Q126 / 180 · NEET 2015 (Phase 1) · Physics

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 e.m.f induced in the frame (when the centre of the frame is at a distance x from the wire) will be proportional to :
Figure

  • A
    1x2
  • B
    12x-a2
  • C
    12x+a2
  • Correct answer
    D
    12x-a2x+a

Solution

Figure
The potential difference across AB is
VA-VB=B1 a.V.
⇒  μi2π x-a2 aV
The potential difference across CD is
VC-VD=B2 a.V
B2=μ0i2πx+a2
VC-VD= μ0i2πx+a2 aV
Net Potential difference =μi aV2 π 1x - a2- 1x + a2
( V A − V B )−( V C − V D )= μia 2π ( 2a x 2 − a 2 4 ) ∝  14x2-a2   ∝   12x+a(2x-a)    

Hence, the correct answer is option: 'D'.

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