To solve the problem of finding the magnitude of the magnetic force on an electron moving through a magnetic field, we will use the formula for the magnetic force acting on a charged particle:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge of the electron,
- [tex]\( v \)[/tex] is the speed of the electron,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity of the electron and the magnetic field.
Given values:
- [tex]\( v = 3.5 \times 10^5 \, \text{m/s} \)[/tex]
- [tex]\( B = 0.701 \, \text{T} \)[/tex]
- [tex]\( \theta = 60.0^\circ \)[/tex]
- [tex]\( q = 1.602 \times 10^{-19} \, \text{C} \)[/tex] (the elementary charge of an electron)
We need to follow these steps to find the magnitude of the magnetic force:
1. Convert the angle to radians:
[tex]\[
\theta_\text{rad} = \frac{60.0 \times \pi}{180} = \frac{\pi}{3}
\][/tex]
2. Calculate the sine of the angle:
[tex]\[
\sin\left( \frac{\pi}{3} \right) = \sin(60^\circ) = \frac{\sqrt{3}}{2}
\][/tex]
3. Substitute the known values into the magnetic force formula:
[tex]\[
F = (1.602 \times 10^{-19} \, \text{C}) \cdot (3.5 \times 10^5 \, \text{m/s}) \cdot (0.701 \, \text{T}) \cdot \frac{\sqrt{3}}{2}
\][/tex]
4. Perform the multiplication and simplification to find [tex]\( F \)[/tex]:
[tex]\[
F = 1.602 \times 10^{-19} \cdot 3.5 \times 10^5 \cdot 0.701 \cdot 0.866
\][/tex]
Putting all constants together:
[tex]\[
1.602 \times 3.5 \times 0.701 \times 0.866 \approx 3.403918911752562 \times 10^{-14} \, \text{N}
\][/tex]
Thus, the magnitude of the magnetic force on the electron is approximately
[tex]\[ 3.4 \times 10^{-14} \, \text{N} \][/tex]
Therefore, the correct answer is:
[tex]\[ 3.4 \times 10^{-14} \, \text{N} \][/tex]
