Question

An electric field of $300V/m$ is confined to a circular area $10cm$ in diameter. If the field is increasing at the rate of $20V/m/s$, the magnetic field at a point $15cm$ from the center of the circle will be:

Answer 1

Step 1: Given data

The magnitude of the electric field is $E=300V/m$.

The diameter of the circular area is $D=10cm=1\times {10}^{-1}m.$

The electric flux increases at a rate $20V/m/s$

The distance of the point where the magnetic field has to be calculated is $s=15cm.$

The velocity of light $c=3\times {10}^{8}m$

Step 2: Formulae and concept

1. The number of electric lines of force passes through a unit area is called electric flux.
2. We know that, $d\phi =dE\times ds$, $d\phi$ is the change in electric flux, $dE$ is the change in an electric field, and $ds$ is any surface.
3. According to Maxwell's electromagnetic theory, $\oint \overrightarrow{B}.\overrightarrow{dl}={\epsilon }_o{\mu }_o\frac{d\phi }{dt}$, where, B is the magnetic field, dl is the small length, ${\mu }_o$and ${\epsilon }_o$ are the permeability and permittivity of free space.
4. Again, $c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}$, c is the velocity of light in free space.

Step 3: Calculation of electric flux

We know, electric flux is,

$$\begin{gathered} d\phi =dE\times ds \\ ord\phi =20\times \pi {\left(\frac{1\times {10}^{-1}}{2}\right)}^2 \\ ord\phi =20\times \pi \times 5^2\times {10}^{-4} \\ ord\phi =500\pi \times {10}^{-4} \\ ord\phi =5\pi \times {10}^{-2} \end{gathered}$$ (since the area of the circle is $\pi {\left(\frac{D}{2}\right)}^2$ and the radius of the circle is $\frac{D}{2}=0.05m$).

Now, flux change in unit time is

$\frac{d\phi }{dt}=5\pi \times {10}^{-2}..............\left(1\right)$

Step 4: Calculation of magnetic field field

Now, from Maxwell's electromagnetic theory,

$$\begin{gathered} \oint \overrightarrow{B}.\overrightarrow{dl}=\frac{1}{c^2}\frac{d\phi }{dt} \\ orB\oint dl=\frac{1}{c^2}\times 5\pi \times {10}^{-2} \\ orB\times \pi \times {\left(0.05\right)}^2=\frac{1}{c^2}\times 5\pi \times {10}^{-2}(\sin ce,radius,\frac{D}{2}=0.05m) \\ orB=\frac{1}{c^2}\times \frac{5\pi \times {10}^{-2}}{2\pi \times \left(0.15\right)}=\frac{5\pi \times {10}^{-2}}{{\left(3\times {10}^8\right)}^2\times 2\mathrm{\pi }\times \left(0.15\right)}=1.851\times {10}^{-18} \\ orB=1.851\times {10}^{-18}T \end{gathered}$$

Therefore, the magnetic field at a point 15cm from the center of the circle will be $1.851\times {10}^{-18}T$.