A point charge of [tex]$5.0 \times 10^{-7} \, \text{C}$[/tex] moves to the right at [tex]2.6 \times 10^5 \, \text{m/s}[/tex] in a magnetic field that is directed into the screen and has a field strength of [tex]1.8 \times 10^{-2} \, \text{T}$[/tex].

What is the magnitude of the magnetic force acting on the charge?

A. 0 N

B. [tex]2.3 \times 10^{-3} \, \text{N}[/tex]

C. 23 N

D. [tex]2.3 \times 10^{11} \, \text{N}[/tex]



Answer :

To determine the magnitude of the magnetic force acting on a point charge, we use the following formula:

[tex]\[ F = q v B \sin(\theta), \][/tex]

where:
- [tex]\( F \)[/tex] is the magnitude of 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 direction of the velocity and the magnetic field.

Since the charge is moving perpendicular to the magnetic field, the angle [tex]\( \theta \)[/tex] is 90 degrees. The sine of 90 degrees is 1, so the equation simplifies to:

[tex]\[ F = q v B. \][/tex]

Given the values:
- Charge, [tex]\( q = 5.0 \times 10^{-7} \, \text{C} \)[/tex],
- Velocity, [tex]\( v = 2.6 \times 10^{5} \, \text{m/s} \)[/tex],
- Magnetic field strength, [tex]\( B = 1.8 \times 10^{-2} \, \text{T} \)[/tex],

the magnetic force can be calculated by substituting these values into the equation:

[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]

After calculating this, we get:

[tex]\[ F = 0.00234 \, \text{N}. \][/tex]

Therefore, the magnitude of the magnetic force acting on the charge is [tex]\( 2.34 \times 10^{-3} \, \text{N} \)[/tex].

Among the given options, this matches:

[tex]\[ 2.3 \times 10^{-3} \, \text{N}. \][/tex]