What is the force exerted on a charge of [tex]2.5 \mu C[/tex] moving perpendicular through a magnetic field of [tex]3.0 \times 10^2 \, T[/tex] with a velocity of [tex]5.0 \times 10^3 \, m/s[/tex]?

A. 3.8 N
B. 38 N
C. [tex]3.8 \times 10^5 \, N[/tex]
D. [tex]3.8 \times 10^6 \, N[/tex]



Answer :

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]