A metal rod PQ is resting on the rails A'B' and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = `9.0 mOmega`. Assume the field to be uniform.

Suppose K is open and the rod is moved with a speed of 12 cm. `s^(-1)` in the direction shown. Give the polarity and magnitude of the induced emf.

Explanation

Induced emf, `e=Bvl`.
`therefore e=0.50xx12xx10^(-2)xx15xx10^(-2)=9xx10^(-3)V " "[because B=0.50T,v=12xx10^(-2)m.s^(-1),l=15xx10^(-2)m]`
The electrons of the rod will experience a force along PQ. Therefore, P end is positive and Q end is negative.