The potential different across the Coolidge tube is `20 kV and 10 m A` current flows through the voltage supply. Only `0.5%` of the energy carried by the electrons striking the target is converted into X-ray. The power carried by the X-ray beam is `p`. Then

Options

A. `P = 0.1 W`
B. `P = 1 W`
C. `P = 2 W`
D. `P = 10 W`

Explanation

To solve the problem, we need to calculate the power carried by the X-ray beam (denoted as \( P \)) based on the given information about the Coolidge tube.

### Step 1: Calculate the total power supplied to the Coolidge tube
The power \( P \) supplied to the Coolidge tube can be calculated using the formula:
\[ P = V \times I \]
where:
- \( V \) is the potential difference (in volts),
- \( I \) is the current (in amperes).

Given:
- \( V = 20 \, \text{kV} = 20 \times 10^3 \, \text{V} \)
- \( I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} \)

Substituting the values into the formula:
\[ P = (20 \times 10^3) \times (10 \times 10^{-3}) = 200 \, \text{W} \]

### Step 2: Calculate the power converted into X-rays
According to the problem, only \( 0.5\% \) of the total power is converted into X-rays. We can calculate the power of the X-ray beam using the formula:
\[ P_{\text{X-ray}} = \left( \frac{0.5}{100} \right) \times P \]

Substituting the total power calculated in Step 1:
\[ P_{\text{X-ray}} = \left( \frac{0.5}{100} \right) \times 200 = 1 \, \text{W} \]

### Conclusion
The power carried by the X-ray beam is:
\[ P_{\text{X-ray}} = 1 \, \text{W} \]