To solve this problem, we need to calculate the magnitude of the magnetic force acting on a moving point charge. The relevant formula to use here is [tex]\( F = qvB \)[/tex], where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge,
- [tex]\( v \)[/tex] is the velocity of the charge, and
- [tex]\( B \)[/tex] is the magnetic field strength.
We are given:
- [tex]\( q = 5.0 \times 10^{-7} \)[/tex] Coulombs,
- [tex]\( v = 2.6 \times 10^5 \)[/tex] meters per second,
- [tex]\( B = 1.8 \times 10^{-2} \)[/tex] Teslas.
Step-by-step, we substitute these values into the formula:
1. Write down the formula:
[tex]\[
F = qvB
\][/tex]
2. Substitute the given values:
[tex]\[
F = (5.0 \times 10^{-7} \, \text{C}) \times (2.6 \times 10^5 \, \text{m/s}) \times (1.8 \times 10^{-2} \, \text{T})
\][/tex]
3. Calculate the product of the values:
[tex]\[
F = 5.0 \times 2.6 \times 1.8 \times 10^{-7+5-2} \, \text{N}
\][/tex]
4. Simplify the power of ten:
[tex]\[
F = 5.0 \times 2.6 \times 1.8 \times 10^{-4} \, \text{N}
\][/tex]
5. Multiply the constants:
[tex]\[
5.0 \times 2.6 = 13.0
\][/tex]
[tex]\[
13.0 \times 1.8 = 23.4
\][/tex]
6. Combine the constant with the power of ten:
[tex]\[
F = 23.4 \times 10^{-4} \, \text{N}
\][/tex]
7. Convert to scientific notation:
[tex]\[
F = 2.34 \times 10^{-3} \, \text{N}
\][/tex]
Therefore, the magnitude of the magnetic force acting on the charge is:
[tex]\[
F = 2.34 \times 10^{-3} \, \text{N}
\][/tex]
Out of the provided answer choices:
- 0 N
- [tex]\(2.3 \times 10^{-3}\)[/tex] N (approx)
- 23 N
- [tex]\(2.3 \times 10^{11}\)[/tex] N
The closest option to [tex]\(2.34 \times 10^{-3}\)[/tex] N is [tex]\( 2.3 \times 10^{-3} \)[/tex] N.
Hence, the correct answer is:
[tex]\[
2.3 \times 10^{-3} \, \text{N}
\][/tex]
