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 find the energy stored in a very long metallic hollow pipe of length \( l \) and radius \( r \) that is rotating with a constant angular speed \( \omega \) and carrying a uniformly distributed charge \( Q \). We specifically want to find the energy stored in a segment of the pipe of length \( \frac{l}{100} \).

### Step-by-Step Solution:

1. **Identify the Equivalent Current**:
The charge \( Q \) is uniformly distributed along the length of the pipe. When the pipe rotates with angular speed \( \omega \), the charge on the pipe can be thought of as creating an equivalent current \( I \). The time period \( T \) for one complete rotation is given by:
\[ T = \frac{2\pi}{\omega} \]
Therefore, the equivalent current \( I \) can be expressed as:
\[ I = \frac{Q}{T} = \frac{Q \omega}{2\pi} \]

2. **Calculate the Magnetic Field**:
The rotating charge creates a magnetic field similar to that of a solenoid. For a long solenoid, the magnetic field \( B \) is given by:
\[ B = \frac{\mu_0 n I}{L} \]
Here, \( n \) is the number of turns per unit length. In our case, since we have a continuous charge distribution, we can relate it to the equivalent current:
\[ B = \frac{\mu_0 Q \omega}{2\pi l} \]

3. **Energy Density of the Magnetic Field**:
The energy density \( u \) (energy per unit volume) stored in a magnetic field is given by:
\[ u = \frac{B^2}{2\mu_0} \]
Substituting the expression for \( B \):
\[ u = \frac{1}{2\mu_0} \left(\frac{\mu_0 Q \omega}{2\pi l}\right)^2 = \frac{\mu_0 Q^2 \omega^2}{8\pi^2 l^2} \]

4. **Calculate the Volume of the Segment**:
We need to find the energy stored in a segment of length \( \frac{l}{100} \). The volume \( V \) of this segment is:
\[ V = \pi r^2 \left(\frac{l}{100}\right) \]

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

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