Chapter 32, Problem 60AP

(a)

To determine

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

Answer to Problem 60AP

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

Explanation of Solution

Write the expression to calculate the electric fields between the plates.

    $E=\frac{\sigma }{{\epsilon }_0}$

Here, $E$ is the electric field, $\sigma$ is the stress and ${\epsilon }_0$ is the permittivity of free space.

Write the expression to calculate the force.

    $F={\epsilon }_0{E}^2$

Here, $F$ is the force.

Write the expression to calculate the force on one plate.

    $\begin{array}{c}F\text{'}=\frac{F}{2}\\ F\text{'}=\frac{{\epsilon }_0{E}^2}{2}\end{array}$                                                                                                                  (I)

Write the expression to calculate the energy density of the electric field.

    $U_{\text{E}}=-\frac{1}{2}{\epsilon }_0{E}^2$                                                                                                           (II)

Here, $U_{\text{E}}$ is the energy density.

Negative sign indicates that the plates are drawn towards each other.

Equate equation (I) and (II).

    $\begin{array}{c}F\text{'}=U_E\\ \frac{{\epsilon }_0{E}^2}{2}=\frac{{\epsilon }_0{E}^2}{2}\end{array}$

Conclusion:

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

(b)

To determine

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

Answer to Problem 60AP

The force per area acting on the sheet due to the magnetic field is $\left(\frac{{\mu }_0{J}_{\text{x}}^2}{2}\right)$.

Explanation of Solution

Write the expression for magnetic field due to one plate using Ampere’s law.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space and $J_{\text{x}}$ is the current density.

Write the expression to calculate the force per unit area.

    $F=B{J}_{\text{x}}$

Here, $F$ is the force.

Substitute $\left(\frac{{\mu }_0{J}_{\text{x}}}{2}\right)$ for $B$ in the above equation to calculate $F$.

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

Conclusion:

Therefore, the force per area acting on the sheet due to the magnetic field is $\left(\frac{{\mu }_0{J}_{\text{x}}^2}{2}\right)$.

(c)

To determine

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

Answer to Problem 60AP

The net magnitude field between the sheets is $\left({\mu }_0{J}_{\text{x}}\right)$  and the field outside of the volume is $0$.

Explanation of Solution

Write the expression for magnetic field due to one plate using Ampere’s law.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space and $J_{\text{x}}$ is the current density.

Write the expression to calculate the total magnetic field.

    $B\text{'}=2B$

Substitute $\left(\frac{{\mu }_0{J}_{\text{x}}}{2}\right)$ for $B$ in the above equation to calculate $B\text{'}$.

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

The magnetic field between the sheets is perpendicular to the plane of the page and magnetic field outside is zero.

Conclusion:

Therefore, the net magnitude field between the sheets is $\left({\mu }_0{J}_{\text{x}}\right)$  and the field outside of the volume is $0$.

(d)

To determine

The energy density in the magnetic field between the sheets.

Answer to Problem 60AP

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

Explanation of Solution

Write the expression for energy density in the magnetic field.

    $E_{\text{d}}=\frac{B^2}{2{\mu }_0}$

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

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

Conclusion:

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

(e)

To determine

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

Answer to Problem 60AP

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

Explanation of Solution

Write the expression total magnetic field.

    $\begin{array}{l}B\text{'}={\mu }_0{J}_{\text{x}}\\ J_{\text{x}}=\frac{B\text{'}}{{\mu }_0}\end{array}$

Write the expression for force on one sheet of paper.

    $F=B\times J_{\text{x}}$

Substitute $\left(\frac{B\text{'}}{{\mu }_0}\right)$ for $J_{\text{x}}$ and $\left(\frac{B\text{'}}{2}\right)$ for $B$ in the above equation to calculate $F$.

    $\begin{array}{c}F=\frac{B\text{'}}{2}\left(\frac{B\text{'}}{{\mu }_0}\right)\\ F=\frac{1}{2{\mu }_0}B{\text{'}}^2\end{array}$                                                                                                          (III)

Write the expression for energy density.

    $U_B=\frac{1}{2{\mu }_0}B{\text{'}}^2$                                                                                                          (IV)

Here, $U_B$ is the energy density.

Conclusion:

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