Two parallel wires AL and KM placed at a distance I are connected by a resistor R and placed in a magnetic field B which is perpendicular to the plane containing the wire as shown in figure-5.340. Another wire CD now connects the two wires perpendicularly and made to slide with velocity v. Calculate the workdone needed to slide the wire CD. Neglect the resistance of all the wires.

Explanation

When a rod of length l moves in a magnetic field with velocity v as shown in figure, an emf `epsi=Bvl` will be induced in it. Due to this induced emf, a current

`i=(epsi)/(R)=(Bvl)/(R)` will flow in the circuit as shown in figure. Due to this induced current, the wire will experience a magnetic force
`F_(M)=Bil=(B^(2)l^(2)v)/(R)`
which will oppose its motion, So to maintain the motion of the wire CD, a force `F=F_(M)` must be applied in the direction of motion.
The work done per second, i.e., power needed to slide the wire is given by
`P=(dW)/(dt)=Fv=F_(M)" "v=(B^(2)v^(2)l^(2))/(R)`
The power delivered by the external agent is converted into joule heating in a by using the circuit(as shwon above). It means magnetic field helps in converting the mechanical energy into joule heating.