Electromagnetic Induction

$ \begin{aligned} e & =-M \frac{d i}{d t} \\ \frac{d i}{d t} & =10 \times 100 \pi \cos (100 \pi t) \end{aligned} $

Maximum value of $\frac{d i}{d t}$ is $10 \times 100 \pi \cdot[\because \cos (100 \pi t)=1]$

$ \text { } \begin{aligned} Now,\, M & =\frac{e}{\frac{d i}{d t}}=\frac{5 \pi}{10 \times 100 \pi} \\ & =5 \times 10^{-3} \mathrm{H}=5 \mathrm{mH} \\ M & =5 \mathrm{mH} \end{aligned} $

The changing magnetic field induces an electric field, which accelerates the electron.

To determine the acceleration of the electron, we use the formula:

$ a = \frac{eE}{m} $

where $ F = qE $.

To calculate the electric field $ E $, we apply Faraday's law of electromagnetic induction:

$ \varepsilon = -\frac{d\Phi}{dt} = -B \cdot A $

Here, $ B $ is the magnetic field, and $ A $ is the cross-sectional area. Therefore, the electromotive force (emf) is:

$ \varepsilon = -A \cdot \frac{dB}{dt} = -\pi r^2 \cdot \frac{dB}{dt} $

The electric field $ E $ is then given by:

$ E = -\frac{\varepsilon}{\text{distance}} = -\frac{\varepsilon}{d} $

$ E = \frac{\pi r^2}{d} \cdot \frac{dB}{dt} $

Finally, the acceleration $ a $ is:

$ a = \frac{e}{m} \cdot \frac{\pi r^2}{d} \cdot \frac{dB}{dt} $

Substituting the given values:

$ \begin{aligned} a &= \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \times \frac{22}{7} \times \frac{(0.5)^2}{0.3} \times 50 \times 10^{-6} \\ &= 23 \times 10^6 \, \mathrm{ms}^{-2} \end{aligned} $

Induced emf, $\varepsilon = - M{{di} \over {dt}}$

where, i is given as, $i = {i_0}\sin \omega t$

$ \Rightarrow \varepsilon = - M{d \over {dt}}[{i_0}\sin \omega t]$

$ = M\,{i_0}\omega \cos \omega t$

For maximum value of emf in second coil, cos $\omega$t is maximum, i.e., 1.

${E_{\max }} = M\,{i_0}\omega $ [$\because$ $\cos \omega t = 1$]

$ \Rightarrow 0.001 \times 10 \times 10\pi $

$ \Rightarrow {E_{\max }} = 0.1$ $\pi$V

Hence, maximum emf value induced in second coil is 0.1 $\pi$V.

Given, reverse voltage, E = 120 V

Change in current, $\Delta$I = 0.50 $-$ 0.20 = 0.30 A

Time, $\Delta$t = 0.030 ms = 0.030 $\times$ 10^$-$3 s

In an inductor (choke),

Emf induced, ${E_{induced}} = L{{dl} \over {dt}}$

$\therefore$ Self inductance of choke, $L = {{E \times \Delta t} \over {\Delta I}}$

$ = {{120 \times 0.030 \times {{10}^{ - 3}}} \over {0.30}}$

$ = 12 \times {10^{ - 3}}$ H