Two parallel long straight conductors lie on | vedclass

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Question

(A) Let the side length of the square at any time $t$ be $x$. The area of the square is $A = x^2$.The magnetic flux linked with the loop is $\phi = B

Vedclass · Accepted Answer

$\frac{Bv}{r}$

Vedclass · Answer

(A) Let the side length of the square at any time $t$ be $x$. The area of the square is $A = x^2$.The magnetic flux linked with the loop is $\phi = B \cdot A = B x^2$.The induced electromotive force $(EMF)$ is given by Faraday's law: $e = -\frac{d\phi}{dt} = -\frac{d}{dt}(B x^2) = -2Bx \frac{dx}{dt}$.Since each side moves outward with velocity $v$,the rate of change of the side length is $\frac{dx}{dt} = 2v$ (because both ends of the side move away from the center).Thus,the magnitude of the induced $EMF$ is $e = 2Bx(2v) = 4Bxv$.The total length of the wire forming the square is $L = 4x$.The total resistance of the circuit is $R = L \cdot r = 4xr$.The induced current is $I = \frac{e}{R} = \frac{4Bxv}{4xr} = \frac{Bv}{r}$.

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