To determine the magnitude of the charge on the particle, we can use the formula for magnetic force on a moving charge:
[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 particle,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity and the magnetic field.
Given:
- The magnetic force [tex]\( F = 7.5 \times 10^{-2} \, \text{N} \)[/tex],
- The velocity [tex]\( v = 2.5 \times 10^4 \, \text{m/s} \)[/tex],
- The magnetic field strength [tex]\( B = 8.1 \times 10^{-2} \, \text{T} \)[/tex],
- The angle [tex]\( \theta = 25^\circ \)[/tex].
Let's calculate the magnitude of the charge [tex]\( q \)[/tex].
First, we need to convert the angle from degrees to radians since the trigonometric functions in physics typically use radians.
[tex]\[ \theta = 25^\circ = 25 \times \frac{\pi}{180} \approx 0.436 \, \text{radians} \][/tex]
Now, we can substitute these values into the magnetic force equation and solve for [tex]\( q \)[/tex]:
[tex]\[ q = \frac{F}{v \cdot B \cdot \sin(\theta)} \][/tex]
Let's plug in the known values:
[tex]\[ q = \frac{7.5 \times 10^{-2} \, \text{N}}{2.5 \times 10^4 \, \text{m/s} \times 8.1 \times 10^{-2} \, \text{T} \times \sin(0.436)} \][/tex]
Using the value we have for [tex]\(\sin(0.436) \approx 0.4226\)[/tex]:
[tex]\[ q = \frac{7.5 \times 10^{-2}}{2.5 \times 10^4 \times 8.1 \times 10^{-2} \times 0.4226} \][/tex]
[tex]\[ q \approx \frac{7.5 \times 10^{-2}}{8.52315 \times 10^{-1}} \][/tex]
[tex]\[ q \approx 8.763709567231476 \times 10^{-5} \, \text{C} \][/tex]
Therefore, the magnitude of the charge is approximately [tex]\( 8.8 \times 10^{-5} \, \text{C} \)[/tex].
Among the given options, the correct answer is:
[tex]\[ 8.8 \times 10^{-5} \, \text{C} \][/tex]
