Correct Option is (B)

As the rectangular conductor moves in field area, so flux is increasing up to x = b, then flux is generated on return journey from x = b to x = 0. The flux is shown by plot A of the graph.

As, emf, $e=-\frac{d\varphi }{dt}$, which is shown by curve B and power dissipated, P = VI which is shown by curve C.

Correct answer is 60

Maximum emf $\epsilon =N\omega AB$

N = 20, $\omega$ = 50, B = 3 $\times$ 10^$-$2 T

$\epsilon$ = 20 $\times$ 50 $\times$ $\pi$ $\times$ (0.08)² $\times$ 3 $\times$ 10^$-$2 = 60.28 $\times$ 10^$-$2

Rounded off to nearest integer = 60

Correct answer is 60

$|\in |=\frac{d\varphi }{dt}$ = 20t + 20 mV

$\left|i\right|=\frac{|\in |}{R}$ = 10t + 10 mA

at t = 5

$\left|i\right|$ = 60 mA

Correct answer is 80

$\varphi =\overrightarrow{B}.\overrightarrow{S}$

$\varphi =\frac{4}{\pi }\times {10}^{-3}\left(1-\frac{t}{100}\right).\pi R^2$

$\varphi =4\times {10}^{-3}\times (1{)}^2\left(1-\frac{t}{100}\right)$

$\epsilon =\frac{-d\varphi }{dt}$

$\epsilon =\frac{-d}{dt}\left(4\times {10}^{-3}\left(1-\frac{t}{100}\right)\right)$

$\epsilon =4\times {10}^{-3}\left(\frac{1}{100}\right)=4\times {10}^{-5}V$

When B = 0

$1-\frac{t}{100}=0$

t = 100 sec

Heat $=\frac{{\epsilon }^2}{R}t$

Heat $=\frac{{(4\times {10}^{-5})}^2}{2\times {10}^{-6}}\times 100$ J

Heat $=\frac{16\times {10}^{-10}\times 100}{2\times {10}^{-6}}$ J

Heat = 0.08 J

Heat = 80 mJ

When bar magnet is entering with constant speed, flux will change and an e.m.f. is induced, so galvanometer will deflect in positive direction. When magnet is completely inside, flux will not change, so reading of galvanometer will be zero. When bar magnet is making on exit, again flux will change and on e.m.f. is induced in opposite direction to not of (a), so galvanometer will deflect in negative direction.