To determine the magnitude of the magnetic force acting on a moving charge in a magnetic field, we use the formula:
[tex]\[ F = q \cdot v \cdot B \][/tex]
where:
- [tex]\( F \)[/tex] is the magnitude of the magnetic force,
- [tex]\( q \)[/tex] is the charge of the particle,
- [tex]\( v \)[/tex] is the velocity of the particle, and
- [tex]\( B \)[/tex] is the magnetic field strength.
Given values:
- The charge, [tex]\( q \)[/tex], is [tex]\( 5.0 \times 10^{-7} \)[/tex] Coulombs.
- The velocity, [tex]\( v \)[/tex], is [tex]\( 2.6 \times 10^5 \)[/tex] meters per second.
- The magnetic field strength, [tex]\( B \)[/tex], is [tex]\( 1.8 \times 10^{-2} \)[/tex] Tesla.
Next, we substitute these values into the formula to find [tex]\( F \)[/tex]:
[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]
When we calculate the above expression step-by-step:
1. Calculate the multiplication of the charge and velocity:
[tex]\[ 5.0 \times 10^{-7} \times 2.6 \times 10^5 = 1.3 \times 10^{-1} \][/tex]
(Here we have combined [tex]\( 5.0 \times 2.6 \)[/tex] which equals 13. After dividing by 10 position shifts to left [tex]\(1.3\times10^0 \)[/tex] reduces to [tex]\(1.3 \times 10^{-1}\)[/tex])
2. Multiply the result by the magnetic field strength:
[tex]\[ 1.3 \times 10^{-1} \times 1.8 \times 10^{-2} = 2.34 \times 10^{-3} \][/tex]
(Here, using the power rule [tex]\(a^b c^d = (ac)^(b+d)\)[/tex] and normal multiplication [tex]\( 1.3 \times 1.8 = 2.34\)[/tex])
Finally, we have:
[tex]\[ F = 2.34 \times 10^{-3} \, \text{N} \][/tex]
Therefore, the magnitude of the magnetic force acting on the charge is:
[tex]\[ 2.34 \times 10^{-3} \, \text{N} \][/tex]
From the given options, this corresponds to [tex]\( \boxed{2.3 \times 10^{-3} \, \text{N}} \)[/tex].
