A copper rod of length `0.19m` is moving with uniform velocity `10ms^(-1)` parallel to a long straight wire carrying a current of `5.0A`. The rod is perpendicular to the wire with its ends at distances `0.01` and `0.2m` from it. Calculate the emf induced in the rod.

Explanation

As shown in Fig.3.55, consider an element of length `dy` at a distance `y` from the wire, then at this position of the element, the magnetic field due to the current - carrying wire`PQ` will be `B = (mu_(0))/(4pi)(2I)/(y)` into the plane of the paper.
So, the emf induced in the element `d epsilon = Bv dy = (mu_(0))/(4pi)(2I)/(y) vdy` and hence the emf induced across the ends of the rod due to its motion in the field of the wire,
`epsilon = int_(a)^(b) depsilon = (mu_(0))/(4pi)2Iv int_(a)^(b)(dy)/(y)`, i.e., `epsilon = (mu_(0))/(4pi)2Iv log_(e) ((b)/(a))`
Substituting the given data with `b = (a + l)`,
`epsilon = 10^(-7) xx 2 xx 5 xx 10 log_(e) = (0.20)/(0.01) = 10^(-5) xx log_(e) 20`
`= 30muV`