Question

A $10cm$diameter parallel-plate capacitor has a $1.0mm$spacing. The electric field between the plates is increasing at the rate$1.0x{10}^{6}V/ms$. What is the magnetic field strength (a) on the axis, (b) $3.0cm$from the axis, and (c) $7.0cm$from the axis

Short answer

(a) On the axis =$0T$

(b) $3.0cm$from the axis = $1.668x{10}^{-13}T$

(c) $7.0cm$from the axis = $3.89x{10}^{-13}T$

Step-by-step solution

Part (a) Step 1 : Given Information

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.

Part (a) Step 2 : Simplification

We Know that (using Maxwell Equation)

$$\begin{gathered} \nabla \times B={\mu }_{0}J+{\mu }_0{\epsilon }_0\frac{\partial E}{\partial t} \\ \int \left(\nabla \times B\right)da={\mu }_0\int Jda+{\mu }_0{\epsilon }_0\int \left(\frac{\partial E}{\partial t}\right)da \\ \oint B.dl={\mu }_0{\epsilon }_0\int \left(\frac{\partial E}{\partial t}\right)da \end{gathered}$$

Inside the capacitor Current =$0.$

$B2\pi r={\mu }_0{\epsilon }_0+\left(\frac{\partial E}{\partial t}\right)\pi r^2$

$B=\frac{{\mu }_0{\epsilon }_0}{2}\left(\frac{\partial E}{\partial t}\right)r$

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

On the axis $r=0.$

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

Part (b) Step 1 : Given Information

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.

Part (b) Step 2 : Simplification

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

For $3.0cm$away from the axis,

$B=\frac{{\mu }_0{\epsilon }_0}{2}x{10}^{6}x3x{10}^{-2}$$({\mu }_0=4\pi x{10}^{-7}H/m,{\epsilon }_0=8.854x{10}^{-12}F/m)$

After Solving the mathematical part we get,

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

Part (c) Step 1 : Given Information

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.

Part (c) Step 2 : Simplification

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

For $7.0cm$away from the axis ,

$B=\frac{{\mu }_0{\epsilon }_0}{2}x{10}^{6}x7x{10}^{-2}$ $({\mu }_0=4\pi x{10}^{-7}H/m,{\epsilon }_0=8.854x{10}^{-12}F/m)$

After solving the Mathematical part we get,

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