Question

A small, 2.0-mm-diameter circular loop with $R=0.020\Omega$is at the center of a large 100-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from $+1.0A$to $-1.0A$in $0.10s$. What is the induced current in the inner loop?

Short answer

In the inner loop, the induced current is,$I_{\text{induced}}=39\mathrm{nA}.$

Step-by-step solution

Step: 1 Equating Ohm's law: 

The outer loop causes a current flowing through the inner loop. The magnetic field generated by the outer loop is calculated as follows:

$B=\frac{{\mu }_{o}I}{2r_{\text{out}}}$

The magnetic flux via the inner loop is calculated as follows:

localid="1648903004243" ${\mathrm{\Phi }}_{\mathrm{m}}=BA$

The induced emf, as defined by Faraday's law, is the change in magnetic flux inside the loop, and it is provided by equation.

$$\begin{gathered} \epsilon =\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt} \\ =A\frac{dB}{dt} \\ =\frac{{\mu }_{o}A}{2r_{\text{out}}}\frac{dI}{dt} \end{gathered}$$

We utilise Ohm's law to determine the induced current I via the inner loop, as indicated in the next equation.

localid="1648903010275" $I_{\text{induced}}=\frac{\epsilon }{R}.$

Step: 2  Finding Area of inner loop: 

To get$I_{\text{induced}}$by using expression of$\xi$

$I_{\text{induced}}=I_{\text{induced}}=\frac{{\mu }_{o}A}{2R{r}_{\text{out}}}\frac{dI}{dt}$

The rate of change of current by

localid="1648903022923" $\frac{dI}{dt}=\frac{1\mathrm{A}-(-1\mathrm{A})}{0.10\mathrm{s}}=20\mathrm{A}/\mathrm{s}$

The area of inner loop is

localid="1648903027017" $$\begin{gathered} A=\pi {\left(\frac{d}{2}\right)}^2 \\ A=\pi {\left(\frac{2\times {10}^{-3}\mathrm{m}}{2}\right)}^2 \\ A=3.14\times {10}^{-6}{\mathrm{m}}^2. \end{gathered}$$

Step: 3  Induced current in inner loop: 

Inserting the values of ${\mu }_o,A,R,r_{\text{out}}$and $\left(\raisebox{1ex}{$dI$}\!\left/ \!\raisebox{-1ex}{$dT$}\right.\right)$into the equation,

$\begin{array}{l}{I}_{\text{induced}}=\frac{{\mu }_{o}A}{2R{r}_{\text{out}}}\frac{dI}{dt}\\ =\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(3.14\times {10}^{-6}{\mathrm{m}}^2\right)}{2\left(0.020\mathrm{\Omega }\right)\left(50\times {10}^{-3}\mathrm{m}\right)}(20\mathrm{A}/\mathrm{s})\\ =39\times {10}^{-9}\mathrm{A}\\ =39\mathrm{nA}.\end{array}$