The acceleration of any electron due to the magnetic fields is

Options

A. `3.14xx10^(14)m//s^(2)`
B. `4.28xx10^(14)m//s^(2)`
C. `5.56xx10^(12)m//s^(2)`
D. `6.28xx10^(14)m//s^(2)`

Answer

To find the acceleration of an electron due to a magnetic field, we can follow these steps:

### Step 1: Understand the Force on a Charged Particle in a Magnetic Field
The force experienced by a charged particle (like an electron) moving in a magnetic field is given by the Lorentz force equation:
\[ F = q(\mathbf{v} \times \mathbf{B}) \]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( \mathbf{v} \) is the velocity vector of the particle,
- \( \mathbf{B} \) is the magnetic field vector.

### Step 2: Analyze Different Cases of Electron Projection
1. **Parallel to the Magnetic Field**: If the electron is projected parallel to the magnetic field, the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{B} \) is 0 degrees. Thus:
\[ F = qvB \sin(0) = 0 \]
Therefore, the acceleration \( a \) is also 0.

2. **Perpendicular to the Magnetic Field**: If the electron is projected perpendicular to the magnetic field, the angle \( \theta \) is 90 degrees. Thus:
\[ F = qvB \sin(90) = qvB \]
This force causes the electron to move in a circular path, and the centripetal force required for circular motion is provided by this magnetic force:
\[ \frac{mv^2}{r} = qvB \]
Rearranging gives:
\[ r = \frac{mv}{qB} \]

3. **At an Angle \( \theta \)**: If the electron is projected at an angle \( \theta \) to the magnetic field, the force can be split into two components: one parallel and one perpendicular to the magnetic field. The parallel component does not contribute to acceleration, while the perpendicular component does.

### Step 3: Calculate the Acceleration
The acceleration \( a \) due to the magnetic force can be expressed as:
\[ a = \frac{F}{m} = \frac{qvB}{m} \]

### Step 4: Substitute Known Values
Using the known values:
- Charge of an electron, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of an electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Assuming a constant velocity \( v \) and magnetic field \( B \), we can express acceleration as:
\[ a = \frac{qBv}{m} \]

### Step 5: Example Calculation
Assuming \( v/B \) is a constant value, we can substitute values to find acceleration:
\[ a = \frac{(1.6 \times 10^{-19})(B)(v)}{9.1 \times 10^{-31}} \]
Assuming \( \frac{v}{B} = 35.7 \times 10^2 \):
\[ a = \frac{(1.6 \times 10^{-19})(35.7 \times 10^2)}{9.1 \times 10^{-31}} \approx 6.28 \times 10^{14} \, \text{m/s}^2 \]

### Conclusion
The acceleration of the electron due to the magnetic field is approximately:
\[ a \approx 6.28 \times 10^{14} \, \text{m/s}^2 \]