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

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a

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b

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c

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d

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answer is B.

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Detailed Solution

Let 2a be the side of the triangle and b the length AE.
$\begin{array}{l} \frac{AH}{AE} = \frac{GH}{EC} \\ ∴ GH = (\frac{AH}{AE}) EC = \frac{b - vt}{b} \cdot a = a - (\frac{a}{b}) vt ∴ FG = 2 GH = 2 [a - \frac{a}{b} vt] \end{array}$

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Induced e.m.f., $e = Bv (FG) = 2 Bv (a - \frac{a}{b} vt)$
$∴$ Induced current, $I = \frac{e}{R} = \frac{2 Bv}{R} [a - \frac{a}{b} vt] or I = k_{1} - k_{2} I$
Thus, $I - l$ graph is a straight line with negative slope and positive intercept.

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