91Ó°ÊÓ

We are given that,

Diameter of Parallel plate capacitor = $10cm$

Spacing between the capacitor = $1.0mm$

Electric field Increasing Rate = $1.0x{10}^{6}V/ms$

We have to find magnetic field strength on the axis.

We Know that (using Maxwell Equation)

$$\begin{gathered} ∇×B={\mathrm{μ}}_{0}J+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t} \\ ∫\left(∇×B\right)da={\mathrm{μ}}_0∫Jda+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0∫\left(\frac{∂E}{∂t}\right)da \\ âˆ\circledR B.dl={\mathrm{μ}}_0{\mathrm{Î\mu }}_0∫\left(\frac{∂E}{∂t}\right)da \end{gathered}$$

Inside the capacitor Current =$0.$

$B2\mathrm{Ï€}r={\mathrm{μ}}_0{\mathrm{Î\mu }}_0+\left(\frac{∂E}{∂t}\right)\mathrm{Ï€}{r}^2$

$B=\frac{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}{2}\left(\frac{∂E}{∂t}\right)r$

Using the above derived result :-$B=\frac{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}{2}\left(\frac{∂E}{∂t}\right)r$

On the axis $r=0.$

So, Magnetic field strength on the axis is$0T.$

We are given that,

Diameter of Parallel plate capacitor = $10cm$

Spacing between the capacitor = $1.0mm$

Electric field Increasing Rate = $1.0x{10}^{6}V/ms$

We have to find magnetic field strength $3.0cm$away from the axis.

Again using the same result derives in Part (a) :- $B=\frac{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}{2}\left(\frac{∂E}{∂t}\right)r$

For $3.0cm$away from the axis,

$B=\frac{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}{2}x{10}^{6}x3x{10}^{-2}$$({\mathrm{μ}}_0=4\mathrm{Ï€}x{10}^{-7}H/m,{\mathrm{Î\mu }}_0=8.854x{10}^{-12}F/m)$

After Solving the mathematical part we get,

$B=1.668x{10}^{-13}T$

We are given that,

Diameter of Parallel plate capacitor = $10cm$

Spacing between the capacitor = $1.0mm$

Electric field Increasing Rate = $1.0x{10}^{6}V/ms$

We have to find magnetic field strength $7.0cm$away from the axis.

Again using the same result derived in Part (a) :- $B=\frac{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}{2}\left(\frac{∂E}{∂t}\right)r$

For $7.0cm$away from the axis ,

$B=\frac{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}{2}x{10}^{6}x7x{10}^{-2}$ $({\mathrm{μ}}_0=4\mathrm{Ï€}x{10}^{-7}H/m,{\mathrm{Î\mu }}_0=8.854x{10}^{-12}F/m)$

After solving the Mathematical part we get,

$B=3.89x{10}^{-13}T$