A $20$cm$*20$cm square loop of wire lies in the $xy$-plane with its bottom edge on the $x$-axis. The resistance of the loop is$0.50\Omega$. A magnetic field parallel to the $z$-axis is given by$B=0.80y2t$, where$B$ is in tesla, $y$in meters, and$t$ in seconds. What is the size of the induced current in the loop at $t=0.50s$?

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.

$I_{\text{induced}}=\frac{\epsilon }{R}$

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 the equation

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

Where ${\Phi }_{\mathrm{m}}$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

${\Phi }_{\mathrm{m}}=BA$

use this expression of ${\Phi }_m$into equation to induce $\epsilon$by

$\epsilon =\frac{d\left(BA\right)}{dt}=A\frac{dB}{dt}$

By using expression of $\epsilon$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

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

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

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

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

$=2.56\times {10}^{-3}\mathrm{A}$

$=2.56\mathrm{mA}$