In a cylindrical region having radius R, magnetic field varies with time as $B = a + b t,$ where a and b are positive constant. OPQ is a triangular loop made of wire having resistance per unit length $λ .$ The current induced in the loop is found to be $\frac{π b R (\sqrt{3} - 1)}{3 (n) \sqrt{3} λ} .$ Find the value of ‘n’

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answer is 4.

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Detailed Solution

$P Q = R \tan 60^{0} = \sqrt{3} R, O P = \frac{R}{\cos 60^{0}} = 2 R$

Total resistance of the loop, $r = (3 + \sqrt{3}) R λ$

Area of the loop enclosing the field, $A = \frac{π R^{2}}{6}$

Induced current, $i = \frac{A}{r} \frac{d B}{d t}$ $= \frac{{πR}^{2}}{6 (3 + \sqrt{3}) R λ} . (b) = \frac{πRb}{12 \sqrt{3} λ}$

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