Step-by-step Derivations

1. Analyze the problem and apply Ampere's Law We are given an infinitely long hollow conducting cylinder with inner radius a = R/2 and outer radius b = R. It carries a uniform current density J parallel to its axis. We need to find the magnetic field B as a function of the radial distance r from the axis.

   Due to the cylindrical symmetry of the current distribution, the magnetic field lines will be concentric circles around the axis. The magnitude of the magnetic field B will only depend on the radial distance r. We will use Ampere's circuital law to find the magnetic field in different regions.

   Ampere's Law states: $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I_{enc}$ For a circular Amperian loop of radius r concentric with the cylinder, the left side of the equation becomes: $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = B \oint dl = B (2\pi r)$ So, $B(2\pi r) = \mu_0 I_{enc} \implies B = \frac{\mu_0 I_{enc}}{2\pi r}$ We need to find the enclosed current $I_{enc}$ for different radial distances r.

2. Region 1: Inside the hollow part (r < R/2) For an Amperian loop with radius r < R/2, the loop is inside the hollow region where there is no current. $I_{enc} = 0$ Therefore, the magnetic field is zero in this region. $B = 0 \quad \text{for} \quad 0 \le r < R/2$

3. Region 2: Inside the conductor ($R/2 \le r \le R$) For an Amperian loop with radius $R/2 \le r \le R$, the loop encloses the current flowing through the cross-sectional area between R/2 and r. The area of the conductor enclosed by the loop is $A_{enc} = \pi r^2 - \pi (R/2)^2 = \pi(r^2 - R^2/4)$. The enclosed current is $I_{enc} = J \cdot A_{enc} = J \pi (r^2 - R^2/4)$.

   Applying Ampere's Law: $B(2\pi r) = \mu_0 [J \pi (r^2 - R^2/4)]$ $B = \frac{\mu_0 J}{2r} (r^2 - \frac{R^2}{4}) = \frac{\mu_0 J}{2} (r - \frac{R^2}{4r})$ Let's check the values at the boundaries: At r = R/2, $B = \frac{\mu_0 J}{2} (\frac{R}{2} - \frac{R^2}{4(R/2)}) = \frac{\mu_0 J}{2} (\frac{R}{2} - \frac{R}{2}) = 0$. This is continuous with Region 1. At r = R, $B = \frac{\mu_0 J}{2} (R - \frac{R^2}{4R}) = \frac{\mu_0 J}{2} (R - \frac{R}{4}) = \frac{3\mu_0 J R}{8}$. The function is non-linear and increasing in this region. The slope $dB/dr = \frac{\mu_0 J}{2}(1 + \frac{R^2}{4r^2})$ is positive and decreases as r increases, so the curve is concave down.

4. Region 3: Outside the cylinder (r > R) For an Amperian loop with radius r > R, the loop encloses the total current flowing through the cylinder. The total cross-sectional area of the conductor is $A_{total} = \pi R^2 - \pi (R/2)^2 = \frac{3\pi R^2}{4}$. The total current is $I_{total} = J \cdot A_{total} = J \frac{3\pi R^2}{4}$.

   Applying Ampere's Law: $B(2\pi r) = \mu_0 I_{total} = \mu_0 (J \frac{3\pi R^2}{4})$ $B = \frac{3 \mu_0 J R^2}{8r}$ In this region, B is proportional to 1/r. At r = R, $B = \frac{3 \mu_0 J R^2}{8R} = \frac{3\mu_0 J R}{8}$. This is continuous with Region 2.

Summary of Magnetic Field Behavior:

• For $0 \le r < R/2$: B = 0. The graph is a horizontal line on the r-axis.
• For $R/2 \le r \le R$: B increases non-linearly from 0 to a maximum value at r=R. The curve $B(r) = \frac{\mu_0 J}{2} (r - \frac{R^2}{4r})$ is concave down.
• For r > R: B decreases hyperbolically, as $B \propto 1/r$.

Evaluate the Options:

• Option A: Incorrect. Shows B is non-zero for r < R/2.
• Option B: Incorrect. Shows B is non-zero for r < R/2.
• Option C: Incorrect. Shows a linear increase for $R/2 \le r \le R$. Our derived function is non-linear.
• Option D: Correct.
  • B=0 for r < R/2.
  • B increases non-linearly from r=R/2 to r=R.
  • B decreases as 1/r for r > R. This graph correctly represents the derived behavior of the magnetic field.