Question

A circular wire loop (radius r , resistance R ) encloses a region of uniform magnetic field, B , perpendicular to its plane. The field (occupying the shaded region in Fig. 7.56) increases linearly with time$(B=\propto t)$An ideal voltmeter (infinite internal resistance) is connected between points P and Q.

(a) What is the current in the loop?

(b) What does the voltmeter read? Answer:$[\propto r^2/2]$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{current}\mathrm{in}\mathrm{the}\mathrm{loop}\mathrm{is}\mathrm{I}=\frac{\mathrm{\pi }\propto {\mathrm{r}}^2}{\mathrm{R}}. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{voltmeter}\mathrm{reading}\mathrm{is}\frac{\propto {\mathrm{r}}^2}{2}. \end{gathered}$$

Step-by-step solution

Given information

The radius of circular wire loop is, r .

The resistance of circular wire loop is, R .

The uniform magnetic field inside the wire loop is, B .

The relation between the magnetic field and time is, $\mathrm{B}=\propto \mathrm{t}$.

Magnetic flux

The magnetic flux inside the wire loop having magnetic field B and r radius is given by,

$\mathrm{\Phi }={\mathrm{B\pi r}}^2$

If the radius of the circular wire loop is increased then the magnetic flux produced also increases.

The current in the loop

(b)

The formula for the emf generated in the loop due to magnetic flux is given by,

$$\begin{gathered} \mathrm{\epsilon }=-\frac{\mathrm{d\Phi }}{\mathrm{dt}} \\ \mathrm{\epsilon }=-\frac{\mathrm{d}\left(\mathrm{B}.{\mathrm{\pi r}}^2\right)}{\mathrm{dt}} \\ \mathrm{\epsilon }=-{\mathrm{\pi r}}^2\frac{\mathrm{dB}}{\mathrm{dt}} \\ \mathrm{\epsilon }=-{\mathrm{\pi r}}^2\frac{\mathrm{d}\left(\propto \mathrm{t}\right)}{\mathrm{dt}} \\ \mathrm{Solve}\mathrm{further}\mathrm{as}: \\ \mathrm{\epsilon }=-\propto {\mathrm{\pi r}}^2\frac{\mathrm{d}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ \mathrm{\epsilon }=-\propto {\mathrm{\pi r}}^2 \end{gathered}$$

The negative sign indicates the emf value is decreasing.

Also, the emf using Ohm’s law,

$\epsilon =IR$

Then equating both values,

$$\begin{gathered} \mathrm{IR}=\propto {\mathrm{\pi r}}^2 \\ \mathrm{I}=\frac{\mathrm{\pi }\propto {\mathrm{r}}^2}{\mathrm{R}} \end{gathered}$$

Hence, the current in the loop is$\mathrm{I}=\frac{\mathrm{\pi }\propto {\mathrm{r}}^2}{\mathrm{R}}$.

Determine the voltmeter reading value

(b)

Assume a small elemental region $dI$of radius s inside the given inside the given region between points P and Q.

For a circle of radius s , applying Faraday’s law for a closed area, the formula for the measured emf is given by,

$$\begin{gathered} \oint \mathrm{E}.\mathrm{dI}=-\frac{\partial }{\partial \mathrm{t}}\int \mathrm{B}.\mathrm{ds} \\ \mathrm{E}.2\mathrm{\pi }.\mathrm{s}=-{\mathrm{\pi s}}^2\propto \\ \mathrm{E}=-\frac{\propto \mathrm{s}}{2}\hat{\mathrm{\varphi }} \end{gathered}$$

In polar form,

$$\begin{gathered} \mathrm{E}=-\frac{\propto \mathrm{s}}{2}\left(-\mathrm{sin\varphi }\hat{\mathrm{x}}+\mathrm{c}\mathrm{os\varphi }\hat{\mathrm{y}}\right) \\ \mathrm{E}=\frac{\propto }{2}\left(\mathrm{s}\mathrm{sin\varphi }\hat{\mathrm{x}}-\mathrm{s}\mathrm{cos\varphi }\hat{\mathrm{y}}\right) \\ \mathrm{E}=\frac{\propto }{2}\left(\mathrm{y}\hat{\mathrm{x}}-\mathrm{x}\hat{\mathrm{y}}\right) \end{gathered}$$

Along the line from P to Q,

$\mathrm{dI}=\mathrm{dx}.\hat{\mathrm{x}}\mathrm{and}\mathrm{y}=\frac{\mathrm{r}}{\sqrt{2}},$

Then the voltage reading between points P and Q can be calculated as,

role="math" localid="1658300624375" $$\begin{gathered} \mathrm{V}=-\int \mathrm{E}.\mathrm{dI} \\ \mathrm{V}=-\frac{\propto }{2}\int \mathrm{ydx} \\ \mathrm{V}=-\frac{\propto }{2}\frac{\mathrm{r}}{\sqrt{2}}\left(\sqrt[\mathrm{r}]{2}\right) \\ V=\frac{\propto r^2}{2} \end{gathered}$$

Hence, the voltmeter reading is$\frac{\propto r^2}{2}$.