A plane loop is shaped in the form as shown in Fig. with radii $a = 20 cm$ and $b = 10 cm$ and is placed in a uniform time varying magnetic field $B = B_{0} \sin ω t$ where $B_{0} = 10 mT and ω = 100 rad/s$ . Find the amplitude for the current induced in the loop if its resistance per unit length is equal to $50 \times 10^{- 3} Ω / m$ . The inductance of the loop is negligible.

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$1 A$

$2 A$

$8 A$

Zero

answer is A.

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Detailed Solution

Instantaneous flux through loop is
$ϕ = π a^{2} B \cos 0 ° + π b^{2} B \cos 180 °$
$= π (a^{2} - b^{2}) B$
$= π (a^{2} - b^{2}) B_{0} \sin ω t$
differentiating with respect to time,
$\frac{d ϕ}{d t} = π (a^{2} - b^{2}) B_{0} ω \cos ω t$
$i = \frac{π (a^{2} - b^{2}) B_{0} ω \cos ω t}{R}$ (only the modulus)
$i_{m a x} = \frac{π (a^{2} - b^{2}) B_{0} ω}{2 π (a + b) ρ}$ , $ρ =$ resistance per unit length
$= \frac{1}{2 ρ} (a - b) B_{0} ω$
$= \frac{1}{2} (20 - 10) \times 10^{2} \times 10 \times 10^{- 3} \times \frac{100}{50 \times 10^{- 3}}$
$i_{m a x} = 1 A$ Ans.