NEETPhysicsMoving Charges and MagnetismMCQ+4 / −1
A conducting square frame of side 'a' and a long straight wire carrying current 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


- A
- B
- C
- D
View written solutionFree
Correct answer: B

The emf in AD:
e1 = (a × μ0iv)/2π(x −a/2)
The emf in EF :
e2 = (a × μ0i × v)/2π(x + a/2)
Net emf = e1 – e2
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