Question

Current density in a cylindrical wire of radius \(R\) varies with radial distance as \(\beta(r+r_0)^2\). The current through the section of the wire shown in the figure is

• A. \(\displaystyle \pi\beta\left[\frac{R^4}{12}+\frac{r_0^2R^2}{6}+\frac{2r_0R^3}{9}\right]\)
• B. \(\displaystyle \pi\beta\left[\frac{R^4}{6}+\frac{r_0^2R^2}{12}+\frac{r_0R^3}{9}\right]\)
• C. \(\displaystyle \pi\beta\left[\frac{R^4}{12}+\frac{r_0^2R^2}{12}+\frac{r_0R^3}{9}\right]\)
• D. \(\displaystyle \pi\beta\left[\frac{R^4}{8}+\frac{r_0^2R^2}{6}+\frac{r_0R^3}{12}\right]\)

Hint:

For current through a non-uniform current density region, use \(I=\int J\,dA\). In circular regions, polar coordinates make the calculation easier.

Answer 1

The Correct Option is A

Step 1: Write the current element.
Current density is
\[ J=\beta(r+r_0)^2 \]
For a small area element in polar coordinates,
\[ dA=r\,dr\,d\theta \]
So, small current through the element is
\[ dI=J\,dA \]
\[ dI=\beta(r+r_0)^2r\,dr\,d\theta \]

Step 2: Use the angular region from the figure.
From the figure, the total shaded angular region is
\[ \frac{\pi}{3} \]
Therefore, current through the shaded section is
\[ I=\int_0^{\pi/3}\int_0^R \beta(r+r_0)^2r\,dr\,d\theta \]

Step 3: Expand the integrand.
\[ (r+r_0)^2r=(r^2+2rr_0+r_0^2)r \]
\[ =r^3+2r_0r^2+r_0^2r \]
Thus,
\[ I=\beta\int_0^{\pi/3}d\theta\int_0^R(r^3+2r_0r^2+r_0^2r)\,dr \]

Step 4: Integrate with respect to \(r\).
\[ \int_0^R r^3\,dr=\frac{R^4}{4} \]
\[ \int_0^R 2r_0r^2\,dr=\frac{2r_0R^3}{3} \]
\[ \int_0^R r_0^2r\,dr=\frac{r_0^2R^2}{2} \]
So,
\[ I=\beta\left(\frac{\pi}{3}\right) \left[ \frac{R^4}{4}+\frac{2r_0R^3}{3}+\frac{r_0^2R^2}{2} \right] \]

Step 5: Simplify.
\[ I=\pi\beta\left[ \frac{R^4}{12}+\frac{2r_0R^3}{9}+\frac{r_0^2R^2}{6} \right] \]
\[ I=\pi\beta\left[ \frac{R^4}{12}+\frac{r_0^2R^2}{6}+\frac{2r_0R^3}{9} \right] \]

Step 6: Final conclusion.
Hence,
\[ \boxed{\pi\beta\left[\frac{R^4}{12}+\frac{r_0^2R^2}{6}+\frac{2r_0R^3}{9}\right]} \]