Question

A long conducting cylinder having a radius ‘b’ is placed along the z axis. The current density is \( J = J_a r^3 \hat{z} \) for the region \( r<b \) where \( r \) is the distance in the radial direction. The magnetic field intensity (H) for the region inside the conductor (i.e. for \( r<b \)) is

• A. ( \frac{J_a}{4} r^4 \)
• B. ( \frac{J_a}{3} r^3 \)
• C. ( \frac{J_a}{5} r^4 \)
• D. ( J_a r^3 \)

Hint:

To calculate the magnetic field inside a conductor, apply Ampère’s law using the current density distribution and integrate over the radial distance.

Answer 1

The Correct Option is C

Step 1: Understanding the Magnetic Field Intensity.
To find the magnetic field intensity \( H \) inside the conductor, we can use Ampère’s Law, which states that the integral of the magnetic field intensity around a closed loop is proportional to the current enclosed by the loop. For a conducting cylinder with a radial current density \( J = J_a r^3 \), we can apply the integral form of Ampère’s Law. Step 2: Applying Ampère’s Law.
By applying Ampère's law for the circular path inside the conductor, the magnetic field intensity at a distance \( r \) is given by: \[ H(2\pi r) = \int_0^r J_a r^3 2\pi r \, dr = J_a \int_0^r r^4 \, dr \] This results in: \[ H = \frac{J_a}{5} r^4 \] Step 3: Conclusion.
The correct answer is (C) \( \frac{J_a}{5} r^4 \), which is the magnetic field intensity inside the conductor.