A long, hollow insulating cylinder of radius R and negligible thickness has a uniform surface charge density `sigma`. The cylinder rotates about its central axis at a constant angular speed `omega`. Choose the correct option(s):

Options

A. The charge crossing an imaginary line of length L on the surface of the cylinder, parallel to its axis, per unit time, is `omega sigma RL`.
B. The magnetic moment of a length L of the cylinder is `pi omega sigma R^(3) L`
C. The magnetic field intensity at a point on the axis of the cylinder, inside the cylinder, is `mu_(0)omega sigma R^(2)`
D. If the direction of rotation of the cylinder is reversed, the magnetic field at a point on the axis of the cylinder will also reverse direction

Explanation

To solve the problem regarding the long hollow insulating cylinder with a uniform surface charge density that rotates about its central axis, we will analyze the situation step by step.

### Step 1: Understanding the System
We have a long hollow insulating cylinder of radius \( R \) with a uniform surface charge density \( \sigma \). The cylinder rotates about its central axis at a constant angular speed \( \omega \).

### Step 2: Charge Crossing an Imaginary Line
To find the charge crossing an imaginary line of length \( L \) on the surface of the cylinder per unit time, we can use the following reasoning:

1. The total charge \( dq \) on the surface of the cylinder can be expressed as:
\[ dq = \sigma \cdot dA \]
where \( dA \) is the area element.

2. The area of the cylindrical surface for a length \( L \) is given by:
\[ dA = 2\pi R \cdot L \]

3. The total charge \( Q \) on the surface of length \( L \) is:
\[ Q = \sigma \cdot 2\pi R \cdot L \]

4. The time period \( T \) of one complete rotation is:
\[ T = \frac{2\pi}{\omega} \]

5. The charge crossing the imaginary line per unit time is:
\[ \text{Charge per unit time} = \frac{Q}{T} = \frac{\sigma \cdot 2\pi R \cdot L}{\frac{2\pi}{\omega}} = \sigma \cdot \omega \cdot R \cdot L \]

### Step 3: Magnetic Moment Calculation
Next, we need to find the magnetic moment \( M \) of the cylinder:

1. The magnetic moment \( M \) is given by:
\[ M = I \cdot A \]
where \( I \) is the current and \( A \) is the area.

2. The current \( I \) can be expressed as:
\[ I = \frac{dq}{dt} \]
where \( dq = \sigma \cdot dA \).

3. The area \( A \) for the circular cross-section of the cylinder is:
\[ A = \pi R^2 \]

4. Substituting \( I \):
\[ I = \sigma \cdot \omega \cdot R \cdot L \]

5. Therefore, the magnetic moment \( M \) becomes:
\[ M = I \cdot A = \left(\sigma \cdot \omega \cdot R \cdot L\right) \cdot \pi R^2 = \pi \sigma \omega R^3 L \]

### Step 4: Magnetic Field Intensity Calculation
Now, we need to find the magnetic field intensity \( B \) at a point on the axis of the cylinder:

1. The magnetic field intensity \( B \) at a distance \( x \) from the center of the cylinder can be calculated using the formula:
\[ B = \frac{\mu_0 I}{2R^2 + x^2} \cdot \frac{R^2}{2} \]

2. Integrating over the entire length of the cylinder gives:
\[ B = \frac{\mu_0 \sigma \omega R^3}{2} \int \frac{dx}{(R^2 + x^2)^{3/2}} \]

3. The limits of integration are from \(-\infty\) to \(+\infty\), leading to:
\[ B = \mu_0 \sigma \omega R \]

### Step 5: Effect of Reversing Rotation
If the direction of rotation of the cylinder is reversed, the magnetic field direction will also reverse. This is due to the right-hand rule applied to the direction of current and the resulting magnetic field.

### Conclusion
The correct options based on the calculations are:
1. The charge crossing an imaginary line of length \( L \) is \( \sigma \omega R L \).
2. The magnetic moment is \( \pi \sigma \omega R^3 L \).
3. The magnetic field intensity at a point on the axis is \( \mu_0 \sigma \omega R \).
4. Reversing the cylinder's rotation reverses the magnetic field direction.