Question

A conducting square frame of side ‘a’ and a long staight
wire carrying current I are located in the same plane as
shown in the figure. The frame moves to the right with a
constant velocity ‘V’. The emf induced in the frame will be
proportional to [2015]
(a) 1/(2x – a)² (b) 1/(2x + a)²
(c) 1/(2x – a)(2x + a) (d) 1/x²

Answer 1

Induced emf ε=B1av−B2av=2π(x−2aμ0Iav−2π(x+2a)μ0I)av=2πμ0Iav(x−2a1−x+2a1)=2πμ0Iav(2x−a)(2x+a)a

⇒B∝(2x−a)(2x+a)1