A long straight wire is parallel to one edge | vedclass

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Question

(A) Consider a strip of width $dx$ in the loop at a distance $x$ from the wire. The magnetic field $B$ at this distance is $B = \frac{\mu_0 I}{2\pi x}

Vedclass · Accepted Answer

$\frac{\mu_0 b I}{2\pi 	au} \ln \left( \frac{d+a}{d} ight)$

Vedclass · Answer

(A) Consider a strip of width $dx$ in the loop at a distance $x$ from the wire. The magnetic field $B$ at this distance is $B = \frac{\mu_0 I}{2\pi x}$.The magnetic flux $d\phi$ through this strip of area $dA = b \cdot dx$ is:$d\phi = B \cdot dA = \frac{\mu_0 I}{2\pi x} (b \cdot dx) = \frac{\mu_0 I b}{2\pi x} dx$Integrating from $x = d$ to $x = d+a$ to find the total flux $\phi$ through the loop:$\phi = \int_{d}^{d+a} \frac{\mu_0 I b}{2\pi x} dx = \frac{\mu_0 I b}{2\pi} [\ln x]_{d}^{d+a} = \frac{\mu_0 I b}{2\pi} \ln \left( \frac{d+a}{d} ight)$Given $I = I_0 e^{-t/	au}$,the flux is $\phi(t) = \frac{\mu_0 b I_0}{2\pi} \ln \left( \frac{d+a}{d} ight) e^{-t/	au}$.The induced $emf$ is given by Faraday's law,$\varepsilon = -\frac{d\phi}{dt}$:$\varepsilon = -\frac{d}{dt} \left[ \frac{\mu_0 b I_0}{2\pi} \ln \left( \frac{d+a}{d} ight) e^{-t/	au} ight]$$\varepsilon = -\frac{\mu_0 b I_0}{2\pi} \ln \left( \frac{d+a}{d} ight) \left( -\frac{1}{	au} ight) e^{-t/	au}$$\varepsilon = \frac{\mu_0 b I_0 e^{-t/	au}}{2\pi 	au} \ln \left( \frac{d+a}{d} ight) = \frac{\mu_0 b I}{2\pi 	au} \ln \left( \frac{d+a}{d} ight)$

$A$ long straight wire is parallel to one edge of a rectangular loop as shown in the figure. If the current in the long wire varies with time as $I = I_0 e^{-t/\tau}$,what will be the induced $emf$ in the loop?

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