Q54P - Introduction to Electrodynamics

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:$[\propto r^2/2]$

Step-by-Step Solution

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 1: 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, $\mathrm{B}=\propto \mathrm{t}$.

Step 2: 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.

Step 3: 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 $\mathrm{I}=\frac{\mathrm{\pi }\propto {\mathrm{r}}^2}{\mathrm{R}}$.

Step 4: 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,

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

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