A conducting square frame of side 'a' and a long straight 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 [figure]
- 1(2x+a)2
- 1(2x-a)(2x+a)
- 1x2
- 1(2x-a)2
Answer
(2) 1(2x-a)(2x+a)
Solution
The left side is at distance x - a2 and the right side at x + a2 from the wire.
Field of the long wire: B = 0 I2 r
Net emf = B1 a v - B2 a v
= 0 I a v2 [ 1x - a2 - 1x + a2 ]
= 0 I a v2 4a(2x-a)(2x+a)
emf 1(2x-a)(2x+a)
Answer: (2)