Question

The rectangular loop in FIGURE has $0.020\Omega$resistance. What is the induced current in the loop at this instant?

Short answer

Inductance,$I_{\text{induced}}=1\mathrm{mA}$

Step-by-step solution

Introduction

The Biot-Savart law permits us to estimate the magnitude and direction of a magnetic field produced by a current-carrying wire. at any point where, as a function of the segment, the magnetic flux$\Delta \overrightarrow{s}$of current-carrying wire is given by equation

$B=\frac{{\mu }_o}{2\pi }\frac{I}{x}$

To get the induced current $I$through the loop, we use Ohm's law as shown within the next equation

$I_{\text{induced}}=\frac{\epsilon }{R_{\text{loop}}}$

Where $\epsilon$is that the induced emf within the loop. From Faraday's law, the induced emf is that the change of the magnetic flux inside the loop and it's given by equation within the form

$\epsilon =\left|\frac{d{\Phi }_{\mathrm{m}}}{dt}\right|$

Where${\Phi }_{\mathrm{m}}$is that the flux through the loop which is that the amount of magnetic field that flows through a loop of area $A$and it's given by

$d{\Phi }_{\mathrm{m}}=d\left(BA\right)=BdA$

$=\left(\frac{{\mu }_o}{2\pi }\frac{I}{x}\right)\left(ldx\right)$

$=\frac{{\mu }_{o}l}{2\pi }\frac{I}{x}dx$

Find flux

The flux changes from point $x$to point $x+1$. Where $l$is that the length of the loop. So, we integrate the flux over $x$to$x+1$

$\int d{\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}lI}{2\pi }{\int }_x^{x+1\mathrm{cm}}\frac{1}{x}dx$

${\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}lI}{2\pi }[\ln x{]}_x^{x+1\mathrm{cm}}$

${\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}lI}{2\pi }\left[\ln \right(x+1\mathrm{cm})-\ln (x\left)\right]$

${\Phi }_{\mathrm{m}}=\frac{{\mu }_{o}lI}{2\pi }\ln \left(\frac{x+1\mathrm{cm}}{x}\right)$

Find Induced Current

Let us use this expression of${\Phi }_m$into equation to urge$\epsilon$by

$\epsilon =\frac{d}{dt}\left(\frac{{\mu }_{o}lI}{2\pi }\ln \left(\frac{x+1\mathrm{cm}}{x}\right)\right)$

$=\frac{{\mu }_{o}lI}{2\pi }\left(\frac{x}{x+1\mathrm{cm}}\right)\left[\left(-x^{-2}\right)(x+1)+\frac{1}{x}\right]\frac{dx}{dt}$

$=\frac{{\mu }_{o}lI}{2\pi }\left[\frac{1\mathrm{cm}}{x(x+1\mathrm{cm})}\right]$

Where $v$is that the velocity of the loop. We use this expression of $\epsilon$into equation to induce the induced current

$I_{\text{induced}}=\frac{{\mu }_{o}lI}{2\pi R_{\text{loop}}}\left[\frac{1\mathrm{cm}}{x(x+1\mathrm{cm})}\right]$

$=\frac{\left(4\pi \times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}\right)(4\mathrm{cm})(15\mathrm{A})}{2\pi (0.020\Omega )}\left[\frac{1\mathrm{cm}}{(2\mathrm{cm})(2\mathrm{cm}+1\mathrm{cm})}\right](10\mathrm{m}/\mathrm{s})$

$=1\times {10}^{-3}\mathrm{A}$

$=1\mathrm{mA}$