The bob of a simple pendulum is replaced by a magnet. The oscillations are set along the length of the magnet. $A$ copper coil is added so that one pole of the magnet passes in and out of the coil. The coil is short-circuited. Then which one of the following happens?

In $SI$,Henry is the unit of

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

The variation of induced emf $(E)$ with time $(t)$ in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as

Two different coils have self-inductance $L_1 = 8 \, mH$ and $L_2 = 2 \, mH$. The current in both coils is increased at the same constant rate. At a certain instant of time,the power supplied to the two coils is the same. At that time,the current,induced voltage,and energy stored in the first coil are $i_1, V_1$,and $W_1$ respectively. Corresponding values for the second coil at the same instant are $i_2, V_2$,and $W_2$ respectively. Then:

$A$ coil having $N$ turns and resistance $R$ $\Omega$ is connected to a galvanometer of resistance $6R$ $\Omega$. The magnetic flux linked with this coil changes from $\phi_1$ weber to $\phi_2$ weber in time $t$ second. The induced current in the circuit is