A charged particle moves at [tex]2.5 \times 10^4 \, \text{m/s}[/tex] at an angle of [tex]25^{\circ}[/tex] to a magnetic field that has a field strength of [tex]8.1 \times 10^{-2} \, \text{T}[/tex].

If the magnetic force is [tex]7.5 \times 10^{-2} \, \text{N}[/tex], what is the magnitude of the charge?

A. [tex]3.7 \times 10^{-5} \, \text{C}[/tex]

B. [tex]4.1 \times 10^{-5} \, \text{C}[/tex]

C. [tex]8.8 \times 10^{-5} \, \text{C}[/tex]

D. [tex]1.0 \times 10^{-4} \, \text{C}[/tex]



Answer :

To solve this problem, we need to determine the magnitude of the charge on the particle given the following data:
- Velocity ([tex]\( v \)[/tex]) = [tex]\( 2.5 \times 10^4 \)[/tex] m/s
- Angle ([tex]\( \theta \)[/tex]) = [tex]\( 25^\circ \)[/tex]
- Magnetic field strength ([tex]\( B \)[/tex]) = [tex]\( 8.1 \times 10^{-2} \)[/tex] T
- Magnetic force ([tex]\( F \)[/tex]) = [tex]\( 7.5 \times 10^{-2} \)[/tex] N

Let's use the formula for the magnetic force acting on a charged particle moving in a magnetic field:

[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,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity and the magnetic field.

First, we need to convert the angle from degrees to radians because the trigonometric functions in scientific calculations typically use radians.

[tex]\[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \left( \frac{\pi}{180} \right) \][/tex]

Therefore:

[tex]\[ \theta_{\text{rad}} = 25^\circ \times \left( \frac{\pi}{180} \right) \approx 0.436 \text{ radians} \][/tex]

Now, rearrange the formula to solve for [tex]\( q \)[/tex]:

[tex]\[ q = \frac{F}{v \cdot B \cdot \sin(\theta_{\text{rad}})} \][/tex]

Plugging the known values into the formula:

[tex]\[ q = \frac{7.5 \times 10^{-2}}{2.5 \times 10^4 \cdot 8.1 \times 10^{-2} \cdot \sin(0.436)} \][/tex]

Here, [tex]\(\sin(0.436)\)[/tex] is approximately 0.4226. Now calculate the charge:

[tex]\[ q = \frac{7.5 \times 10^{-2}}{2.5 \times 10^4 \cdot 8.1 \times 10^{-2} \cdot 0.4226} \][/tex]

Simplifying this:

[tex]\[ q = \frac{7.5 \times 10^{-2}}{856.2} \][/tex]
[tex]\[ q \approx 8.76 \times 10^{-5} \, \text{C} \][/tex]

Comparing this result with the given options:
- [tex]\( 3.7 \times 10^{-5} \)[/tex] C
- [tex]\( 4.1 \times 10^{-5} \)[/tex] C
- [tex]\( 8.8 \times 10^{-5} \)[/tex] C
- [tex]\( 1.0 \times 10^{-4} \)[/tex] C

The closest value to [tex]\( 8.76 \times 10^{-5} \)[/tex] C is [tex]\( 8.8 \times 10^{-5} \)[/tex] C.

Therefore, the magnitude of the charge is approximately [tex]\( 8.8 \times 10^{-5} \)[/tex] C.