$$\eqalign{ & {e_{{\text{back}}}} = L\frac{{di}}{{dt}} \cr & {\text{where}}\,L = \frac{{{\mu _0}{N^2}A}}{l} \cr & \therefore {e_{{\text{back}}}} = \frac{{{\mu _0}{N^2}A}}{l}\left[ {\frac{{1.5 - 0}}{{1 \times {{10}^3}}}} \right] \cr & = \frac{{4\pi \times {{10}^{ - 7}}{{\left( {400} \right)}^2} \times 20 \times {{10}^{ - 4}}}}{{20 \times {{10}^{ - 2}}}} \times \left( {1.5 \times {{10}^3}} \right) \cr & = 0.3\,V \cr} $$

From $$KVL$$  at any time $$t$$

$$\eqalign{ & \frac{q}{c} - iR - L\frac{{di}}{{dt}} = 0 \cr & i = - \frac{{dq}}{{dt}} \Rightarrow \frac{q}{c} + \frac{{dq}}{{dt}}R + \frac{{L{d^2}q}}{{d{t^2}}} = 0 \cr & \frac{{{d^2}q}}{{d{t^2}}} + \frac{R}{L}\frac{{dq}}{{dt}} + \frac{q}{{Lc}} = 0 \cr} $$
From damped harmonic oscillator, the amplitude is given by $$A = {A_0}e - \frac{{dt}}{{2m}}$$
Double differential equation $$\frac{{{d^2}x}}{{d{t^2}}} + \frac{b}{m}\frac{{dx}}{{dt}} + \frac{k}{m}x = 0$$
$${Q_{\max }} = {Q_o}e - \frac{{Rt}}{{2L}} \Rightarrow Q_{\max }^2 = Q_o^2e - \frac{{Rt}}{L}$$
Hence damping will be faster for lesser self inductance.

$$\eqalign{ & {e_{\operatorname{ind} }} = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {BA} \right) = A\frac{{dB}}{{dt}} \cr & = A\frac{d}{{dt}}\left( {4{t^2} + 2t + 5} \right) = A\left( {8t + 2} \right) = 0.06\left( {8t + 2} \right) \cr & = 2.52\,{\text{at}}\,t = 5 \cr & i = \frac{{2 - {e_{{\text{ind}}}}}}{2} = 0.26\,A \cr} $$

Induced emf $$ = v{B_H}l = 1.5 \times 5 \times {10^{ - 5}} \times 2$$
$$ = 15 \times {10^{ - 5}} = 0.15\,mV$$

Apply Lenz's law such that the net force on loop will be towards the infinite long wire.

Consider a small element of length $$dx$$ of the rod at a distance $$x$$ and $$\left( {x + dx} \right)$$   from the wire.
The emf induced across the element
$$de = B\,v\,dx\,.......\left( {\text{i}} \right)$$
We know that magnetic field $$B$$ at a distance $$x$$ from a wire carrying a current $$i$$ is given by
$$B = \frac{{{\mu _0}}}{{2\pi }}.\frac{i}{x}\,.....\left( {{\text{ii}}} \right)$$
From eqs. (i) and (ii),
$$de = \frac{{{\mu _0}i}}{{2\pi x}}v\,dx\,......\left( {{\text{iii}}} \right)$$
The emf $$e$$ induced in the entire length of the rod $$PQ$$  is given by
$$\eqalign{ & e = \int {de} = \int_P^Q {\frac{{{\mu _0}}}{{2\pi }}\frac{i}{x}} v\,dx \cr & = \int_r^{r + L} {\frac{{{\mu _0}}}{{2\pi }}\frac{i}{x}} v\,dx \cr & = \frac{{{\mu _0}}}{{2\pi }}i\,v\int_r^{r + L} {\frac{{dx}}{x}} \cr & = \frac{{{\mu _0}iv}}{{2\pi }}{\log _e}\left( {\frac{{r + L}}{r}} \right) \cr} $$

The induced emf across the sliding wire
$$e = Bv\ell = 0.5 \times 4 \times 0.25 = 0.5\,V$$

The effective circuit is shown in figure.
The equivalent resistance of the circuit
$$\eqalign{ & r = \frac{{4 \times 2}}{{4 + 2}} + 0.5 = 1.83\,\Omega \cr & {\text{Now,}}\,i = \frac{V}{R} = \frac{{0.5}}{{1.83}} = 0.27\,A \cr} $$

Given, length of conductor $$\left( l \right) = 0.4\,m$$
speed $$\left( v \right) = 7\,m/s$$
Magnetic field $$\left( B \right) = 0.9\,Wb/{m^2}$$
Induced emf, $$e = Blv\sin \theta \,\,\left[ {\theta = {{90}^ \circ }\,{\text{as}}\,B\,{\text{is}}\, \bot \,v} \right]$$
$$\eqalign{ & = 0.9 \times 0.4 \times 7 \times \sin {90^ \circ } \cr & = 2.52\,V \cr} $$

Eddy currents are the currents induced in the body of a conductor when the amount of magnetic flux linked with the conductor changes.
e.g. when we move a metal plate out of a magnetic field, the relative motion of the field and the conductor again induces a current in the conductor for which conduction electrons move in closed loops forming circular eddy currents in such a way that it opposes the magnetic field that created it as if the electrons are caught in an eddy or whirlpool. It is also called Foucault current.

$$B = \frac{{{\mu _0}}}{{2\pi }}\frac{M}{{{x^3}}}$$
$$\therefore $$ Flux passing through the coil $$ = B{a^2}$$
$$\therefore $$ The induced emf in the coil
$$\eqalign{ & e = \frac{{ - d\phi }}{{dt}} = - \frac{d}{{dt}}\left[ {\frac{{{\mu _0}M{a^2}}}{2} \times \frac{1}{{{x^3}}}} \right] \cr & = \frac{{3{\mu _0}M{a^2}}}{{2{x^4}}}\frac{{dx}}{{dt}}\left[ {\because v = \frac{{dx}}{{dt}}} \right] \cr & = \frac{{3{\mu _0}M{a^2}}}{{2{x^4}}} \times v \cr} $$
$$\therefore $$ Current in the coil $$ = I = \frac{e}{R} = \frac{{3{\mu _0}M{a^2}v}}{{2{x^4}R}}$$
The magnetic moment of the loop
$$M' = I \times A = \frac{{3{\mu _0}M{a^2}v}}{{2{x^4}R}} \times {\left( {\pi a} \right)^2}$$
Now the potential energy of the loop placed in the magnetic field is
$$\eqalign{ & U = - M'B\cos {180^ \circ } = \frac{{3{\mu _0}M{a^2}v \times \pi {a^2}}}{{2{x^4}R}} \times \frac{{{\mu _0}M}}{{2\pi {x^3}}} \cr & \therefore U = \frac{{3\mu _0^2{M^2}{a^4}v}}{{4R{x^7}}} \cr & {\text{Now,}}\,\,\left| {\vec F} \right| = - \frac{{dU}}{{dx}} \cr & \therefore F = \frac{{21}}{4}\frac{{\mu _0^2{M^2}{a^4}v}}{{R{x^8}}} \cr} $$
Since by Newton’s third law, action and reaction are equal. Therefore, the above calculated force is acting on the magnet. The direction of the force is in $$ - \hat i$$ direction by Lenz’s law.