$A$ flat coil of $500$ turns,each of area $50 \,cm^2$,rotates in a uniform magnetic field of $0.14 \,Wb/m^2$ about an axis normal to the field at an angular speed of $150 \,rad/s$. The coil has a resistance of $5 \,\Omega$. The induced $e.m.f.$ is applied to an external resistance of $10 \,\Omega$. The peak current through the resistance is .......... $A$. (in $.5$)

$A$ very long solenoid of radius $R$ is carrying current $I(t) = kt e^{-\alpha t}$ $(k > 0)$,as a function of time $(t \geq 0)$. Counter-clockwise current is taken to be positive. $A$ circular conducting coil of radius $2R$ is placed in the equatorial plane of the solenoid and is concentric with the solenoid. The induced current in the outer coil as a function of time is correctly depicted by:

In the $LR$ circuit shown,what is the variation of the current $I$ as a function of time? The switch is closed at time $t = 0 \, s.$

$A$ rectangular loop of sides $12 \text{ cm}$ and $5 \text{ cm}$,with its sides parallel to the $x$-axis and $y$-axis respectively,moves with a velocity of $5 \text{ cm/s}$ in the positive $x$-axis direction in a space containing a variable magnetic field in the positive $z$-direction. The field has a gradient of $10^{-3} \text{ T/cm}$ along the negative $x$-direction and it is decreasing with time at the rate of $10^{-5} \text{ T/s}$. If the resistance of the loop is $6 \text{ m}\Omega$,the power dissipated by the loop as heat is . . . . . . $\times 10^{-9} \text{ W}$.

Consider the conducting square loop shown in the figure. If the switch is closed and after some time it is opened again,then the square loop will show:

$A$ source of constant voltage $V$ is connected to a resistance $R$ and two ideal inductors $L_1$ and $L_2$ through a switch $S$ as shown. There is no mutual inductance between the two inductors. The switch $S$ is initially open. At $t=0$,the switch is closed and current begins to flow. Which of the following options is/are correct?
$[A]$ After a long time,the current through $L_1$ will be $\frac{V}{R} \frac{L_2}{L_1+L_2}$
$[B]$ After a long time,the current through $L_2$ will be $\frac{V}{R} \frac{L_1}{L_1+L_2}$
$[C]$ The ratio of the currents through $L_1$ and $L_2$ is fixed at all times $(t>0)$
$[D]$ At $t=0$,the current through the resistance $R$ is $\frac{V}{R}$