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 find `p` .

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the given values into standard units
- The potential difference across the Coolidge tube is given as \( 20 \, \text{kV} \). We convert this to volts:
\[ V = 20 \, \text{kV} = 20 \times 10^3 \, \text{V} = 20000 \, \text{V} \]
- The current flowing through the voltage supply is given as \( 10 \, \text{mA} \). We convert this to amperes:
\[ I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} = 0.01 \, \text{A} \]

### Step 2: Calculate the total power supplied to the tube
The power \( P_T \) supplied to the tube can be calculated using the formula:
\[ P_T = V \times I \]
Substituting the values we have:
\[ P_T = 20000 \, \text{V} \times 0.01 \, \text{A} = 200 \, \text{W} \]

### Step 3: Calculate the power converted into X-rays
We know that only \( 0.5\% \) of the total power is converted into X-rays. To find the power \( P \) of the X-ray beam, we use the formula:
\[ P = \frac{0.5}{100} \times P_T \]
Substituting the value of \( P_T \):
\[ P = 0.005 \times 200 \, \text{W} = 1 \, \text{W} \]

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