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

In Fig. length of rod `AB = 0.17 m`.
Velocity or rod, `upsilon = 10 m//s`
`i= 5 A`
Consider a small element of the rod with width dx at a distance x form the wire. Magnetic field intensity at a distance x from the wire,
`B = (mu_(0) i)/(2 pi x)`
Small amount of e.m.f. induced in the element,
`dE = Blv = (mu_(0) i)/(2 pi x) (dx) v`
`:.` e.m.f. induced in the rod,
`E = int_(x = 0.01 m)^(x = 0.2 m) (mu_(0) i upsilon)/(2 pi x) dx = (mu_(0) i upsilon)/(2 pi) [log_(e) x]_(0.01 m)^(0.2 m)`
`= (mu_(0) i upsilon)/(2 pi) ["log"_(e) (0.20)/(0.01)]`
`= (4 pi xx 10^(-7) xx 5 xx 10)/(2 pi) log_(e) 20`
`= 10^(-5) xx 2.303 xx 1.3010`
`E = 2.996 xx 10^(-5)` volt