Two fixed long straight wires carry the same current `i` in opposite directions as shown in . A square loop of side `b` is fixed in the plane of the wires with its length parallel to one wire at a distance a shown in the figure.

(a) alculate at the induced emf in the loop if the current in both the wires is changing at the rate `di//dt`.
(b) What is the direction of force on the loop if `di//dt` is positive?

Answer

(a) `(mu_(0)b)/(2pi)1n(((a + d)(b + a))/(a(b + a + d)))(di)/(dt)`; (b) away from wires

Explanation

(a)`phi = (mu_(0))/(2pi)ib 1n((b + a)/(a)) - (mu_(0)i)/(2pi)b 1n((b + a + d)/(a + d))`
`phi = (mu_(0)i)/(2pi)b 1n((b + a)(b + a))/(a(b + a + d))`
`e = (dphi)/(dt) = (mu_(0)b)/(2pi)1n(((a + d)(b + a))/(a(b + a + d)))(di)/(dt)`
(b) as `(di)/(dt) gt 0` the direction of induced current is clockwise Therefore, force on the loop will be away from wires.