91Ó°ÊÓ

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

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

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

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

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

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

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

localid="1648903010275" $I_{\text{induced}}=\frac{\mathrm{Î\mu }}{R}.$

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

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

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" $$\begin{gathered} A=\mathrm{π}{\left(\frac{d}{2}\right)}^2 \\ A=\mathrm{π}{\left(\frac{2×{10}^{−3}\mathrm{m}}{2}\right)}^2 \\ A=3.14×{10}^{−6}{\mathrm{m}}^2. \end{gathered}$$

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

$\begin{array}{l}{I}_{\text{induced}}=\frac{{\mathrm{μ}}_{o}A}{2R{r}_{\text{out}}}\frac{dI}{dt}\\ =\frac{\left(4\mathrm{Ï€}×{10}^{−7}\mathrm{T}â‹\dots \mathrm{m}/\mathrm{A}\right)\left(3.14×{10}^{−6}{\mathrm{m}}^2\right)}{2\left(0.020\mathrm{Î\copyright }\right)\left(50×{10}^{−3}\mathrm{m}\right)}(20\mathrm{A}/\mathrm{s})\\ =39×{10}^{−9}\mathrm{A}\\ =39\mathrm{nA}.\end{array}$