Emf induced in the coil due to change in magnetic flux
$$e = - \frac{{d\phi }}{{dt}} = - \frac{{\left( {{\phi _2} - {\phi _1}} \right)}}{{dt}}$$
When magnetic field is perpendicular to coil
$${\phi _1} = NBA$$
When coil is removed, $${\phi _2} = 0$$
So, $$e = - \frac{{\left( {0 - NBA} \right)}}{{dt}}\,\,{\text{or}}\,\,dt = \frac{{NBA}}{e}$$
Here, $$N = 50,B = 2 \times {10^{ - 2}}T,A = 100\,c{m^2}$$
$$\eqalign{ & = {10^{ - 2}}{m^2}\,{\text{and}}\,\,e = 0.1\,V \cr & \therefore dt = \frac{{50 \times 2 \times {{10}^{ - 2}} \times {{10}^{ - 2}}}}{{0.1}} \cr & = 0.1\,s \cr} $$

$$\eqalign{ & \phi \left( {{\text{flux}}\,{\text{linked}}} \right) = {a^2}B\cos {0^ \circ } - {b^2}B\cos {180^ \circ } \cr & E = - \frac{{d\phi }}{{dt}} = - \left( {{a^2} - {b^2}} \right)\frac{{dB}}{{dt}} \cr & = \left( {{a^2} - {b^2}} \right){B_0}\,\omega \cos \omega t \cr} $$
where $$B = {B_0},\sin \omega t,{B_0} = {10^{ - 3}}T,\omega = 100$$
$$\eqalign{ & \therefore {I_{\max }} = \left( {{a^2} - {b^2}} \right)\frac{{{B_0}\omega }}{R} \cr & {\text{and}}\,R = \left( {4a + 4b} \right)r = 4\left( {a + b} \right)r \cr & \therefore {I_{\max }} = \frac{{\left( {a - b} \right){B_0}\omega }}{{4r}} = \frac{{\left( {1 - 0.4} \right) \times {{10}^{ - 3}} \times 100}}{{4 \times 5 \times {{10}^{ - 3}}}} = 3A \cr} $$

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

These three inductors are connected in parallel. The equivalent inductance $${L_p}$$ is given by
$$\eqalign{ & \frac{1}{{{L_p}}} = \frac{1}{{{L_1}}} + \frac{1}{{{L_2}}} + \frac{1}{{{L_3}}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \cr & \therefore {L_p} = 1 \cr} $$

$${\text{For}}\,A:{e_A} = \frac{1}{2}B\omega {\left( {3\ell } \right)^2}\,{\text{and}}\,{e_B} = \frac{1}{2}B\left( { - \omega } \right){\left( {2\ell } \right)^2}$$
Resultant induced emf will be
$$e = {e_A} + {e_B} = \frac{1}{2}B\omega {\ell ^2}\left( {9 - 4} \right)\,{\text{or}}\,e = \frac{5}{2}B\omega {\ell ^2}$$

As we know that,
emf induced, $$e = - L\frac{{di}}{{dt}}$$
During 0 to $$\frac{T}{4},\frac{{di}}{{dt}} = {\text{constant}}$$
$$\eqalign{ & {\text{So,}}\,\,e = {\text{negative}} \cr & {\text{For}}\,\,\frac{T}{4}\,{\text{to}}\,\,\frac{T}{2},\frac{{di}}{{dt}} = 0 \cr & {\text{So,}}\,\,e = 0 \cr & {\text{For}}\,\,\frac{T}{4}\,\,{\text{to}}\,\frac{{3T}}{4},\frac{{di}}{{dt}} = {\text{constant}} \cr & {\text{So,}}\,\,e = {\text{positive}} \cr} $$

Magnetic flux $$\phi $$ linked with magnetic field $$B$$ and area $$A$$ is given by
$$\eqalign{ & \phi = B \cdot A = \left| B \right|\left| A \right|\cos \theta \cr & {\text{Here,}}\,\,\theta = {0^ \circ } \cr & {\text{So,}}\,\,\phi = BA = B\pi {r^2} \cr} $$
Now, Induced emf, $$\left| e \right| = \left| {\frac{{ - d\phi }}{{dt}}} \right| = B\pi \left( {2r} \right)\frac{{dr}}{{dt}}$$
$$\eqalign{ & = 0.025 \times \pi \times 2 \times 2 \times {10^{ - 2}} \times 1 \times {10^{ - 3}} \cr & = \pi \mu V \cr} $$

It is given that emf is zero i.e.,
$$\eqalign{ & e = - L\frac{{di}}{{dt}} = 0 \cr & {\text{or}}\,\,L\frac{{di}}{{dt}} = 0 \cr & {\text{or}}\,\,\frac{d}{{dt}}\left( {{t^2}{e^{ - t}}} \right) = 0\,\,\left( {{\text{As,}}\,\,i = {t^2}{e^{ - t}}} \right) \cr & {\text{or}}\,\,2t \times {e^{ - t}} + {t^2} \times \left( { - 1} \right){e^{ - t}} = 0 \cr & {\text{or}}\,\,t{e^{ - t}}\left( {2 - t} \right) = 0\,\,{\text{or}}\,\,t = 2\,s\,\,\,\left( {\because t{e^{ - t}} \ne 0} \right) \cr} $$

Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage out of these loops, from the field region will not remain constant.

Parallel $$ = \frac{{{L_1}{L_2}}}{{{L_1} + {L_2}}}$$