A-cm-diameter coil has 20 turns and a resistance of CALC$0.50\mathrm{\Omega }$. A magnetic field perpendicular to the coil is $B=0.02t+0.010t^2$where B is in tesla and t is in seconds.

a. Find an expression for the induced current I t as a function of time.

b. Evaluate I at $t=5\mathrm{s}$and $t=10\mathrm{s}$.

Magnetic field strength is one of two ways to express a magnetic field's intensity. Magnetic field strength is measured in amperes per metre $A/m$ and is distinguished from magnetic flux density $B$, which is measured in Newton-meters per ampere $(Nm/A)$also known as teslas $T$

The magnetic field in terms of time is given to in the form is

$B=0.02t+0.010t^2$

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

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

Where is the induced emf in the loop? According to Faraday's law, the induced emf is the change in magnetic flux inside the loop, and it is given by equation (30.14) in the form

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

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

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

Use this expression of$\epsilon$and B into equation (l) to get $I_{\text{induced}}$by

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

$\begin{array}{r}=\left(\frac{NA}{R}\right)\frac{d}{dt}\left(0.02t+0.010t^2\right)\\ I_{induced}=\left(\frac{NA}{R}\right)(0.02+0.02t)\end{array}$

$I_{induced}=\left(\frac{20}{0.50}\right)(0.02+0.02t)$

$I_{induced}=40(0.02+0.02t)$

(b) The area of the loop is calculated by

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{0.05\mathrm{m}}{2}\right)}^2=1.96\times {10}^{-3}{\mathrm{m}}^2$

At $t=10\mathrm{s}$, we use equation (3) in part (a) to get the induced current by

$I_{10\mathrm{s}}=\left(\frac{NA}{R}\right)(0.02+0.02t)$

$\begin{array}{r}=\left(\frac{\left(20\right)\left(2\times {10}^{-3}{\mathrm{m}}^2\right)}{0.5\mathrm{\Omega }}\right)(0.02+0.02(10\mathrm{s}\left)\right)\end{array}$

$=17.6\times {10}^{-3}\mathrm{A}$

$I_{10s}=17.6\mathrm{mA}$

For $t=5s$

${\mathrm{I}}_{5\mathrm{s}}=\left(\frac{\mathrm{NA}}{\mathrm{R}}\right)(0.02+0.02\mathrm{t})$

$=\left(\frac{\left(20\right)\left(2\times {10}^{-3}{\mathrm{m}}^2\right)}{0.5\mathrm{\Omega }}\right)(0.02+0.02(5\mathrm{s}\left)\right)$

$=9.6\times {10}^{-3}\mathrm{A}$

$=9.6\mathrm{mA}$