$$\eqalign{ & i = \frac{e}{R} = \frac{{\frac{{nAdB}}{{dt}}}}{R} \cr & = \frac{{20 \times \left( {25 \times {{10}^{ - 4}}} \right) \times 1000}}{{100}} = 0.5\,A \cr} $$

$$\eqalign{ & \oint {\overrightarrow E } .\overrightarrow {d\ell } = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {\overrightarrow B .\overrightarrow A } \right) = \frac{d}{{dt}}\left( {BA\cos {0^ \circ }} \right) = A\frac{{dB}}{{dt}} \cr & \Rightarrow E\left( {2\pi r} \right) = \pi {a^2}\frac{{dB}}{{dt}}{\text{ for }}r \geqslant a \cr & \Rightarrow E = \frac{{{a^2}}}{{2r}}\frac{{dB}}{{dt}} \Rightarrow E \propto \frac{1}{r} \cr} $$

Induced emf. $$\left| e \right| = \frac{{d\phi }}{{dt}} = A\left( {\frac{{dB}}{{dt}}} \right)$$
$$\eqalign{ & = \frac{{2 \times 2}}{2} \times \frac{d}{{dt}}\left( {0.042 - 0.87\,t} \right) \cr & = 2 \times 0.87 = 1.74\,V \cr} $$
The net induced emf $$ = 20 + 1.74 = 21.74\,V$$

When switch $$S$$ is closed, a magnetic field is set-up in the space around $$P.$$ The field lines threading $$Q$$ increases in the direction from right to left. According to Lenz's law, $${I_{{Q_1}}}$$ will flow so as to oppose the cause and flow in anticlockwise direction as seen by $$E.$$ Reverse is the case when $$S$$ is opened. $${I_{{Q_2}}}$$  will be clockwise.

$$BI\ell = mg\,\,{\text{or}}\,\,B\frac{{Bv\ell }}{R}\ell = mg\,\,{\text{or}}\,\,v = \frac{{mgR}}{{{B^2}{\ell ^2}}}$$

KEY CONCEPT : Initially, $${\phi _B}$$ increases as magnet approaches the solenoid
∴ $$\varepsilon = - ve$$  and increasing in magnitude. When magnet is moving inside the solenoid, increase in $${\phi _B}$$ slow down and finally $${\phi _B}$$ starts decreasing
∴ emf is positive and increasing.
Only graph (C) shows these characteristic.

From Faraday’s second law, emf induced in the circuit $$e = \frac{{\Delta \phi }}{{\Delta t}}$$
If $$R$$ is the resistance of the circuit, then
$$i = \frac{e}{R} = \frac{{\Delta \phi }}{{R\Delta t}}$$
Thus, charge passing through the circuit,
$$\eqalign{ & q = i \times \Delta t \cr & \Rightarrow q = \frac{{\Delta \phi }}{{R\Delta t}} \cdot \Delta t \cr & \Rightarrow q = \frac{{\Delta \phi }}{R} \cr} $$

$$\theta = \omega t.$$  Only half circular part will be involved in inducing emf, so effective area $$A = \frac{{\pi {a^2}}}{2}$$

$$\eqalign{ & \phi = BA\cos \theta \cr & e = - \frac{{d\phi }}{{dt}} \cr & = + BA\sin \theta \left( {\frac{{d\theta }}{{dt}}} \right) \cr & \Rightarrow e = \frac{{B\pi {a^2}}}{2}\omega \sin \theta \cr & I = \frac{e}{R} = \frac{{B\pi {a^2}\omega }}{{2R}}\sin \theta \cr} $$

Initially, when steady state is achieved,
$$i = \frac{E}{R}$$
Let $$E$$ is short circuited at $$t = 0.$$  Then
At $$t = 0,{i_0} = \frac{E}{R}$$
Let during decay of current at any time the current flowing is $$ - L\frac{{di}}{{dt}} - iR = 0$$
$$\eqalign{ & \Rightarrow \frac{{di}}{i} = - \frac{R}{L}dt \Rightarrow \int\limits_{{i_0}}^i {\frac{{di}}{i}} = \int\limits_0^t { - \frac{R}{L}} dt \cr & \Rightarrow {\log _e}\frac{i}{{{i_0}}} = - \frac{R}{L}t \Rightarrow i = {i_0}{e^{ - \frac{R}{L}t}} \cr & \Rightarrow i = \frac{E}{R}{e^{ - \frac{R}{L}t}} = \frac{{100}}{{100}}{e^{\frac{{ - 100 \times {{10}^{ - 3}}}}{{100 \times {{10}^{ - 3}}}}}} = \frac{1}{e} \cr} $$

First current develops in direction of $$abcd$$  but when electron moves away then magnetic field inside loop decreases and current changes its direction.