To calculate the force acting on an electron moving through a magnetic field, we can use the formula for the magnetic force on a charged particle, which is given by:
\[ F = qvB \sin(\theta) \]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( v \) is the speed of the particle,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity of the particle and the magnetic field.
In this case, the electron is moving at right angles to the magnetic field, so \( \theta = 90^\circ \), and \( \sin(90^\circ) = 1 \).
Given:
- \( q = 1.6 \times 10^{-19} , \text{C} \) (charge of the electron),
- \( v = 4.0 \times 10^{6} , \text{m/s} \),
- \( B = 1 , \text{T} \).
Now, substituting these values into the formula:
\[ F = (1.6 \times 10^{-19} , \text{C})(4.0 \times 10^{6} , \text{m/s})(1 , \text{T}) \]
Calculating it:
\[ F = 1.6 \times 10^{-19} \times 4.0 \times 10^{6} \]
\[ F = 6.4 \times 10^{-13} , \text{N} \]
Thus, the force acting on the electron is:
\[ \boxed{6.4 \times 10^{-13} , \text{N}} \]
This matches one of the answer choices.
2 answers