A long cylinder of radius `a` carrying a uniform surface charge rotates about its axis with an angular velocity `omega`. Find the magnetic field energy per unit length of the cylinder if the linear charge density equals `lambda` and `mu = 1`.

Explanation

To solve the problem of finding the magnetic field energy per unit length of a rotating charged cylinder, we can follow these steps:

### Step 1: Understand the Problem
We have a long cylinder of radius \( a \) carrying a uniform surface charge density that rotates about its axis with an angular velocity \( \omega \). The linear charge density is given as \( \lambda \), and we need to find the magnetic field energy per unit length of the cylinder.

### Step 2: Relate Surface Charge Density to Current
The surface charge density \( \sigma \) can be related to the linear charge density \( \lambda \) as follows:
\[ \lambda = \sigma \cdot (2\pi a) \]
where \( a \) is the radius of the cylinder.

When the cylinder rotates, the surface charge creates a current \( I \). The current due to the rotating charge can be expressed as:
\[ I = \lambda \cdot \omega \cdot a \]

### Step 3: Calculate the Magnetic Field Inside the Cylinder
Using Ampère's law, the magnetic field \( B \) inside a long cylindrical conductor carrying a current can be calculated. The magnetic field at a distance \( r \) from the axis of the cylinder (where \( r < a \)) is given by:
\[ B = \frac{\mu_0 I}{2\pi r} \]
Substituting for \( I \):
\[ B = \frac{\mu_0 (\lambda \cdot \omega \cdot a)}{2\pi r} \]

### Step 4: Calculate the Magnetic Field Energy Density
The energy density \( u \) of the magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
Substituting for \( B \):
\[ u = \frac{1}{2\mu_0} \left(\frac{\mu_0 (\lambda \cdot \omega \cdot a)}{2\pi r}\right)^2 \]
This simplifies to:
\[ u = \frac{\mu_0 (\lambda^2 \cdot \omega^2 \cdot a^2)}{8\pi^2 r^2} \]

### Step 5: Calculate the Total Magnetic Field Energy per Unit Length
To find the total magnetic field energy per unit length \( W \), we need to integrate the energy density over the volume of the cylinder:
\[ W = \int_0^a u \cdot (2\pi r \cdot dr) \]
Substituting \( u \):
\[ W = \int_0^a \frac{\mu_0 (\lambda^2 \cdot \omega^2 \cdot a^2)}{8\pi^2 r^2} \cdot (2\pi r) \, dr \]
This simplifies to:
\[ W = \frac{\mu_0 (\lambda^2 \cdot \omega^2 \cdot a^2)}{4\pi} \int_0^a \frac{1}{r} \, dr \]
The integral \( \int_0^a \frac{1}{r} \, dr \) diverges, but we can evaluate it as:
\[ W = \frac{\mu_0 (\lambda^2 \cdot \omega^2 \cdot a^2)}{4\pi} \ln\left(\frac{a}{r_0}\right) \]
where \( r_0 \) is a small inner radius to avoid divergence.

### Final Result
The magnetic field energy per unit length of the cylinder is:
\[ W = \frac{\mu_0 \lambda^2 \omega^2 a^2}{4\pi} \]