To determine the magnetic force acting on an electron, we will use the formula for the magnetic force [tex]\( F \)[/tex] on a charge moving perpendicular to a magnetic field, which is given by:
[tex]\[ F = qvB \][/tex]
where:
- [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.
Given:
- The speed of the electron [tex]\( v = 3.0 \times 10^6 \)[/tex] meters/second.
- The magnetic field strength [tex]\( B = 0.020 \)[/tex] teslas.
- The charge of the electron is [tex]\( q = -1.6 \times 10^{-19} \)[/tex] coulombs.
We substitute these values into the formula:
[tex]\[ F = (-1.6 \times 10^{-19} \, \text{C}) \times (3.0 \times 10^6 \, \text{m/s}) \times 0.020 \, \text{T} \][/tex]
Let's calculate this step-by-step:
1. Multiply the charge [tex]\( q \)[/tex] by the speed [tex]\( v \)[/tex]:
[tex]\[ q \times v = (-1.6 \times 10^{-19}) \times (3.0 \times 10^6) \][/tex]
[tex]\[ = -4.8 \times 10^{-13} \, \text{C} \cdot \text{m/s} \][/tex]
2. Now multiply the result by the magnetic field [tex]\( B \)[/tex]:
[tex]\[ F = (-4.8 \times 10^{-13}) \times 0.020 \][/tex]
[tex]\[ = -9.6 \times 10^{-15} \, \text{N} \][/tex]
So, the magnetic force acting on the electron is [tex]\( -9.6 \times 10^{-15} \)[/tex] newtons.
The correct answer is:
C. [tex]\( F = -9.6 \times 10^{-15} \)[/tex] newtons
