A rod of length b moves with a constant velocity v in the magnetic field of a infinitely long straight conducting wire that carries a current I as shown in the figure. The induced emf in the rod is

Options

A. `(mu_(0)iv)/(2pi)tan^(-1)((a)/(b))`
B. `(mu_(0)iv)/(2pi)ln(1+(b)/(a))`
C. `(mu_(0)ivsqrt(ab))/(4pi(a+b))`
D. `(mu_(0)iv(a+b))/(4piab)`

Answer

B

Explanation

The induced emf between two ends of a segment dx, de = Bv.dx
[B = magnetic field due to current I in the wire at perpendicular distance `x=(mu_(0)i)/(2pix)`]
`therefore e=intde=(mu_(0)iv)/(2pi)int_(a)^(a+b)(dx)/(x)`
`=(mu_(0)iv)/(2pi)ln(1+(b)/(a))`