To find the force exerted on a charge moving through a magnetic field, we can use the formula for the magnetic Lorentz force:
[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.
Given parameters:
- Charge, [tex]\( q = 2.5 \times 10^{-6} \)[/tex] Coulombs
- Magnetic field strength, [tex]\( B = 3.0 \times 10^2 \)[/tex] Tesla
- Velocity, [tex]\( v = 5.0 \times 10^3 \)[/tex] meters per second
- Since the charge is moving perpendicular to the magnetic field, [tex]\( \theta = 90^\circ \)[/tex]. Therefore, [tex]\( \sin(90^\circ) = 1 \)[/tex].
Let's substitute these values into the formula:
[tex]\[ F = (2.5 \times 10^{-6} C) \cdot (5.0 \times 10^3 \frac{m}{s}) \cdot (3.0 \times 10^2 T) \cdot 1 \][/tex]
First, calculate the product of [tex]\( q \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ 2.5 \times 10^{-6} \, C \times 5.0 \times 10^3 \, \frac{m}{s} = 12.5 \times 10^{-3} \][/tex]
Next, multiply by [tex]\( B \)[/tex]:
[tex]\[ 12.5 \times 10^{-3} \times 3.0 \times 10^2 = 3.75 \][/tex]
Therefore, the force [tex]\( F \)[/tex] is:
[tex]\[ F = 3.75 \, \text{N} \][/tex]
So, the force exerted on the charge is approximately [tex]\( 3.75 \, \text{N} \)[/tex].
Among the given options, the closest answer to [tex]\( 3.75 \, \text{N} \)[/tex] and represented correctly with significant figures is:
[tex]\[ 3.8 \, \text{N} \][/tex]
Hence, the correct option is:
[tex]\[ 3.8 \, \text{N} \][/tex]
