Find the total field energy of magnetic field stored per unit length inside a long cylindrical wire of radius R and carrying a current I.

Answer

To find the total magnetic field energy stored per unit length inside a long cylindrical wire of radius \( R \) carrying a current \( I \), we can follow these steps:

### Step 1: Define the Current Density
The current density \( J \) is defined as the current \( I \) flowing through the wire per unit area. For a cylindrical wire of radius \( R \), the cross-sectional area \( A \) is given by:
\[ A = \pi R^2 \]
Thus, the current density \( J \) can be expressed as:
\[ J = \frac{I}{\pi R^2} \]

### Step 2: Consider an Elemental Shell
Consider an elemental cylindrical shell of radius \( x \) and thickness \( dx \). The current flowing through this shell, denoted as \( I' \), can be expressed in terms of the current density:
\[ I' = J \cdot A' = J \cdot (2\pi x \, dx) = \frac{I}{\pi R^2} \cdot (2\pi x \, dx) = \frac{2I x}{R^2} \, dx \]

### Step 3: Calculate the Magnetic Field
The magnetic field \( B \) at a distance \( x \) from the center of the wire due to the current \( I' \) flowing in the shell is given by Ampère's Law:
\[ B = \frac{\mu_0 I'}{2\pi x} = \frac{\mu_0}{2\pi x} \cdot \frac{2I x}{R^2} = \frac{\mu_0 I}{\pi R^2} \]

### Step 4: Calculate the Elemental Volume
The volume \( dV \) of the elemental shell is given by:
\[ dV = 2\pi x \, dx \cdot 1 \quad (\text{considering unit length}) \]

### Step 5: Calculate the Elemental Magnetic Field Energy
The energy density \( u \) of the magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
Substituting for \( B \):
\[ u = \frac{1}{2\mu_0} \left( \frac{\mu_0 I}{\pi R^2} \right)^2 = \frac{\mu_0 I^2}{2\pi^2 R^4} \]
The elemental magnetic field energy \( du \) is then:
\[ du = u \cdot dV = \frac{\mu_0 I^2}{2\pi^2 R^4} \cdot (2\pi x \, dx) = \frac{\mu_0 I^2}{\pi R^4} x \, dx \]

### Step 6: Integrate to Find Total Energy
To find the total energy \( U \) stored in the magnetic field per unit length, integrate \( du \) from \( x = 0 \) to \( x = R \):
\[ U = \int_0^R \frac{\mu_0 I^2}{\pi R^4} x \, dx = \frac{\mu_0 I^2}{\pi R^4} \cdot \left[ \frac{x^2}{2} \right]_0^R = \frac{\mu_0 I^2}{\pi R^4} \cdot \frac{R^2}{2} = \frac{\mu_0 I^2}{2\pi R^2} \]

### Final Result
Thus, the total magnetic field energy stored per unit length inside the cylindrical wire is:
\[ U = \frac{\mu_0 I^2}{2\pi R^2} \]