PQ and MN are two parallel conductors at a distance l apart and connected by a resistance R as shown in figure. They are placed in a magnetic field B which is perpendicular to the plane of the conductors. A wire XY is placed over PQ and MN and then made to slide over PQ and MN with a velocity v. Neglecting the resistance of PQ, MN and wire XY, calculate the work done per second to slide the wire XY.

Explanation

When a wire slides in a magnetic field on two conducting wires an emf is induced in the wire and current flows through the wire, hence a force acts which is opposite to the motion of the wire. Consider the diagram in which wire XY moves as shown in figure.
By Faraday.s law of induced emf, `E=vBl`
If resistance of loop is R, then `I=(vBl)/(R )`
`therefore` Magnetic force `=IBlsin90^(@)`
`=((vBl)/(R))Bl=(vB^(2)l^(2))/(R)`
`therefore` External force must be equal to magnetic force and in opposite directions.
`therefore` External force = `(vB^(2)l^(2))/(R)`
Due to flow of current, there is a loss of energy per unit time power in the circuit.
Power loss,
`P=I^(2)R=((vBl)/(R))^(2)xxR=(v^(2)B^(2)l^(2))/(R^(2))xxR`
`P=(v^(2)B^(2)l^(2))/(R)`
This loss of power energy per unit time must be compensated by external agent to maintain constant speed of the rod.