91影视

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

Calculate the induced emf by using the slope of the line form Fig. 30-42(b)inFaraday鈥檚 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鈥檚 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:

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

Where,${桅}_B$ is magnetic flux,dt is time, dq is charge, i is current, R is resistance, 饾渶 is emf,

First, find the emf by using Faraday鈥檚 law.

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

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

Faraday鈥檚 law is given as,

$蔚=-\frac{d{蠁}_B}{dt}$

As magnetic field is perpendicular to the area vector,

${蠁}_B=BA$

Therefore,

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

Substituting the values,

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

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

The slope of the graph gives dq/dt.

Therefore,

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

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

$R=\frac{\left|蔚\right|}{i}$

Substituting the values,

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

Hence, resistance of the loop is, $R=0.0012惟$

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