91影视

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, $\mathrm{B}=鈭�\mathrm{t}$.

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

$\mathrm{桅}={\mathrm{叠蟺谤}}^2$

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

(b)

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

$$\begin{gathered} \mathrm{蔚}=-\frac{\mathrm{诲桅}}{\mathrm{dt}} \\ \mathrm{蔚}=-\frac{\mathrm{d}\left(\mathrm{B}.{\mathrm{蟺谤}}^2\right)}{\mathrm{dt}} \\ \mathrm{蔚}=-{\mathrm{蟺谤}}^2\frac{\mathrm{dB}}{\mathrm{dt}} \\ \mathrm{蔚}=-{\mathrm{蟺谤}}^2\frac{\mathrm{d}\left(鈭�\mathrm{t}\right)}{\mathrm{dt}} \\ \mathrm{Solve}\mathrm{further}\mathrm{as}: \\ \mathrm{蔚}=-鈭�{\mathrm{蟺谤}}^2\frac{\mathrm{d}\left(\mathrm{t}\right)}{\mathrm{dt}} \\ \mathrm{蔚}=-鈭�{\mathrm{蟺谤}}^2 \end{gathered}$$

The negative sign indicates the emf value is decreasing.

Also, the emf using Ohm鈥檚 law,

$蔚=IR$

Then equating both values,

$$\begin{gathered} \mathrm{IR}=鈭�{\mathrm{蟺谤}}^2 \\ \mathrm{I}=\frac{\mathrm{蟺}鈭�{\mathrm{r}}^2}{\mathrm{R}} \end{gathered}$$

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

(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鈥檚 law for a closed area, the formula for the measured emf is given by,

$$\begin{gathered} 鈭�\mathrm{E}.\mathrm{dI}=-\frac{鈭�}{鈭�\mathrm{t}}鈭�\mathrm{B}.\mathrm{ds} \\ \mathrm{E}.2\mathrm{蟺}.\mathrm{s}=-{\mathrm{蟺蝉}}^2鈭� \\ \mathrm{E}=-\frac{鈭�\mathrm{s}}{2}\widehat{\mathrm{蠒}} \end{gathered}$$

In polar form,

$$\begin{gathered} \mathrm{E}=-\frac{鈭�\mathrm{s}}{2}\left(-\mathrm{蝉颈苍蠒}\hat{\mathrm{x}}+\mathrm{c}\mathrm{辞蝉蠒}\hat{\mathrm{y}}\right) \\ \mathrm{E}=\frac{鈭�}{2}\left(\mathrm{s}\mathrm{蝉颈苍蠒}\hat{\mathrm{x}}-\mathrm{s}\mathrm{c辞蝉蠒}\hat{\mathrm{y}}\right) \\ \mathrm{E}=\frac{鈭�}{2}\left(\mathrm{y}\hat{\mathrm{x}}-\mathrm{x}\hat{\mathrm{y}}\right) \end{gathered}$$

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

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