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/answer is:
$\frac{J_a}{5}{r}^4$

Step-by-step Derivation:

1. Understand the Given Information:
We have a long conducting cylinder of radius $b$ aligned along the z-axis.
The current density inside the cylinder (for $r<b$) is given as:
$J=J_a{r}^3\hat{z}$
where $\hat{z}$ represents the unit vector along the z-axis, which is the direction of the current flow.

2. Apply Ampere's Circuital Law:
Ampere's Circuital Law states that the line integral of the magnetic field intensity $H$ around a closed path is equal to the enclosed current $I_{\mathrm{enclosed}}$:
$\oint H·dl=I_{\mathrm{enclosed}}$

3. Determine the Left-Hand Side (LHS) of Ampere's Law:
Due to the cylindrical symmetry of the system, the magnetic field intensity vector $H$ points in the azimuthal direction ($\hat{\varphi }$) and its magnitude depends only on the radial distance $r$.
For a circular Amperian loop of radius $r$ (where $r<b$) centered on the z-axis:
$\oint H·dl=H·\left(2\pi r\right)$

4. Calculate the Enclosed Current (RHS of Ampere's Law):
The current enclosed by the Amperian loop of radius $r$ is found by integrating the current density over the cross-sectional area of the loop:
$I_{\mathrm{enclosed}}={\int }_0^{r}J·dA$
For a cylindrical coordinate system, the differential area element $dA$ perpendicular to the z-axis is $dA=2\pi r\text{'}dr\text{'}$. Substituting $J=J_a{r\text{'}}^3$:
$I_{\mathrm{enclosed}}={\int }_0^r\left(J_a{r\text{'}}^3\right)·(2\pi r\text{'}dr\text{'})$
$I_{\mathrm{enclosed}}=2\pi J_a{\int }_0^r{r\text{'}}^{4}dr\text{'}$
Integrating ${r\text{'}}^4$ yields:
$I_{\mathrm{enclosed}}=2\pi J_a{\left[}_{\frac{{r\text{'}}^5}{5}}^{\right]}=2\pi J_a\frac{r^5}{5}$

5. Solve for H:
Equating the LHS and the RHS of Ampere's Law:
$H·\left(2\pi r\right)=2\pi J_a\frac{r^5}{5}$
Dividing both sides by $2\pi r$:
$H=\frac{J_a}{5}{r}^4$

Thus, the magnitude of the magnetic field intensity $H$ inside the region $r<b$ is $(J_a/5)r^4$.