}$.

From part (d) the energy density in the magnetic field between the sheets is $\frac{{\mu }_0{J}_{{}_{\text{x}}}^2}{2}$.

Both the energy density in the magnetic field and the force per area acting on one sheet due to the magnetic field are equal. Hence, the force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density.

Conclusion:

Therefore, the force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density.