Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity `v=10m.s^(-1)`. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Explanation

Induced emf,
`e=(mu_(0))/(2pix).(Ia^(2)v)/(a+x)=(2xx10^(-7)xx50xx(0.1)^(2)xx10)/(0.2(0.1+0.2))`
`=1.67xx10^(-5)~~1.7xx10^(-5)V`