A long, thick straight conductor of radius `R` carries current `I` uniformly distributed in its cross section area.The ratio of energy density of the magnetic field at distance `R//2` from surface inside the conductor and outside the conductor is:

Options

A. `1:16`
B. `1:1`
C. `1:4`
D. `9//16`

Explanation

To solve the problem, we need to find the ratio of the energy density of the magnetic field at a distance \( R/2 \) from the surface inside the conductor and outside the conductor.

### Step-by-Step Solution:

1. **Identify the Regions**:
- Inside the conductor at a distance \( R/2 \) from the surface means we are at a distance \( R - R/2 = R/2 \) from the center of the conductor.
- Outside the conductor, we will consider a point at a distance \( R + R/2 = 3R/2 \) from the center.

2. **Use Ampere's Law**:
- For a cylindrical conductor, the magnetic field \( B \) at a distance \( r \) from the center can be calculated using Ampere's law.
- Inside the conductor (at \( r = R/2 \)):
\[ B_1 = \frac{\mu_0 I_1}{2 \pi r} \]
- Outside the conductor (at \( r = 3R/2 \)):
\[ B_2 = \frac{\mu_0 I_2}{2 \pi r} \]

3. **Calculate the Current Enclosed \( I_1 \)**:
- The current \( I \) is uniformly distributed. The area of the cross-section of the conductor is \( \pi R^2 \).
- The area of the circle at \( r = R/2 \) is \( \pi (R/2)^2 = \frac{\pi R^2}{4} \).
- The fraction of the total current \( I \) enclosed within this area is:
\[ I_1 = I \cdot \frac{\text{Area at } R/2}{\text{Total Area}} = I \cdot \frac{\frac{\pi R^2}{4}}{\pi R^2} = \frac{I}{4} \]

4. **Substitute \( I_1 \) into \( B_1 \)**:
\[ B_1 = \frac{\mu_0 \cdot \frac{I}{4}}{2 \pi \cdot \frac{R}{2}} = \frac{\mu_0 I}{4 \pi R/2} = \frac{\mu_0 I}{2 \pi R} \]

5. **Calculate the Current Enclosed \( I_2 \)**:
- For the outside region, the total current \( I_2 = I \) since we are considering the entire conductor.
- Thus, for \( r = 3R/2 \):
\[ B_2 = \frac{\mu_0 I}{2 \pi \cdot \frac{3R}{2}} = \frac{\mu_0 I}{3 \pi R} \]

6. **Calculate the Ratio of Magnetic Fields**:
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2 \pi R}}{\frac{\mu_0 I}{3 \pi R}} = \frac{3}{2} \]

7. **Energy Density Relation**:
- The energy density \( u \) of the magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
- Therefore, the ratio of energy densities \( \frac{u_1}{u_2} \) is:
\[ \frac{u_1}{u_2} = \left(\frac{B_1}{B_2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

8. **Final Ratio of Energy Densities**:
- The ratio of the energy density of the magnetic field at distance \( R/2 \) from the surface inside the conductor to that outside the conductor is:
\[ \frac{u_1}{u_2} = \frac{9}{16} \]

### Conclusion:
The ratio of the energy density of the magnetic field at distance \( R/2 \) from the surface inside the conductor and outside the conductor is \( \frac{9}{16} \).