A copper rod of length 0.19 m is moving with uniform velocity 10 m/s parallel to a long straight wire carrying a current of 5.0 ampere. There rod itself is perpendicular to the wite with 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 from the wrie, then at this position of the element, the field due to the current carrying wire PQ will be

`B=(mu_(0))/(4pi)(2I)/y` into the page
So the emf induced in the element
`de=Bvdh=(mu_(0))/(4pi)(2I)/y vdy`
and hence the emf induced acros 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`
Substituting the given data with `b=(a+l)`,
`e=10^(-7)xx2xx5xx10log_(e)0.20/0.01=10^(-5)xxlog_(e)20`
So `e=10^(-5)xx2.3026xx13010=30muV`