91Ó°ÊÓ

A induced current when a wire coil is put in a changing flux. Something creates an electrical field that causes current to flow by moving the costs around the wire.

The force field in terms of your time within the form

$B=4t-2t^2$

We apply Ohm's law to calculate the induced current across the loop, as illustrated in the following equation.

$I_{\text{induced}}=\frac{\mathrm{Î\mu }}{R}$

Where $\mathrm{Î\mu }$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 the equation

$\mathrm{Î\mu }=\left|\frac{d{\mathrm{Φ}}_{\mathrm{m}}}{dt}\right|$

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

${\mathrm{Φ}}_{\mathrm{m}}=BA$

use this expression of ${\mathrm{Φ}}_m$into equation to induce $\mathrm{Î\mu }$by

$\mathrm{Î\mu }=\frac{d\left(BA\right)}{dt}=A\frac{dB}{dt}$

By using expression of $\mathrm{Î\mu }$and $B$into equation to induce $I_{\text{induced}}$by

$I_{\text{induced}}=\left(\frac{A}{R}\right)\frac{dB}{dt}$

$=\left(\frac{NA}{R}\right)\frac{d}{dt}\left(0.80y^{2}t\right)$

$=\left(\frac{0.80y^{2}A}{R}\right)$

$y$are going to be the side of the square loop $y=20\mathrm{cm}$. The area of the loop is

$A=0.20\mathrm{m}×0.20\mathrm{m}=0.04{\mathrm{m}}^2$

plug the values for $y,A$and $R$into equation (3) to get $I_{\text{induced}}$

$I_{\text{induced}}=\left(\frac{0.80y^{2}A}{R}\right)$

$=\left(\frac{0.80(0.20\mathrm{m}{)}^2\left(0.04{\mathrm{m}}^2\right)}{0.50\mathrm{Î\copyright }}\right)$

$=2.56×{10}^{-3}\mathrm{A}$

$=2.56\mathrm{mA}$