Question

A current flows through a cylindrical conductor of radius \( R \). The current density at a point in the conductor is \( j = \alpha r \) (along its axis), where \( \alpha \) is a constant and \( r \) is the distance from the axis of the conductor. The current flowing through the portion of the conductor from \( r = 0 \) to \( r = \frac{R}{2} \) is proportional to:

• A. \( R \)
• B. \( R^2 \)
• C. \( R^3 \)
• D. \( R^4 \)

Hint:

For variable current density, always integrate the current density over the area using cylindrical coordinates: \( dA = 2\pi r\, dr \).

Answer 1

The Correct Option is C

Given current density: \( j = \alpha r \)
To find the current from \( r = 0 \) to \( r = \frac{R}{2} \), we integrate over the circular cross-section in cylindrical coordinates: \[ I = \int_0^{\frac{R}{2}} j \cdot dA = \int_0^{\frac{R}{2}} (\alpha r) \cdot (2\pi r \, dr) \] \[ I = 2\pi \alpha \int_0^{\frac{R}{2}} r^2 \, dr = 2\pi \alpha \left[ \frac{r^3}{3} \right]_0^{\frac{R}{2}} = 2\pi \alpha \cdot \frac{1}{3} \cdot \left( \frac{R}{2} \right)^3 \] \[ I = \frac{2\pi \alpha}{3} \cdot \frac{R^3}{8} = \frac{\pi \alpha R^3}{12} \] So, the current is proportional to \( R^3 \). Final answer: \( R^3 \)