Lenz's law is expressed by the following formula (here $e$ = induced $e.m.f.$,$\phi$ = magnetic flux in one turn and $N$ = number of turns).

What will be the direction of current in the coil $A$ as the switch $S$ is closed?

The magnetic flux through a circuit of resistance $R$ changes by an amount $\Delta \phi$ in the time $\Delta t$. The total quantity of electric charge $Q$ which passes during this time through any point of the circuit is

$A$ conducting coil having $500$ turns has a cross-sectional area of $0.15 \ m^2$. $A$ magnetic field of strength $0.2 \ T$ linked perpendicular to this area changes to $1.0 \ T$ in $0.4 \ s$. The induced emf produced in the coil will be . . . . . . $V$.

$A$ long solenoid of radius $2\, cm$ has $100\, turns/cm$ and carries a current of $5\, A$. $A$ coil of radius $1\, cm$ having $100\, turns$ and a total resistance of $20\, \Omega$ is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction,find the charge flown through the galvanometer.

There are two square loops $A$ and $B$. When $A$ moves towards $B$,a current starts flowing in $B$ as shown in the figure,and the current in $B$ stops when $A$ stops moving. From this,we can infer that (Assume loop $B$ is at rest):