91Ó°ÊÓ

Write the expression of Maxwell-Ampere Law:

$âˆ\circledR Bâ‹\dots ds={\mathrm{μ}}_{0}I+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{d{\mathrm{Ï•}}_E}{dt}\text{ }$….. (1)

Here, B is the magnetic field produced by the moving charges, I is the current due to moving charges, is the permeability of free space and $\frac{d{\mathrm{Ï•}}_E}{dt}$is the change in electric flux due to the change in velocity of the charged particles.

(a)

For closed curved paths, the integration of the magnetic field is zero. Hence, equation (1) becomes,

$\begin{array}{l}âˆ\circledR Bâ‹\dots ds={\mathrm{μ}}_{0}I+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{d{\mathrm{Ï•}}_E}{dt}\\ 0={\mathrm{μ}}_{0}I+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{d{\mathrm{Ï•}}_E}{dt}\end{array}$

Rearrange the above equation for I.

$I=−{\mathrm{Î\mu }}_0\frac{d{\mathrm{Ï•}}_E}{dt}$….. (2)

Here, a negative sign indicates the direction of a current.

Write the formula for the electric flux.

${\mathrm{Ï•}}_E=Eâ‹\dots A\text{ }$….. (3)

Here, A is the area of cross section.

Write the formula for the area of cross-section of the plates of the capacitor.

$A=\mathrm{π}{r}^2$….. (4)

Substitute the value of equations (3) and (4) in equation (2).

$\begin{array}{l}I={\mathrm{Î\mu }}_0\frac{d\left(EA\right)}{dt}\\ I={\mathrm{Î\mu }}_0\left(\mathrm{Ï€}{r}^2\right)\frac{dE}{dt}\\ dE=\frac{I}{{\mathrm{Î\mu }}_0\left(\mathrm{Ï€}{r}^2\right)}dt\end{array}$

Integrate the above equation.

$\begin{array}{l}∫dE=∫\frac{I}{{\mathrm{Î\mu }}_0\left(\mathrm{Ï€}{r}^2\right)}dt\\ E=\frac{It}{{\mathrm{Î\mu }}_0\left(\mathrm{Ï€}{a}^2\right)}\\ \stackrel{→}{E}=\frac{It}{\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}\widehat{z}\end{array}$

Write the formula for the electric flux through the wire.

${\mathrm{Ï•}}_E=âˆ\circledR Eâ‹\dots da\text{ }$….. (5)

As there is no current enclosed in the gap between the wires, the value of Iwill be zero.

Substitute the value of equation (5) in equation (1).

$\begin{array}{l}âˆ\circledR Bâ‹\dots dl={\mathrm{μ}}_0{I}_{\text{enc}}+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{d{\mathrm{Ï•}}_E}{dt}\\ âˆ\circledR Bâ‹\dots dl={\mathrm{μ}}_0\left(0\right)+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{d\left(Eâ‹\dots da\right)}{dt}\\ B\left(2\mathrm{Ï€}s\right)=0+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\left(\frac{dE}{dt}\right)\left(\mathrm{Ï€}{s}^2\right)\\ B\left(2\mathrm{Ï€}s\right)={\mathrm{μ}}_0{\mathrm{Î\mu }}_0\left(\frac{I}{\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}\right)\left(\mathrm{Ï€}{s}^2\right)\end{array}$

On further solving,

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_{0}Is}{2\mathrm{π}{a}^2}\widehat{\mathrm{ϕ}}$

Therefore, the electric and magnetic fields in the gap is $\stackrel{→}{E}=\frac{It}{\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^2}\widehat{z}$and $\stackrel{→}{B}=\frac{{\mathrm{μ}}_{0}Is}{2\mathrm{Ï€}{a}^2}\widehat{\mathrm{Ï•}}a$respectively.