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 find the ratio of the energy density of the magnetic field at a distance \( R/2 \) from the surface inside the conductor and at a distance \( 3R/2 \) outside the conductor, we can follow these steps:

### Step 1: Understand the Problem
We have a long, thick straight conductor of radius \( R \) carrying a current \( I \) uniformly distributed across its cross-sectional area. We need to find the magnetic field at two points: one inside the conductor at a distance \( R/2 \) from the surface and one outside the conductor at a distance \( 3R/2 \) from the center.

### Step 2: Calculate the Magnetic Field Inside the Conductor
For a point inside the conductor at a distance \( r \) from the center (where \( r = R/2 \)):
The magnetic field \( B \) inside a long straight conductor is given by:
\[ B = \frac{\mu_0 I r}{2 \pi R^2} \]
Here, \( I \) is the total current, and \( R \) is the radius of the conductor. Since the current is uniformly distributed, the current enclosed \( I_{enc} \) at radius \( r \) is:
\[ I_{enc} = I \cdot \frac{r^2}{R^2} \]
Substituting \( r = R/2 \):
\[ I_{enc} = I \cdot \frac{(R/2)^2}{R^2} = I \cdot \frac{1}{4} = \frac{I}{4} \]
Now substituting this into the magnetic field equation:
\[ B\left(\frac{R}{2}\right) = \frac{\mu_0 \left(\frac{I}{4}\right)}{2 \pi \left(\frac{R}{2}\right)} = \frac{\mu_0 I}{4 \pi R} \]

### Step 3: Calculate the Magnetic Field Outside the Conductor
For a point outside the conductor at a distance \( r = 3R/2 \):
The magnetic field \( B \) outside a long straight conductor is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
Substituting \( r = 3R/2 \):
\[ B\left(\frac{3R}{2}\right) = \frac{\mu_0 I}{2 \pi \left(\frac{3R}{2}\right)} = \frac{\mu_0 I}{3 \pi R} \]

### Step 4: Calculate the Energy Density of the Magnetic Field
The energy density \( u \) of the magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
Now, we can calculate the energy density at both points.

1. **Inside the conductor at \( R/2 \)**:
\[ u_{inside} = \frac{B\left(\frac{R}{2}\right)^2}{2\mu_0} = \frac{\left(\frac{\mu_0 I}{4 \pi R}\right)^2}{2\mu_0} = \frac{\mu_0 I^2}{32 \pi^2 R^2} \]

2. **Outside the conductor at \( 3R/2 \)**:
\[ u_{outside} = \frac{B\left(\frac{3R}{2}\right)^2}{2\mu_0} = \frac{\left(\frac{\mu_0 I}{3 \pi R}\right)^2}{2\mu_0} = \frac{\mu_0 I^2}{18 \pi^2 R^2} \]

### Step 5: Find the Ratio of Energy Densities
Now, we can find the ratio of the energy densities:
\[ \frac{u_{inside}}{u_{outside}} = \frac{\frac{\mu_0 I^2}{32 \pi^2 R^2}}{\frac{\mu_0 I^2}{18 \pi^2 R^2}} = \frac{18}{32} = \frac{9}{16} \]

### Conclusion
The ratio of the energy density of the magnetic field at distance \( R/2 \) from the surface inside the conductor to that at distance \( 3R/2 \) outside the conductor is:
\[ \frac{9}{16} \]