Question

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:

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

Let 2a be the side of the triangle and b the length AE.

$\frac{AH}{AE}$ = $\frac{GH}{EC}$

\ GH = $\left( \frac{AH}{AE} \right)$EC

= $\frac{b–vt}{b}$.a = a – $\left( \frac{a}{b} \right)vt$

\ FG = 2GH = 2$\left\lbrack a–\frac{a}{b}vt \right\rbrack$

Induced e.m.f., e = Bv(FG) = 2Bv $\left( a–\frac{a}{b}vt \right)$

\ Induced current, I = $\frac{e}{R}$= $\left\lbrack a–\frac{a}{b}vt \right\rbrack$

or I = k₁ – k₂t

Thus, I – t graph is a straight line with negative slope and positive intercept.