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 a small current element.
The current density varies with radius as $J = \beta(r + r_0)^2$. Through a small area $dA = r\,dr\,d\theta$ the current is \[ dI = J\,dA = \beta(r+r_0)^2\,r\,dr\,d\theta. \]
Step 2: Set the angular limits from the figure.
The shaded wedge subtends an angle of $\dfrac{\pi}{3}$, and the radius runs from $0$ to $R$: \[ I = \int_0^{\pi/3}\!\!\int_0^R \beta(r+r_0)^2\,r\,dr\,d\theta. \]
Step 3: Expand the radial integrand.
\[ (r+r_0)^2\,r = r^3 + 2r_0 r^2 + r_0^2 r. \]
Step 4: Integrate term by term over $r$.
\[ \int_0^R r^3\,dr = \frac{R^4}{4},\quad \int_0^R 2r_0 r^2\,dr = \frac{2r_0 R^3}{3},\quad \int_0^R r_0^2 r\,dr = \frac{r_0^2 R^2}{2}. \]
Step 5: Multiply by the angular integral $\dfrac{\pi}{3}$.
\[ I = \beta\cdot\frac{\pi}{3}\left[\frac{R^4}{4} + \frac{2r_0 R^3}{3} + \frac{r_0^2 R^2}{2}\right]. \] Distributing the $\dfrac{\pi}{3}$, \[ I = \pi\beta\left[\frac{R^4}{12} + \frac{2r_0 R^3}{9} + \frac{r_0^2 R^2}{6}\right]. \]
Step 6: Conclude.
Rearranging the terms to match the option, \[ I = \pi\beta\left[\frac{R^4}{12} + \frac{r_0^2 R^2}{6} + \frac{2r_0 R^3}{9}\right]. \]
\[ \boxed{\pi\beta\left[\dfrac{R^4}{12}+\dfrac{r_0^2 R^2}{6}+\dfrac{2r_0 R^3}{9}\right]} \]