A very long metallic hollow pipe of length l and radius `r (ltltl)` is carrying charge Q , uniformly distributed upon it . The pipe is rotated about its axis with constant angular speed `omega` . The energy stored in the pipe of length `l/100` is

Options

A. `(mu_0Q^2omega^2r^2)/(100pil) `
B. `(mu_0Q^2omega^2r^2)/(200pil) `
C. `(mu_0Q^2omega^2r^2)/(800pil) `
D. `(mu_0Q^2omega^2r^2)/(400pil) `

Answer

To solve the problem, we need to determine the energy stored in a very long metallic hollow pipe of length \( \frac{l}{100} \) that is rotating with a constant angular speed \( \omega \) and carrying a uniformly distributed charge \( Q \).

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a long metallic hollow pipe of length \( l \) and radius \( r \) that is rotating about its axis with an angular speed \( \omega \). The charge \( Q \) is uniformly distributed along the length of the pipe.

2. **Determine the Equivalent Current**:
When the pipe rotates, the charges on the pipe behave like a current. The equivalent current \( I \) can be calculated using the formula:
\[ I = Q \cdot f \]
where \( f \) is the frequency of rotation. The frequency can be expressed in terms of angular speed \( \omega \):
\[ f = \frac{\omega}{2\pi} \]
Thus, the equivalent current becomes:
\[ I = Q \cdot \frac{\omega}{2\pi} \]

3. **Magnetic Field Calculation**:
The pipe can be treated like a solenoid. The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 \cdot n \cdot I \]
where \( n \) is the number of turns per unit length. In our case, since the pipe is long, we can consider the entire length \( l \) as one turn. Therefore:
\[ n = \frac{1}{l} \]
Substituting for \( I \):
\[ B = \mu_0 \cdot \frac{Q}{l} \cdot \frac{\omega}{2\pi} \]

4. **Energy Density Calculation**:
The energy density \( u \) in a magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
Substituting the expression for \( B \):
\[ u = \frac{1}{2\mu_0} \left( \mu_0 \cdot \frac{Q \omega}{2\pi l} \right)^2 = \frac{\mu_0 Q^2 \omega^2}{8\pi^2 l^2} \]

5. **Volume of the Section**:
We need to find the energy stored in a section of the pipe of length \( \frac{l}{100} \). The volume \( V \) of this section is:
\[ V = \pi r^2 \cdot \frac{l}{100} \]

6. **Total Energy Calculation**:
The total energy \( U \) stored in this volume can be calculated by multiplying the energy density by the volume:
\[ U = u \cdot V = \left( \frac{\mu_0 Q^2 \omega^2}{8\pi^2 l^2} \right) \cdot \left( \pi r^2 \cdot \frac{l}{100} \right) \]
Simplifying this expression:
\[ U = \frac{\mu_0 Q^2 \omega^2 \cdot \pi r^2 \cdot \frac{l}{100}}{8\pi^2 l^2} = \frac{\mu_0 Q^2 \omega^2 r^2}{800 \pi l} \]

### Final Answer:
The energy stored in the pipe of length \( \frac{l}{100} \) is:
\[ U = \frac{\mu_0 Q^2 \omega^2 r^2}{800 \pi l} \]