When the North-pole of a bar magnet is moved towards a closed loop like a coil connected to a galvanometer, the magnetic flux through the coil increases. Hence according to Lenz's law, current is induced in the coil in such a direction that it opposes the increase in flux.This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet.Note that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet. Similarly, if the North pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux,the induced current in the coil flows in clockwise direction and its South pole faces the receding North-pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux.

$F=I(l\times B)=\frac{B^2{l}^{2}v}{r}$
where B is the magnetic field,r is the overall resistance of the loop.
Example: A rectangular loop with a sliding connector $CD$ of length $l=1.0$ m is situated in uniform and constant magnetic field $B=2T$ perpendicular to the plane of loop. Resistance of connector $CD$ is $r=2\Omega$. Two resistances of $6\Omega$ and $3\Omega$ are connected as shown in figure. The external force required to keep the connector moving with a constant velocity $v=2m/s$ perpendicular to $CD$ and in the plane of the loop is :
Since resistances 3 $\Omega$ and 6 $\Omega$ are in parallel, tnhe equivalent resistance $R_{eq}=\frac{3\times 6}{3+6}=2\Omega$
Motional emf $\epsilon$ through the rod $CD$ is $\epsilon =Blv$ where $B$ is the magnetic field, $l$ is the length of the rod and $v$ is the velocity of the rod.
Current $I$ through the rod is $I=\frac{Blv}{R_{eq}+r}=\frac{2\times 1\times 2}{2+2}=1A$
Force on the rod $F=Bil=2\times 1\times 1=2N$

$P=\frac{B^2{l}^2{v}^2}{R}$
where B is the magnetic field,
l is the length of the conductor
v is the velocity of the conductor
R is the resistance

Eddy currents are used to advantage in certain applications like:
(i) Magnetic braking in trains: Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth.
(ii) Electromagnetic damping: Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.
(iii) Induction furnace: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it.
(iv) Electric power meters: The shiny metal disc in the electric power meter(analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil.
Disadvantages of eddy current include:
1. Heat is lost in the core of transformers due to eddy currents.
2. The value of eddy current is highly sensitive to the value of permeability. This causes difficulty in the functioning of detectors.

A straight solenoid of length $1$ m has $5000$ turns in the primary and $200$ turns in the secondary. If the area of cross section is $4c{m}^2$, the mutual inductance will be:
$M=\frac{\mu N_1{N}_{2}A}{l}=\frac{4\times \pi \times 10^{-7}T\times 5000\times 200\times 4\times 10^{-4}}{1}=502.4\times {10}^{-6}H$

Example:
A long wire carries a current $5A$. Find the energy stored in the magnetic field inside a volume $1mm$${}^3$ at a distance $10cm$ from the wire.
Solution:
$U$ (energy per unit volume)$=\frac{B^2}{2{\mu }_0}$ and 

Energy $U=\frac{B^2}{2{\mu }_0}\times vol.$
$B=\frac{{\mu }_{0}I}{2\pi d}$

$U={\left(\frac{{\mu }_{0}I}{2\pi d}\right)}^2\times \frac{1}{2{\mu }_0}\times vol.$

$=\frac{{\mu }_0{I}^2}{8{\pi }^2{d}^2}\times vol.$

$=\frac{4\pi \times 10^{-7}\times 25\times 10^{-9}}{8\times 10(10^{-2})}=\frac{\pi }{8}\times 10^{-13}J$

In an ac generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. If coil of $N$ turn and area $A$ is rotated at $v$ revolutions per second in a uniform magnetic field $B$, then the motional emf produced is
$ϵ=NBA(2\pi v)sin(2\pi vt)$
where we have assumed that at time t = 0 s, the coil is perpendicular to the field.