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Q.

An equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the paper (figure). At time t = 0, side DC of the loop is at edge of the magnetic field.The induced current (i) versus time (t) graph will be as

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a

b

c

d

answer is B.

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Detailed Solution

Let 2a be the side CD of the triangle and b be the length AE.    AHAE=GHEC⇒GH=AHAEEC ⇒   GH=(b−vt)b⋅a=a−abvt∴    FG=2GH=2a−abvt∴     Induced emf, e=Bv(FG)=2Bva−abvt∴     Induced current, i=eR=2BvRa−abvt or     i=k1−k2tThus i-t graph is a straight line with negative slope and positive intercept.
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An equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the paper (figure). At time t = 0, side DC of the loop is at edge of the magnetic field.The induced current (i) versus time (t) graph will be as