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.
