To determine the force exerted on a charge moving through a magnetic field, we can use the formula for the magnetic force, which is given by:
[tex]\[ F = q \cdot v \cdot B \][/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.
Given the parameters in the question:
- [tex]\( q = 2.5 \mu C = 2.5 \times 10^{-6} \)[/tex] Coulombs,
- [tex]\( B = 3.0 \times 10^2 \)[/tex] Tesla,
- [tex]\( v = 5.0 \times 10^3 \)[/tex] meters per second.
First, let's plug in the values into the formula to find the force [tex]\( F \)[/tex]:
[tex]\[ F = (2.5 \times 10^{-6} \, \text{C}) \times (5.0 \times 10^3 \, \text{m/s}) \times (3.0 \times 10^2 \, \text{T}) \][/tex]
Now, perform the multiplication step by step to keep track of the significant figures and ensure correct handling of powers of ten:
1. Multiply the charge and velocity:
[tex]\[ 2.5 \times 10^{-6} \, \text{C} \times 5.0 \times 10^3 \, \text{m/s} = 12.5 \times 10^{-3} \, \text{C} \cdot \text{m/s} = 1.25 \times 10^{-2} \, \text{C} \cdot \text{m/s} \][/tex]
2. Multiply the result by the magnetic field strength:
[tex]\[ 1.25 \times 10^{-2} \, \text{C} \cdot \text{m/s} \times 3.0 \times 10^2 \, \text{T} = 3.75 \times 10^0 \, \text{N} = 3.75 \, \text{N} \][/tex]
So, the force [tex]\( F \)[/tex] exerted on the charge is approximately 3.75 N, which rounds up to the nearest significant figure given in the multi-choice options.
Thus, the nearest choice closest to our computed answer is:
[tex]\[ \boxed{3.8 \text{ N}} \][/tex]
