In Figure (a), a circular loop of wire is concentric with a solenoid and lies in a plane perpendicular to the solenoid’s central axis. The loop has radius 6.00 cm.The solenoid has radius 2.00 cm, consists of $\mathbf{8000}\frac{\mathbf{t}\mathbf{u}\mathbf{r}\mathbf{n}\mathbf{s}}{\mathbf{m}}$, and has a current i_sol varying with time tas given in Figure (b), where the vertical axis scale is set by is ${\mathit{i}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{A}$and the horizontal axis scale is set by ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{s}$. Figure (c) shows, as a function of time, the energy E_th that is transferred to thermal energy of the loop; the vertical axis scale is set by ${\mathbf{E}}_{\mathbf{s}}\mathbf{=}\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{nJ}$. What is the loop’s resistance?

1. Radius of loop,$r_{loop}=6cm$
2. Radius of solenoid,$r_{sol}=2cm$
3. Turn density$n=8000turns/m$
4. Vertical scale,$i_s=1A$
5. Horizontal scale,$t_s=2s$
6. Vertical scale,$E_s=100nJ$

Use the magnetic flux formula to find the flux through the area of solenoid. By substituting this value in Faraday’s law, find the magnitude of the induced emf. By using the rate of thermal energy dissipation formula, find the resistance of the coil.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

$$\begin{gathered} B_{sol}={\mu }_{0}n{i}_{sol} \\ {\varphi }_B=BA \\ P=i\left|\epsilon \right| \end{gathered}$$

Where,role="math" localid="1661855936116" $\Phi$is magnetic flux, B is magnetic field, A is area, i is current, P is power, 𝜀 is emf, nis number of turns, 𝜇₀is permeability.

The magnetic field inside a solenoid is,

$B_{sol}={\mu }_{0}n{i}_{sol}$

For the long solenoid, assume that the field outside the solenoid is zero and the field inside the solenoid is constant.

Then, the magnetic flux through the area of a solenoid is,

$$\begin{gathered} {\varphi }_B=B_{sol}{A}_{sol} \\ {\varphi }_B={\mu }_{0}n{i}_{sol}\times \tau \tau r_{{}_{sol}}^2 \end{gathered}$$

Using the Faraday’s law, the magnitude of the induced emf is,

$$\begin{gathered} \left|\epsilon \right|=\frac{d{\varphi }_B}{dt} \\ \left|\epsilon \right|={\mu }_{0}n\frac{d{i}_{sol}}{dt}\times \tau \tau r_{sol}^2 \\ n=8000\frac{turns}{m},{\mu }_0=4\tau \tau \times {10}^{-7}\frac{T.m}{A},r=0.02m \end{gathered}$$

From fig.30-53 (b) , the slope of the line gives$\frac{d{i}_{sol}}{dt}$

Therefore,

$\frac{d{i}_{sol}}{dt}=0.5A/s$

Substituting the values,

$$\begin{gathered} \left|\epsilon \right|=\left(4\tau \tau \times {10}^{-7}\frac{T.m}{A}\right)\left(8000turns/m\right)\left(0.5A/s\right)\times \tau \tau {\left(0.02\right)}^2 \\ \left|\epsilon \right|=6.3\times {10}^{-6}6.3\mu V \end{gathered}$$

The rate of the dissipation of thermal energy is given as,

$P=\frac{d{E}_{th}}{dt}$

Also, from equation 26 - 28,

$P=\frac{{\epsilon }^2}{R}$

Thus,

$\frac{d{E}_{th}}{dt}=\frac{{\epsilon }^2}{R}$

Rearranging the equation for R,role="math" localid="1661856500413" $R=\frac{{\epsilon }^2}{\frac{d{E}_{th}}{dt}}$

From$\mathrm{Fi}g.30-53\left(c\right)$, the slope of the line gives$\frac{d{E}_{th}}{dt}$

Therefore,

$\frac{d{E}_{th}}{dt}=40nJ/s$

Thus,

$$\begin{gathered} R=\frac{{\left(6.3\mu V\right)}^2}{40nJ/s} \\ R=0.992\times {10}^{-3}\Omega \approx 1.0m\Omega \end{gathered}$$

Hence, the loop’s resistance is$1.0m\Omega$.

Therefore, by using the magnetic flux formula to find the flux through the area of solenoid and substituting this value in Faraday’s law, and by using the rate of thermal energy dissipation formula, the resistance of the coil can be determined.