To solve for the magnitude of the magnetic force acting on a moving charge in a magnetic field, we use the formula:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge,
- [tex]\( v \)[/tex] is the velocity of the charge,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\(\theta\)[/tex] is the angle between the velocity and the magnetic field direction.
Given the values:
- [tex]\( q = 5.7 \mu C = 5.7 \times 10^{-6} \, C \)[/tex],
- [tex]\( v = 4.5 \times 10^5 \, m/s \)[/tex],
- [tex]\( B = 3.2 \, mT = 3.2 \times 10^{-3} \, T \)[/tex],
- [tex]\( \theta = 90^\circ \)[/tex] (since this angle will maximize the force, and [tex]\(\sin(90^\circ) = 1\)[/tex]),
we can plug these values into the formula. Since the angle [tex]\(\theta\)[/tex] is [tex]\(90^\circ\)[/tex]:
[tex]\[ \sin(90^\circ) = 1 \][/tex]
Now we compute:
[tex]\[
F = (5.7 \times 10^{-6} \, C) \cdot (4.5 \times 10^5 \, m/s) \cdot (3.2 \times 10^{-3} \, T) \cdot 1
\][/tex]
Let's multiply the values step-by-step:
1. Calculate [tex]\( q \cdot v \)[/tex]:
[tex]\[ 5.7 \times 10^{-6} \, C \cdot 4.5 \times 10^5 \, m/s = 2.565 \times 10^{-6+5} = 2.565 \times 10^{-1} \][/tex]
2. Multiply the result by [tex]\( B \)[/tex]:
[tex]\[ 2.565 \times 10^{-1} \cdot 3.2 \times 10^{-3} = 8.208 \times 10^{-4} \][/tex]
So, the magnitude of the magnetic force is:
[tex]\[ F = 0.008208 \, N \][/tex]
Hence, the correct magnitude of the magnetic force acting on the charge is [tex]\( 0.008208 \, N \)[/tex].
Among the choices given:
[tex]\( 6.6 \times 10^{-3} \, N \)[/tex],
[tex]\( 4.9 \times 10^{-3} \, N \)[/tex],
[tex]\( 4.9 \times 10^3 \, N \)[/tex],
[tex]\( 6.6 \times 10^3 \, N \)[/tex],
the value closest to our computed result of [tex]\( 0.008208 \, N \)[/tex] is:
[tex]\[ 6.6 \times 10^{-3} \, N \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{6.6 \times 10^{-3} \, N} \][/tex]
