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

Explanation

As shown in figure (a) consider an element of length dy at a distance y from the wire, then at this position of the element, the field due to the cuurent - carrying wire PQ will be

`B=(mu_(0))/(4pi)(2I)/(y)` into the page.
So the emf induced in the element
`"de = Bv dy"=(mu_(0))/(4pi)(2I)/(y)" v dy"`
and hence the emf induced across the ends of the rod due to its motion in the field of the wire,
`e=int_(a)^(b)de=(mu_(0))/(4pi)2Ivint_(a)^(b)(dy)/(y)`
i.e., `e=(mu_(0))/(4pi)2Iv" log"_(e)(b/a)`
Subsitituting the given data with `b=(a+l)`,
`e=10^(-7)xx2xx5xx10" log"_(e)(0.20)/(0.01)=10^(-5)xxlog_(e)20`
So, `e=10^(-5)xx2.3026xx1.3010~=30muV`