91Ó°ÊÓ

Write the expression for the electromagnetic momentum density.

$g_{em}={\mathrm{Î\mu }}_0\left(E×B\right)$

Here, E is the electric field, and B is the magnetic field.

(a)

Write the expression for the electric field.

$E=−\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\widehat{z}$

Write the expression for the magnetic field.

$B=−{\mathrm{μ}}_0\mathrm{σ}v\widehat{x}$

Substitute all the known values in equation (1).

$\begin{array}{c}{g}_{\text{em}}={\mathrm{Î\mu }}_0\left(−\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\widehat{z}×\left(−{\mathrm{μ}}_0\mathrm{σ}v\widehat{x}\right)\right)\\ g_{\text{em}}={\mathrm{μ}}_0{\mathrm{σ}}^{2}v\widehat{y}\end{array}$

Write the expression for the electromagnetic momentum in a region of area A.

$p_{\text{em}}=\left(dA\right)g_{\text{em}}$

Substitute the known values in the above expression.

$\begin{array}{c}{p}_{\text{em}}=\left(dA\right)\left(−{\mathrm{μ}}_0{\mathrm{σ}}^{2}v\widehat{y}\right)\\ p_{\text{em}}={\mathrm{μ}}_0{\mathrm{σ}}^{2}vdA\widehat{y}\end{array}$

Therefore, the electromagnetic momentum in a region of area A is $p_{\text{em}}={\mathrm{μ}}_0{\mathrm{σ}}^{2}vdA\widehat{y}$.

(b)

Write the expression for the initial impulse delivered to the system.

$I_1=∫Fâ‹\dots dt$….. (2)

Here, F is the magnetic force which is given as:

Determine the expression for the magnetic force.

$\begin{array}{c}F=q\left(u×B\right)\\ F=\mathrm{σ}A\left[\left(−u\widehat{z}\right)×\left(−\frac{1}{2}{\mathrm{μ}}_0\mathrm{σ}v\widehat{x}\right)\right]\\ F=\frac{1}{2}{\mathrm{μ}}_0{\mathrm{σ}}^{2}Avu\widehat{y}\end{array}$

Substitute the known value in equation (2).

$\begin{array}{c}{I}_1=∫\frac{1}{2}{\mathrm{μ}}_0{\mathrm{σ}}^{2}Avu\widehat{y}dt\\ I_1=\frac{1}{2}{\mathrm{μ}}_0{\mathrm{σ}}^{2}Av\widehat{y}∫udt\\ I_1=\frac{1}{2}d{\mathrm{μ}}_0{\mathrm{σ}}^{2}Av\widehat{y}\end{array}$

Write the expression for the final impulse delivered to the system.

$I_2=∫Fâ‹\dots dt$….. (3)

Here, F is the magnetic force which is given as:

$\begin{array}{c}F=q\left(u×B\right)\\ F=\mathrm{σ}A\left[\left(−u\widehat{z}\right)×\left(−\frac{1}{2}{\mathrm{μ}}_0\mathrm{σ}v\widehat{x}\right)\right]\\ F=\frac{1}{2}{\mathrm{μ}}_0{\mathrm{σ}}^{2}Avu\widehat{y}\end{array}$

Substitute the known value in equation (3).

$\begin{array}{c}{I}_2=∫\frac{1}{2}{\mathrm{μ}}_0{\mathrm{σ}}^{2}Avu\widehat{y}dt\\ I_2=\frac{1}{2}{\mathrm{μ}}_0{\mathrm{σ}}^{2}Av\widehat{y}∫udt\\ I_2=\frac{1}{2}d{\mathrm{μ}}_0{\mathrm{σ}}^{2}Av\widehat{y}\end{array}$

Hence, the total impulse delivered to the sheet will be,

$\begin{array}{c}I=I_1+I_2\\ I=\left(\frac{1}{2}d{\mathrm{μ}}_0{\mathrm{σ}}^{2}Av\widehat{y}\right)+\left(\frac{1}{2}d{\mathrm{μ}}_0{\mathrm{σ}}^{2}Av\widehat{y}\right)\\ I=dA{\mathrm{μ}}_0{\mathrm{σ}}^{2}v\widehat{y}\end{array}$

Hence proved.

Therefore, the impulse delivered to the sheet is equal to the momentum originally stored in the fields.