Magnetic Field Distribution in Hollow Cylinder with Uniform Current Density

To find the magnitude of the magnetic field $|\vec{B}|$ as a function of the radial distance $r$ from the central axis of an infinitely long, hollow cylindrical conductor with inner radius $R/2$ and outer radius $R$, we apply Ampère's Circuital Law:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$

By cylindrical symmetry, the magnetic field lines form concentric circles centered on the axis, and the field magnitude $B(r)$ depends only on $r$. Choosing a circular Amperian loop of radius $r$ centered on the cylinder's axis gives:

$B(r) \cdot (2\pi r) = \mu_0 I_{\text{enclosed}}(r) \implies B(r) = \frac{\mu_0 I_{\text{enclosed}}(r)}{2\pi r}$

We analyze the field in three distinct regions:

1. Region I: $0 \le r < \frac{R}{2}$ (Hollow Inner Core)

Since there is no current-carrying conductor inside this region: $I_{\text{enclosed}}(r) = 0$ $B(r) = 0$

2. Region II: $\frac{R}{2} \le r \le R$ (Inside the Conductor Wall)

Let $J$ be the uniform current density. The enclosed cross-sectional area carrying current within radius $r$ is: $A_{\text{enclosed}}(r) = \pi r^2 - \pi \left(\frac{R}{2}\right)^2 = \pi \left(r^2 - \frac{R^2}{4}\right)$

Thus, the enclosed current is: $I_{\text{enclosed}}(r) = J \cdot \pi \left(r^2 - \frac{R^2}{4}\right)$

Substituting $I_{\text{enclosed}}(r)$ into Ampère's Law: $B(r) = \frac{\mu_0 J \pi \left(r^2 - \frac{R^2}{4}\right)}{2\pi r} = \frac{\mu_0 J}{2} \left(r - \frac{R^2}{4r}\right)$

Evaluating $B(r)$ at key points and checking its slope:

• At $r = \frac{R}{2}$: $B\left(\frac{R}{2}\right) = \frac{\mu_0 J}{2} \left(\frac{R}{2} - \frac{R^2}{4(R/2)}\right) = 0$
• Derivative with respect to $r$: $\frac{dB}{dr} = \frac{\mu_0 J}{2} \left(1 + \frac{R^2}{4r^2}\right) > 0$ This indicates that $B(r)$ is continuous at $r = R/2$ and increases monotonically up to $r = R$.
• At $r = R$: $B(R) = \frac{\mu_0 J}{2} \left(R - \frac{R}{4}\right) = \frac{3 \mu_0 J R}{8}$

3. Region III: $r > R$ (Outside the Conductor)

For any $r > R$, the Amperian loop encloses the total current $I_{\text{total}} = J \cdot \frac{3\pi R^2}{4}$: $I_{\text{enclosed}}(r) = I_{\text{total}}$ $B(r) = \frac{\mu_0 I_{\text{total}}}{2\pi r} \propto \frac{1}{r}$

The magnetic field decreases hyperbolically ($1/r$) for $r > R$.

Conclusion

• $B(r) = 0$ for $0 \le r < \frac{R}{2}$
• $B(r)$ starts from $0$ at $r = \frac{R}{2}$ and increases non-linearly to a maximum value at $r = R$
• $B(r) \propto \frac{1}{r}$ for $r > R$

This behavior corresponds precisely to the graph shown in Option (D).