A long thick conducting cylinder of radius 'R' carries a current uniformly distributed over its cross section :

Options

A. the magnetic field strength is maximum on the surface
B. the strength of the magnetic field inside the conductor will vary as inversely proportional to r where r is the distance from the axis.
C. the strength of the magnetic field outside the conductor varies as inversely e proportional to 1/r^2 where r is the distance from the axis
D. both (2) & (3)

Answer

To solve the problem regarding the magnetic field around a long thick conducting cylinder carrying a uniformly distributed current, we will follow these steps:

### Step 1: Understanding the Problem
We have a long thick conducting cylinder of radius \( R \) that carries a current \( I \) uniformly distributed over its cross-section. We need to find the magnetic field \( B \) at different points: inside the cylinder, at the surface, and outside the cylinder.

### Step 2: Apply Ampère's Law
Ampère's Law states that the line integral of the magnetic field \( B \) around a closed loop is equal to \( \mu_0 \) times the total current \( I \) enclosed by the loop:

\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]

### Step 3: Magnetic Field Inside the Cylinder
For a point inside the cylinder at a distance \( r \) (where \( r < R \)), we consider a circular Amperian loop of radius \( r \). The current enclosed by this loop is given by the fraction of the total current that flows through the area of the loop:

\[ I_{\text{enc}} = I \cdot \frac{A_{\text{loop}}}{A_{\text{cylinder}}} = I \cdot \frac{\pi r^2}{\pi R^2} = \frac{I r^2}{R^2} \]

Now applying Ampère's Law:

\[ B \cdot (2\pi r) = \mu_0 \cdot \frac{I r^2}{R^2} \]

Solving for \( B \):

\[ B = \frac{\mu_0 I r}{2\pi R^2} \]

### Step 4: Magnetic Field at the Surface of the Cylinder
At the surface of the cylinder, where \( r = R \):

\[ B = \frac{\mu_0 I R}{2\pi R^2} = \frac{\mu_0 I}{2\pi R} \]

### Step 5: Magnetic Field Outside the Cylinder
For a point outside the cylinder (where \( r > R \)), the entire current \( I \) is enclosed by the Amperian loop:

\[ B \cdot (2\pi r) = \mu_0 I \]

Solving for \( B \):

\[ B = \frac{\mu_0 I}{2\pi r} \]

### Step 6: Summary of Results
1. Inside the cylinder (\( r < R \)):
\[ B = \frac{\mu_0 I r}{2\pi R^2} \]
(Directly proportional to \( r \))

2. At the surface of the cylinder (\( r = R \)):
\[ B = \frac{\mu_0 I}{2\pi R} \]

3. Outside the cylinder (\( r > R \)):
\[ B = \frac{\mu_0 I}{2\pi r} \]
(Inversely proportional to \( r \))

### Conclusion
From the above calculations, we can conclude that:
- The magnetic field inside the cylinder increases linearly with \( r \).
- The magnetic field at the surface is maximum.
- The magnetic field outside the cylinder decreases inversely with \( r \).