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Q.

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 :

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a

12x-a2

b

12x+a2

c

12x-a2x+a

d

1x2

answer is C.

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Detailed Solution

emf induced in side (I)ε1 = B1 Vlemf induced in side (2)ε2 = B2 Vlemf in the frame = B1 Vl - B2 Vlε=VlB1-B2 ⇒ε∝B1-B2 Since, B∝1r,So, ε∝1x-a2-1x+a2⇒ε∝12x-a-12x+a
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