Question

A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.

Answer 1

Answer:

$B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)$

Explanation:

As the current density is given as

$J = \frac{J_0}{r_0}r$

now we have current inside wire given as

$i = \int J(2\pi r)dr$

$i = \int \frac{J_0}{r_0} r(2\pi r)dr$

$i = 2\pi \frac{J_0}{r_0} \int r^2 dr$

$i = \frac{2}{3} \pi \frac{J_0}{r_0} r^3$

Now by Ampere's law we will have

$\int B. dl = \mu_0 i$

$B. (2\pi r) = \mu_0(\frac{2}{3} \pi \frac{J_0}{r_0} r^3)$

$B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)$