In Figure (a), a uniform magnetic field increases in magnitude with time t as given by Figure (b), where the vertical axis scale is set by ${\mathbf{B}}_{\mathbf{s}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{}\mathbf{T}$and the horizontal scale is set by ${\mathbf{t}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$A circular conducting loop of area $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{m}}^{\mathbf{2}}$lies in the field, in the plane of the page. The amount of charge q passing point A on the loop is given in Figure (c) as a function of t, with the vertical axis scale set by ${\mathit{q}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$ and the horizontal axis scale again set by localid="1661854094654" ${\mathbf{t}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$. What is the loop’s resistance?

1. Cross-sectional area$\mathrm{A}=8\times {10}^{-4}{\mathrm{m}}^2$
2. Magnetic field on vertical axis,${\mathrm{B}}_{\mathrm{s}}=9.0\mathrm{mT}=9\times {10}^{-3}\mathrm{T}$
3. Horizontal axis, ${\mathrm{t}}_{\mathrm{s}}=3.0\mathrm{s}$
4. Vertical axis,${\mathrm{q}}_{\mathrm{s}}=6\mathrm{mC}=6\times {10}^{-3}\mathrm{C}$

Calculate the induced emf by using the slope of the line form Fig. 30-42(b)inFaraday’s law. Find the current by finding the slope of the line from Fig.30-42 (c)tofind the resistance of the loop by using Ohm’s law.

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

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:

$$\begin{gathered} \epsilon =\frac{d{\phi }_B}{dt} \\ i=\frac{dq}{dt} \\ R=\frac{\left|\epsilon \right|}{i} \end{gathered}$$

Where,${\Phi }_B$ is magnetic flux,dt is time, dq is charge, i is current, R is resistance, 𝜀 is emf,

First, find the emf by using Faraday’s law.

From Fig.30 - 42 (b) the slope of the line i.e. dB/dt can be found as,

$\frac{\mathrm{dB}}{\mathrm{dt}}=\frac{9\times {10}^{-3}\mathrm{T}}{3.0\mathrm{s}}=0.003\mathrm{T}/\mathrm{s}$

Faraday’s law is given as,

$\epsilon =-\frac{d{\phi }_B}{dt}$

As magnetic field is perpendicular to the area vector,

${\phi }_B=BA$

Therefore,

role="math" localid="1661853786477" $$\begin{gathered} \epsilon =BA \\ \epsilon =\frac{d\left(BA\right)}{dt} \\ \epsilon =-A\frac{dB}{dt} \end{gathered}$$

Substituting the values,

$$\begin{gathered} \mathrm{\epsilon }=-\left(8.0\mathrm{x}{10}^{-4}\right)\left(0.003\right) \\ \mathrm{\epsilon }=-0.024\times {10}^{-4}\mathrm{V} \end{gathered}$$

Now, the current can be calculated from Fig. 30- 42 (c).

The slope of the graph gives dq/dt.

Therefore,

$$\begin{gathered} \mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{6\times {10}^{-3}\mathrm{C}}{3.0\mathrm{s}} \\ \mathrm{i}=0.002\mathrm{A} \end{gathered}$$

Now, relate the induced emf to resistance and current by using Ohm’s law.

$R=\frac{\left|\epsilon \right|}{i}$

Substituting the values,

$$\begin{gathered} R=\frac{\left|-0.024\times {10}^{-4}\right|}{0.002} \\ R=0.0012\Omega \end{gathered}$$

Hence, resistance of the loop is, $R=0.0012\Omega$

Therefore, we calculated the resistance of the loop by using the slopes from the graph, Faraday’s law, and Ohm’s law.