devarshi-dt-logo

Question:

A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 m/s (see figure). The e.m.f induced in the frame at the time the left arm is at x = 10 cm from the wire is :

2μV

1μV

0.5μV

0.75μV

Solution:

In given fig x = 15 because left arm of the frame is at 10 cm from the wire. and a = 10 cm.
emf in AD → e1 → aμoiv/2π(x−a/2)
emf in EF → e2 → aμoiv/2π(x+a/2)
Net emf = e1 − e2 = aμoiv/2π(x−a/2) − aμoiv/2π(x+a/2)
putting the given values into the equation we get,
emf induced = 1.3μV ≈ 1μV