Due to the movement of resistor $$R,$$ an emf equal to $$Blv$$  will be induced in it as shown in figure clearly,
$$I = {I_1} + {I_2}$$
$${\text{Also,}}\,{I_1} = {I_2}$$
Solving the circuit, we get
$$\eqalign{ & {I_1} = {I_2} = \frac{{Blv}}{{3R}} \cr & {\text{and}}\,I = 2{I_1} = \frac{{2Blv}}{{3R}} \cr} $$

$$\eqalign{ & I\left( 0 \right) = \frac{{15 \times 100}}{{0.15 \times {{10}^3}}} = 0.1A \cr & I\left( \infty \right) = 0 \cr & I\left( t \right) = \left[ {I\left( 0 \right) - I\left( \infty \right)} \right]{e^{\frac{{ - t}}{{\frac{L}{R}}}}} + i\left( \infty \right) \cr & I\left( t \right) = 0.1{e^{^{\frac{{ - t}}{{\frac{L}{R}}}} = 0.1{e^{\frac{R}{L}}}}} \cr & I\left( t \right) = 0.1\quad {e^{\frac{{0.15 \times 1000}}{{0.03}}}} = 0.67mA \cr} $$

$$\eqalign{ & {\varepsilon _{{\text{ind}}}} = Bv\ell = 0.3 \times {10^{ - 4}} \times 5 \times 20 \cr & = 3 \times {10^{ - 3}}V \cr & = 3\,mV. \cr} $$

Potential difference across $$PQ$$  is
$${V_P} - {V_Q} = {B_1}\left( a \right)v = \frac{{{\mu _0}I}}{{2\pi \left( {x - \frac{a}{2}} \right)}}av$$
Potential difference across side $$RS$$  of frame is
$${V_S} - {V_R} = {B_2}\left( a \right)v = \frac{{{\mu _0}I}}{{2\pi \left( {x + \frac{a}{2}} \right)}}av$$
Hence, the net potential difference in the loop will be
$${V_{{\text{net}}}} = \left( {{V_P} - {V_Q}} \right) - \left( {{V_S} - {V_R}} \right)$$
$$\eqalign{ & = \frac{{{\mu _0}iav}}{{2\pi }}\left[ {\frac{1}{{\left( {x - \frac{a}{2}} \right)}} - \frac{1}{{\left( {x + \frac{a}{2}} \right)}}} \right] \cr & = \frac{{{\mu _0}iav}}{{2\pi }}\left( {\frac{a}{{\left( {x - \frac{a}{2}} \right)\left( {x + \frac{a}{2}} \right)}}} \right) \cr} $$
Thus, $${V_{{\text{net}}}} \propto \frac{1}{{\left( {2x - a} \right)\left( {2x + a} \right)}}$$

The induced emf across the ends $$B$$ and $$F$$ due to motion of the loop,
$${e_1} = Bv\left( {BF} \right) = 5 \times 1 \times 2\sin {60^ \circ } = 5\sqrt 3 \,V.$$
The induced emf across the loop due to change in magnetic field
$$\eqalign{ & {e_2} = A\frac{{dB}}{{dt}} = \frac{{\pi {R^2}}}{3}\left( {\frac{{dB}}{{dt}}} \right) = \frac{{\pi {{\left( 1 \right)}^2}}}{3} \times 2 = \frac{{2\pi }}{3}V. \cr & {\text{So}}\,\,e = {e_1} + {e_2} = \left( {5\sqrt 3 + \frac{{2\pi }}{3}} \right)V. \cr} $$

$${\phi _{{\text{total}}}} = {B_{{\text{large}}}}{A_{{\text{small}}}} = \frac{{{\mu _0}}}{{4\pi }}\frac{i}{{\frac{L}{2}}}\left( {2\sin {{45}^ \circ }} \right) \times {\ell ^2}$$
On comparing with $${\phi _{{\text{total}}}} = Mi,$$   we get $$M \propto \frac{{{\ell ^2}}}{L}$$

$$Q = \frac{{\Delta \phi }}{R} = \frac{{{\phi _2} - {\phi _1}}}{R} = \frac{{BA - 0}}{R} = \frac{{BA}}{R}$$

The back e.m.f. in a motor is induced e.m.f., which is maximum, when speed of rotation of the coil is maximum.

By Lenz's law, clockwise current is induced in both loops. Greater the area, large will be the induced emf. Outer loop has greater area.

$$\eqalign{ & M = \frac{{{\mu _0}{\mu _r}{N_1}{N_2}}}{{2{R_1}}}A \cr & = \frac{{4\pi \times {{10}^{ - 4}} \times 800 \times 500 \times 25 \times 12 \times {{10}^{ - 4}}}}{{2 \times 0.1}} = 0.24\,H \cr} $$