An electron after being accelerated through a potential difference of 100 V enters a uniform magnetic field of `0*004T`, perpendicular to its direction of motion. Calculate the radius of the path described by the electron.

Explanation

To solve the problem of finding the radius of the path described by an electron after being accelerated through a potential difference of 100 V and entering a uniform magnetic field of 0.004 T, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between potential difference and kinetic energy**:
When an electron is accelerated through a potential difference (V), it gains kinetic energy (KE) equal to the work done on it by the electric field. The kinetic energy can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, C \)) and \( V \) is the potential difference (100 V).

2. **Calculate the kinetic energy of the electron**:
\[ KE = eV = (1.6 \times 10^{-19} \, C)(100 \, V) = 1.6 \times 10^{-17} \, J \]

3. **Relate kinetic energy to speed**:
The kinetic energy can also be expressed in terms of the mass (m) and speed (v) of the electron:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = eV \]

4. **Solve for the speed (v) of the electron**:
Rearranging the equation:
\[ v^2 = \frac{2eV}{m} \]
Taking the square root:
\[ v = \sqrt{\frac{2eV}{m}} \]

5. **Substitute the known values**:
The mass of the electron (\( m \)) is \( 9.1 \times 10^{-31} \, kg \). Now substituting the values:
\[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19} \, C) \times (100 \, V)}{9.1 \times 10^{-31} \, kg}} \]

6. **Calculate the speed (v)**:
\[ v = \sqrt{\frac{3.2 \times 10^{-17}}{9.1 \times 10^{-31}}} \approx \sqrt{3.51 \times 10^{13}} \approx 5.93 \times 10^6 \, m/s \]

7. **Use the formula for the radius of the circular path**:
The radius (r) of the circular path of the electron in a magnetic field is given by:
\[ r = \frac{mv}{eB} \]
where \( B \) is the magnetic field strength (0.004 T).

8. **Substitute the values into the radius formula**:
\[ r = \frac{(9.1 \times 10^{-31} \, kg)(5.93 \times 10^6 \, m/s)}{(1.6 \times 10^{-19} \, C)(0.004 \, T)} \]

9. **Calculate the radius (r)**:
\[ r = \frac{5.3963 \times 10^{-24}}{6.4 \times 10^{-23}} \approx 0.0843 \, m = 8.43 \, mm \]

### Final Answer:
The radius of the path described by the electron is approximately **8.43 mm**.