AIPMT | 2015Current Electricity and Electrical InstrumentEasy
Question
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 :
Options
A.
B.
C.
D.
Solution
emf Induced in side (1)
ε1 = B1Vl
emf Induced is side (2)
ε2 = B2Vl
emf in the frame = B1Vl − B2Vl
ε = Vl [B1 − B2]
⇒ ε ∝ B1 − B2 Since B ∝
So ε ∝
⇒ ε ∝
Hence Option (3)
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