a long coaxial cable consits of two thin-walled conducting cyclider carries a steady current `0.1 A`, and the outer cylinder provides the return path for that current. The current produced a magnetic field between the two cylinders. Find the energy stored in the magnetic field for length `5m` of the cable. Express answer in Nj (use `1n2 = 0.7`).

Explanation

To find the energy stored in the magnetic field of a coaxial cable, we will follow these steps:

### Step 1: Understand the setup
We have a coaxial cable consisting of two thin-walled conducting cylinders. A steady current \( I = 0.1 \, \text{A} \) flows through the inner cylinder, while the outer cylinder serves as the return path. We need to calculate the energy stored in the magnetic field between these two cylinders over a length \( L = 5 \, \text{m} \).

### Step 2: Magnetic field calculation
The magnetic field \( B \) at a distance \( r \) from the center of the inner cylinder (within the region between the two cylinders) is given by Ampère's Law:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \) is the permeability of free space.

### Step 3: Energy density of the magnetic field
The energy density \( u \) (energy per unit volume) of the magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
Substituting the expression for \( B \):
\[ u = \frac{1}{2\mu_0} \left( \frac{\mu_0 I}{2 \pi r} \right)^2 = \frac{\mu_0 I^2}{8 \pi^2 r^2} \]

### Step 4: Total energy calculation
To find the total energy \( U \) stored in the magnetic field between the two cylinders, we need to integrate the energy density over the volume between the cylinders. The volume element \( dV \) in cylindrical coordinates is given by:
\[ dV = 2 \pi r \, dr \, L \]
Thus, the total energy is:
\[ U = \int_{R_1}^{R_2} u \, dV = \int_{R_1}^{R_2} \frac{\mu_0 I^2}{8 \pi^2 r^2} (2 \pi r \, dr \, L) \]
This simplifies to:
\[ U = \frac{\mu_0 I^2 L}{4 \pi} \int_{R_1}^{R_2} \frac{1}{r} \, dr \]
The integral \( \int \frac{1}{r} \, dr = \ln r \), so we have:
\[ U = \frac{\mu_0 I^2 L}{4 \pi} \left[ \ln R_2 - \ln R_1 \right] = \frac{\mu_0 I^2 L}{4 \pi} \ln \left( \frac{R_2}{R_1} \right) \]

### Step 5: Substitute values
Assuming the inner radius \( R_1 = 0.02 \, \text{m} \) and the outer radius \( R_2 = 0.08 \, \text{m} \):
\[ \ln \left( \frac{R_2}{R_1} \right) = \ln \left( \frac{0.08}{0.02} \right) = \ln(4) \]
Using the approximation \( \ln(4) \approx 2 \ln(2) \) and \( \ln(2) \approx 0.7 \):
\[ \ln(4) \approx 2 \times 0.7 = 1.4 \]

Now substituting all values:
\[ U = \frac{(4 \pi \times 10^{-7}) (0.1)^2 (5)}{4 \pi} \cdot 1.4 \]
\[ U = (10^{-7}) (0.01) (5) \cdot 1.4 = 7 \times 10^{-9} \, \text{J} \]
Converting to nanojoules:
\[ U = 7 \, \text{nJ} \]

### Final Answer
The energy stored in the magnetic field for a length of 5 m of the coaxial cable is \( 7 \, \text{nJ} \).

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