91Ó°ÊÓ

In physics, an electric current that occurs when a second conductor (= substance that carries electricity) is placed in an area where there is already an electric current (Definition of induced current from the Cambridge Academic Content Dictionary Cambridge University Press) Examples of Induced Current

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

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

In the loop, where is the induced emf? The induced emf is the change in magnetic flux inside the loop as defined by Faraday's law, and it is given by equation (30.14) in the form

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

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

${\mathrm{Φ}}_{\mathrm{m}}=BA$

Let us use this expression of ${\mathrm{Φ}}_{\mathrm{m}}$into equation (2) to get$\mathrm{Î\mu }$by

localid="1648899301006" $\mathrm{Î\mu }=A\frac{d\left(BA\right)}{dt}=A\frac{dB}{dt}$

Use this expression of $\mathrm{Î\mu }$into equation (1) to get localid="1648899248140" $I_{\text{induced by}}$

localid="1648899306664" $I_{\text{induced}}=\left(\frac{A}{R}\right)\frac{dB}{dt}$

The area of the loop is calculated by

$A=\mathrm{π}{\left(\frac{d}{2}\right)}^2=\mathrm{π}{\left(\frac{0.1\mathrm{m}}{2}\right)}^2=7.85×{10}^{−3}{\mathrm{m}}^2$

In the first figure (a), the change of the magnetic field is$dB/dt=0.50\mathrm{T}/\mathrm{s}$. So, we use this value into equation (3) to get $I_{\text{induced}}$by

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

$=\left(\frac{7.85×{10}^{−3}{\mathrm{m}}^2}{0.20\mathrm{Î\copyright }}\right)(0.5\mathrm{T}/\mathrm{s})$

$=20×{10}^{−3}\mathrm{A}$

$=20\mathrm{mA}$

The direction is counterclockwise because the induced emf points out of the page in the opposite direction of the applied magnetic field.

In the second figure (b), the change of the magnetic field is $dB/dt=-0.50\mathrm{T}/\mathrm{s}$. So, we use this value into equation (3) to get $I_{\text{induced}}$by

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

$=\left(\frac{7.85×{10}^{−3}{\mathrm{m}}^2}{0.20\mathrm{Î\copyright }}\right)(0.5\mathrm{T}/\mathrm{s})$

$=20×{10}^{−3}\mathrm{A}$

$=20\mathrm{mA}$

The direction is counterclockwise because the induced emf points out of the page in the same direction of the applied magnetic field.

In the third figure $\left(\mathrm{c}\right)$, the change of the magnetic field inside the loop is zero $dB/dt$because there is no flux. So, the induced current is zero.

$I_{\text{induced}}=0$