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(B) The area of an equilateral triangle of side $a$ is $A = \frac{\sqrt{3}}{4} a^2$.The magnetic flux $\phi$ through the triangle is $\phi = B \cdot A

Vedclass · Accepted Answer

$\left( \frac{a^2 \lambda}{6 R} B_0 \right) e^{-\lambda t}$

Vedclass · Answer

(B) The area of an equilateral triangle of side $a$ is $A = \frac{\sqrt{3}}{4} a^2$.The magnetic flux $\phi$ through the triangle is $\phi = B \cdot A = (B_0 e^{-\lambda t}) \left( \frac{\sqrt{3}}{4} a^2 \right)$.According to Faraday's law,the induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} \left( B_0 e^{-\lambda t} \frac{\sqrt{3}}{4} a^2 \right) = B_0 \lambda e^{-\lambda t} \frac{\sqrt{3}}{4} a^2$.The three resistors $R$ are connected in series in the loop,so the total resistance is $R_{eq} = 3R$.The induced current $I$ is given by $I = \frac{\varepsilon}{R_{eq}} = \frac{B_0 \lambda e^{-\lambda t} \frac{\sqrt{3}}{4} a^2}{3R} = \frac{\sqrt{3} a^2 \lambda B_0 e^{-\lambda t}}{12R} = \frac{a^2 \lambda B_0 e^{-\lambda t}}{4\sqrt{3} R}$.Comparing this with the options,the correct expression is $\left( \frac{a^2 \lambda}{4\sqrt{3} R} B_0 \right) e^{-\lambda t}$.

Three resistances of magnitude $R$ each are connected in the form of an equilateral triangle of side $a$. The combination is placed in a magnetic field $B = B_0 e^{-\lambda t}$ perpendicular to its plane. The induced current in the circuit is given by:

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