Question

Current density in a cylindrical wire of radius \( R \) varies with radial distance as \( \beta (r + r^2) \). What is the current through the shaded section (sector angle \( \pi/6 \)) of the wire as shown in the figure?

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

Hint:

For non-uniform current densities, integrate \( J(r) \cdot dA \) in cylindrical coordinates.

Answer 1

The Correct Option is A

Total current \( I = \int J \cdot dA = \int_0^R \int_0^{\pi/6} \beta(r + r^2) \cdot r \, d\theta \, dr \)
\[ = \beta \int_0^{\pi/6} d\theta \int_0^R (r^2 + r^3) \, dr = \beta \cdot \frac{\pi}{6} \cdot \left[ \frac{r^3}{3} + \frac{r^4}{4} \right]_0^R = \frac{\pi \beta}{6} \left( \frac{R^3}{3} + \frac{R^4}{4} \right) \] Simplify the expression using LCM and rewrite in the given option format. Final result matches option (1).