A long conducting cylinder having a radius ‘b’ is placed along the z axis. The current density is J= J_ar³ 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

The correct option is: (J_a/5)r⁴

Step-by-Step Explanation:

We are asked to find the magnetic field intensity (H) inside a long conducting cylinder of radius b. The current density is given as:

$J=J_a{r}^3$

According to Ampere's circuital law, the line integral of the magnetic field intensity H around a closed path is equal to the current enclosed by that path:

$\oint \mathbf{H}·d\mathbf{l}=I_{\text{enclosed}}$

For a circular loop of radius r (where r < b) concentric with the cylinder, the magnetic field intensity H is constant along the path and directed tangentially. Thus, the left-hand side of Ampere's Law becomes:

$\oint \mathbf{H}��d\mathbf{l}=H\left(2\pi r\right)$

Next, we calculate the enclosed current by integrating the current density J over the cross-sectional area of the loop of radius r:

$I_{\text{enclosed}}={\int }_0^{r}J·dA$

Since the differential area element in cylindrical coordinates for a cross-section is $dA=2\pi rdr$, we substitute the given expression for J:

$I_{\text{enclosed}}={\int }_0^r\left(J_a{r}^3\right)\left(2\pi rdr\right)$

We can factor out the constant terms from the integral:

$I_{\text{enclosed}}=2\pi J_a{\int }_0^r{r}^{4}dr$

Evaluating the integral gives:

$I_{\text{enclosed}}=2\pi J_a{[\frac{r^5}{5}}_0^r=\frac{2\pi J_a{r}^5}{5}$

Now, we substitute the expression for the enclosed current back into Ampere's Law:

$H\left(2\pi r\right)=\frac{2\pi J_a{r}^5}{5}$

Solving for H by dividing both sides by $2\pi r$, we obtain:

$H=\frac{J_a}{5}{r}^4$

This matches the correct option.