KEY CONCEPT : The charging of inductance given by, $$i = {i_0}\left( {1 - {e^{ - \frac{{Rt}}{L}}}} \right)$$
$$\frac{{{i_0}}}{2} = {i_0}\left( {1 - {e^{ - \frac{{Rt}}{L}}}} \right) \Rightarrow {e^{\frac{{Rt}}{L}}} = \frac{1}{2}$$
Taking log on both the sides,
$$\eqalign{ & - \frac{{Rt}}{L} = \log 1 - \log 2 \cr & \Rightarrow t = \frac{L}{R}\log 2 = \frac{{300 \times {{10}^{ - 3}}}}{2} \times 0.69 \cr & \Rightarrow t = 0.1\,{\text{sec}} \cr} $$

Because of the Lenz's law of conservation of energy.

Energy stored in a current carrying coil is in the form of magnetic field. Total work done by the external source in building up current from zero to $${i_0}$$ is $$W = \frac{1}{2}Li_0^2$$
where, $$L$$ is self-inductance of coil.

Because of the Lenz's law of conservation of energy.

Induced emf in the region is given by
$$\left| e \right| = \frac{{d\phi }}{{dt}}$$
where, $$\phi = BA = \pi {r^2}B$$
$$ \Rightarrow \frac{{d\phi }}{{dt}} = - \pi {r^2}\frac{{dB}}{{dt}}$$
Rate of change of magnetic flux associated with loop 1
$${e_1} = - \frac{{d{\phi _1}}}{{dt}} = - \pi {r^2}\frac{{dB}}{{dt}}$$
Similarly $${e_2} = $$  emf associated with loop 2
$$ = - \frac{{d{\phi _2}}}{{dt}} = 0\,\,\left[ {\because {\phi _2} = 0} \right]$$

Rate of change of area of the loop
$$\eqalign{ & \frac{{dA}}{{dt}} = A,\beta \frac{{dT}}{{dt}} = A.\left( {2\alpha } \right)\frac{{dT}}{{dt}} = \frac{3}{4} \times 2 \times {10^{ - 6}} \times 1 \cr & = 11.5 \times {10^{ - 6}}{m^2}/s \cr & {\text{emf}} = - \frac{{d\phi }}{{dt}} = - \frac{{\beta .dA}}{{dt}} = - 1.5 \times {10^{ - 6}}V \cr} $$
current in the loop $$ = 1.5 \times {10^{ - 6}}A$$
The direction will be anticlockwise as the induced current will try to negate the increase in fluix due to increase in area.

Magnetic flux, $$\phi = BA$$
and magnetic field due to circular coil is $$B = \frac{{{\mu _0}Ni}}{{2R}}$$
As self-inductance, $$L = \frac{{N\phi }}{i}$$
$$\eqalign{ & \therefore L = \frac{N}{i}\left( {BA} \right) = \frac{N}{i}\left( {\frac{{{\mu _0}Ni}}{{2R}}} \right)A \cr & = \frac{{{\mu _0}{N^2}A}}{{2R}} \cr & \therefore L \propto {N^2} \cr} $$

$$\eqalign{ & {\text{We have, }}V = {V_0}\left( {1 - {e^{\frac{{ - t}}{{RC}}}}} \right) \cr & \Rightarrow 120 = 200\left( {1 - {e^{\frac{{ - t}}{{RC}}}}} \right) \cr & \Rightarrow t = RC{\text{ in }}\left( {2.5} \right) \cr & \Rightarrow R = 2.71 \times {10^6}\Omega \cr} $$

Growth in current in $$L{R_2}$$  branch when switch is closed is given by
$$i = \frac{E}{{{R_2}}}\left[ {1 - {e^{ - \frac{{{R_2}t}}{L}}}} \right] \Rightarrow \frac{{di}}{{dt}} = \frac{E}{{{R_2}}}.\frac{{{R_2}}}{L}.{e^{ - \frac{{{R_2}t}}{L}}} = \frac{E}{L}{e^{ - \frac{{{R_2}t}}{L}}}$$
Hence, potential drop across
$$\eqalign{ & L = \left( {\frac{E}{L}{e^{ - \frac{{{R_2}t}}{L}}}} \right)L = E{e^{ - \frac{{{R_2}t}}{L}}} = 12{e^{ - \frac{{2t}}{{400 \times {{10}^{ - 3}}}}}} \cr & = 12{e^{ - 5t}}V \cr} $$

Let current $${i_1}$$ in straight wire be upward. Then the magnetic field due to the straight wire has magnitude $${B_1} = \frac{{{\mu _0}{i_1}}}{{2\pi r}}$$   at distance $$r.$$ In accrodance with right hand rule, $${B_1}$$ points inward to the plane of page. We consider a differential strip of thickness $$dr,$$  area $$d{A_2} = a\,dr.$$   Magnetic flux through area $$dA,d{\phi _B} = {B_1}\left( {a\,dr} \right).$$
Total flux through the loop,
$$\eqalign{ & {\phi _B} = \int {{B_1}adr = } \int_c^{c + b} {\frac{{{\mu _0}{i_1}}}{{2\pi r}}} a\,dr \cr & = \frac{{{\mu _0}{i_1}a}}{{2\pi }}\int_c^{c + b} {\frac{{dr}}{r}} = \frac{{{\mu _0}{i_1}a}}{{2\pi }}\ln \left( {\frac{{c + b}}{c}} \right) \cr} $$
Therefore mutual inductance,
$$M = \frac{\phi }{{{i_1}}} = \frac{{{\mu _0}a}}{{2\pi }}\ln \left( {1 + \frac{b}{c}} \right)$$