$$\eqalign{ & \phi = \frac{{\sqrt 3 }}{4}{\ell ^2}B \cr & \varepsilon = \left| {\frac{{d\phi }}{{dt}}} \right| = \frac{{\sqrt 3 }}{4}{\ell ^2}\frac{{dB}}{{dt}} \cr & i = \frac{\varepsilon }{R} = \frac{{\sqrt 3 {\ell ^2}}}{{4R}} \cr} $$

Induced emf $$ = \int\limits_a^b {Bv} dx = \int\limits_a^b {\frac{{{\mu _0}I}}{{2\pi x}}} vdx$$
$$ \Rightarrow $$ Induced emf $$ = \frac{{{\mu _0}Iv}}{{2\pi }}\ln \left( {\frac{b}{a}} \right)$$
$$ \Rightarrow $$ Power dissipated $$ = \frac{{{E^2}}}{R}$$
Also, power $$ = F.v \Rightarrow F = \frac{{{E^2}}}{{vR}}$$
$$ \Rightarrow F = \frac{1}{{vR}}{\left[ {\frac{{{\mu _0}Iv}}{{2\pi }}\ln \left( {\frac{b}{a}} \right)} \right]^2}$$

When the contact is suddenly broken, self induced current flow in the direction of main current. Therefore, the bulb $$B$$ will become suddenly bright.

each spoke will behave as a cell induced emf of each spoke $$ = \frac{1}{2}B{l^2}\omega .$$   But all $$\left( n \right)$$  identical spoke behaving as a cell can connected in a parallel fashion, so the resultant emf $$ = e$$
$$ = \frac{1}{2}B{l^2}\omega = $$    same as single cell.

Mutual conductance depends on the relative position and orientation of the two coils.

When a coil of $$N$$ number of turns and area $$A$$ is rotated in external magnetic field $$B,$$ magnetic flux linked with the coil changes and hence an emf is induced in the coil. At this instant $$t,$$ if $$e$$ is the emf induced in the coil, then
$$e = - \frac{{d\phi }}{{dt}} = - \frac{d}{{dt}}\,\,\left( {NBA\cos \omega t} \right)$$
Magnetic flux $$\phi = NBA\cos \omega t$$
$$\eqalign{ & \therefore e = - NBA\frac{d}{{dt}}\left( {\cos \omega t} \right) \cr & = NBA\omega \sin \omega t \cr} $$
The induced emf will be maximum, when
$$\eqalign{ & \sin \omega t = {\text{maximum}} = 1 \cr & \therefore {e_{\max }} = {e_0} = NBA\omega \cr} $$
So, alternating emf induced is
$$e = {e_0} = \sin \omega t$$
Maximum current $${i_0} = \frac{{{e_0}}}{R} = \frac{{NBA\omega }}{R}$$
Given, $$N = 1,B = {10^{ - 2}}T$$
$$\eqalign{ & A = \pi {\left( {0.3} \right)^2}{m^2},R = {\pi ^2}\Omega \cr & f = \frac{{200}}{{60}}{s^{ - 1}}\,\,{\text{and}}\,\,\omega = 2\pi \left( {\frac{{200}}{{60}}} \right) \cr & \therefore {i_0} = \frac{{1 \times {{10}^{ - 2}} \times \pi {{\left( {0.3} \right)}^2} \times 2\pi \times 200}}{{60 \times {\pi ^2}}} \cr & = 6 \times {10^{ - 3}}A \cr & = 6\,mA \cr} $$

Velocity at any time $$t,$$
$$v = a\,t = \frac{F}{m}t$$
Now, $$e = {V_P} - {V_Q} = Bv\ell \sin {30^ \circ } = B\left( {\frac{{Ft}}{m}} \right)\ell \times \frac{1}{2}$$
$$ = \frac{{BF\ell t}}{{2m}}$$

When the current in the loop $$A$$ increases, the magnetic lines of force in loop $$B$$ also increases as loop $$A$$ is placed near loop $$B.$$ This induces an emf in $$B$$ in such a direction that current flows opposite in $$B$$ (as compared to $$A$$).
Since currents are in opposite direction, the loop $$B$$ is repelled by loop $$A.$$