An electron is moving with a speed of `3.5xx10^(5)` m//s when it encounters a magnetic field of `0.60` T. The direction of the magnetic field makes an angle of `60.0^(@)` with respect to the velocity of the electron. What is the magnitude of the magnetic force on the electron ?

Options

A. `4.9xx10^(-13)` N
B. `2.9xx10^(-14)` N
C. `3.2xx10^(-13)` N
D. `3.4xx10^(-14)` N

Answer

To find the magnitude of the magnetic force acting on the electron, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:

\[ F = q \cdot v \cdot B \cdot \sin(\theta) \]

Where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the electron,
- \( v \) is the velocity of the electron,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity and the magnetic field.

### Step 1: Identify the known values
- Charge of the electron, \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Velocity of the electron, \( v = 3.5 \times 10^{5} \, \text{m/s} \)
- Magnetic field strength, \( B = 0.6 \, \text{T} \)
- Angle, \( \theta = 60^\circ \)

### Step 2: Calculate \( \sin(\theta) \)
We need to find \( \sin(60^\circ) \):
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 3: Substitute the values into the formula
Now we can substitute the known values into the magnetic force formula:
\[ F = (1.6 \times 10^{-19} \, \text{C}) \cdot (3.5 \times 10^{5} \, \text{m/s}) \cdot (0.6 \, \text{T}) \cdot \left(\frac{\sqrt{3}}{2}\right) \]

### Step 4: Calculate the force
First, calculate the product of the constants:
1. Calculate \( 1.6 \times 10^{-19} \cdot 3.5 \times 10^{5} \):
\[ 1.6 \times 3.5 = 5.6 \]
\[ 5.6 \times 10^{-14} \, \text{C m/s} \]

2. Now multiply by \( 0.6 \):
\[ 5.6 \times 0.6 = 3.36 \]
\[ 3.36 \times 10^{-14} \, \text{C m/s} \]

3. Finally, multiply by \( \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} \approx 0.866 \]
\[ 3.36 \times 0.866 \approx 2.91 \]
\[ 2.91 \times 10^{-14} \, \text{N} \]

### Final Result
Thus, the magnitude of the magnetic force on the electron is approximately:
\[ F \approx 2.9 \times 10^{-14} \, \text{N} \]