GATE EE 2022 Set 1 — Question 23

To find the magnetic field intensity (H) inside the conductor, we can use Ampere's Law in integral form:$\oint_C H \cdot dl = I_{enc}$Given the current density $J = J_a r^3 \hat{z}$ for $r < b$. The cylinder is placed along the $z$-axis, implying cylindrical symmetry. Therefore, the magnetic field intensity $H$ will have only an azimuthal component, $H = H_{\phi} \hat{\phi}$.Consider a circular Amperian loop $C$ of radius $r'$ (where $r' < b$) in the $xy$-plane, centered at the origin. The path element $dl = r' d\phi \hat{\phi}$.The left side of Ampere's Law becomes:
$\oint_C H \cdot dl = \int_0^{2\pi} (H_{\phi} \hat{\phi}) \cdot (r' d\phi \hat{\phi}) = H_{\phi} r' \int_0^{2\pi} d\phi = H_{\phi} (2\pi r')$ (assuming $H_{\phi}$ is constant along the loop due to symmetry).The enclosed current $I_{enc}$ is the integral of the current density $J$ over the cross-sectional area $S$ enclosed by the Amperian loop:
$I_{enc} = \int_S J \cdot dS$For a cross-sectional area in the $xy$-plane, the differential surface element is $dS = r dr d\phi \hat{z}$.
$I_{enc} = \int_0^{2\pi} \int_0^{r'} (J_a r^3 \hat{z}) \cdot (r dr d\phi \hat{z})$
$I_{enc} = \int_0^{2\pi} \int_0^{r'} J_a r^4 dr d\phi$First, integrate with respect to $r$:
$\int_0^{r'} J_a r^4 dr = J_a \left[ \frac{r^5}{5} \right]_0^{r'} = J_a \frac{(r')^5}{5}$Next, integrate with respect to $\phi$:
$I_{enc} = \int_0^{2\pi} J_a \frac{(r')^5}{5} d\phi = J_a \frac{(r')^5}{5} [\phi]_0^{2\pi} = J_a \frac{(r')^5}{5} (2\pi)$Now, equate the two sides of Ampere's Law:
$H_{\phi} (2\pi r') = J_a \frac{(r')^5}{5} (2\pi)$Divide both sides by $2\pi r'$:
$H_{\phi} = J_a \frac{(r')^5}{5r'} = J_a \frac{(r')^4}{5}$Replacing $r'$ with $r$ for the region inside the conductor ($r < b$):
$H_{\phi} = \frac{J_a}{5} r^4$The magnetic field intensity is $H = \frac{J_a}{5} r^4 \hat{\phi}$. The magnitude of $H$ is $\frac{J_a}{5} r^4$.Comparing this with the given options:
(A) $\frac{J_a}{4} r^4$
(B) $\frac{J_a}{3} r^3$
(C) $\frac{J_a}{5} r^4$
(D) $J_a r^3$The correct option is (C).The final answer is $\boxed{\text{C}}$