Current induced in the coil is given by
$$\eqalign{ & i = \frac{1}{R}\left( {\frac{{d\phi }}{{dt}}} \right) \cr & \Rightarrow \frac{{\Delta q}}{{\Delta t}} = \frac{1}{R}\left( {\frac{{\Delta \phi }}{{\Delta t}}} \right) \cr} $$
Given, resistance of the solenoid,
$$R = 10\,{\pi ^2}\Omega $$
Radius of second and coil $$r = {10^{ - 2}}$$
$$\Delta t = 0.05s,\Delta i = 4 - 0 = 4\,A$$
Charge flowing through the coil is given by
$$\eqalign{ & \Delta q = \left( {\frac{{\Delta \phi }}{{\Delta t}}} \right)\frac{1}{R}\left( {\Delta t} \right) \cr & = {\mu _0}{N_1}{N_2}\pi {r^2}\left( {\frac{{\Delta i}}{{\Delta t}}} \right)\frac{1}{R}\Delta t \cr & = 4\pi \times {10^{ - 7}} \times 2 \times {10^4} \times 100 \times \pi \times {\left( {{{10}^{ - 2}}} \right)^2} \times \left( {\frac{4}{{0.05}}} \right) \times \frac{1}{{10{\pi ^2}}} \times 0.05 \cr & = 32 \times {10^{ - 6}}C = 32\,\mu C \cr} $$

The magnetic field is increasing in the downward direction. Therefore, according to Lenz’s law the current $${I_1}$$ will flow in the direction $$ab$$ and $${I_2}$$ in the direction $$dc.$$

Magnetic flux $$\phi = B \cdot A$$
$$\eqalign{ & \phi = BA\cos \theta \cr & = \frac{1}{\pi } \times \pi {\left( {0.2} \right)^2}\cos {60^ \circ } = 0.02\,Wb \cr} $$

Magnitude of induced electric field due to change in magnetic flux is given by
$$\eqalign{ & \oint {\vec E.d\vec \ell = \frac{{d\phi }}{{dt}} = A\frac{{dB}}{{dt}}\left( {\because N = 1\,{\text{and}}\,\cos \theta = 1} \right)} \cr & {\text{or}}\,\,E.\ell = \pi {R^2}\left( {2{B_0}t} \right)\left( {\frac{{dB}}{{dt}} = 2{B_0}t} \right) \cr} $$
Here, $$E$$ = induced electric field due to change in magnetic flux
$$\eqalign{ & E\left( {2\pi R} \right) = 2\pi {R^2}{B_0}t \cr & {\text{or}}\,\,E = {B_0}Rt \cr} $$
Hence, $$F = QE = {B_0}QRt$$
This force is tangential toring. Ring starts rotating when torque of this force is greater than the torque due to maximum friction $$\left( {{f_{\max }} = \mu mg} \right)$$
or when $${\tau _F} \geqslant {\tau _{{f_{\max }}}}$$
Taking the limiting case, $${\tau _F} \geqslant {\tau _{{f_{\max }}}}\,{\text{or}}\,F.R = \left( {\mu mg} \right)R$$
It is given that ring starts rotating after 2 seconds.
So, putting $$t = 2$$  seconds, we get $$\mu = \frac{{2{B_0}RQ}}{{mg}}$$

When the capacitor is completely charged, the total energy in the $$L.C$$  circuit is with the capacitor and that energy is
$$E = \frac{1}{2}\frac{{{Q^2}}}{C}$$
When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get
$$\frac{E}{2} = \frac{1}{2}\frac{{Q{'^2}}}{C}$$    where $$Q'$$ is the charge on one plate of the capacitor
$$\therefore \frac{1}{2} \times \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}\frac{{Q{'^2}}}{C} \Rightarrow Q' = \frac{Q}{{\sqrt 2 }}$$

$$\phi = Li \Rightarrow NBA = Li$$
Since magnetic field at the centre of circular coil carrying current is given by
$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }},\frac{{2\pi Ni}}{r} \cr & \therefore N.\frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi Ni}}{r}.\pi {r^2} = Li \cr & \Rightarrow L = \frac{{{\mu _0}{N^2}\pi r}}{2} \cr} $$
Hence self inductance of a coil
$$\frac{{4\pi \times {{10}^{ - 7}} \times 500 \times 500 \times \pi \times 0.05}}{2} = 25\,mH$$

The network behaves like a balanced wheatstone bridge.
The induced emf developed is given by
$$\eqalign{ & e = vB\ell = v \times 2 \times 0.1 = 0.2v\,......\left( {\text{i}} \right) \cr & {\text{Now,}}\,\,e = IR \cr & e = {10^{ - 3}} \times 4 = 4 \times {10^3}{\text{amp}}\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii),
$$\eqalign{ & 0.2\,v = 4 \times {10^{ - 3}} \cr & \therefore v = \frac{{4 \times {{10}^{ - 3}}}}{{0.2}} = 0.02\,m/s \cr} $$

Eddy currents are set up inside the plate and the direction of current opposes the motion of the metallic plate swings between the poles of a magnet.

Here, induced e.m.f

$$\eqalign{ & e = \int\limits_{2\ell }^{3\ell } {\left( {\omega x} \right)} Bdx = B\omega \frac{{\left[ {{{\left( {3\ell } \right)}^2} - {{\left( {2\ell } \right)}^2}} \right]}}{2} \cr & = \frac{{5B{\ell ^2}\omega }}{2} \cr} $$

Given, $$N = 50$$
$$\eqalign{ & B = 0.2\,Wb/{m^2},I = 2A \cr & \theta = {60^ \circ },A = 0.12 \times 0.1 = 0.012\,{m^2} \cr} $$

Thus, torque required to keep the coil in stable equilibrium, i.e.
$$\eqalign{ & \tau = NIAB\sin \theta = 50 \times 2 \times 0.012 \times 0.2 \times \sin {60^ \circ } \cr & = 50 \times 2 \times 0.12 \times 0.2 \times \frac{{\sqrt 3 }}{2} = 0.20\,Nm \cr} $$