To determine the magnitude of the charge of a particle moving in a magnetic field, we use the formula for the magnetic force exerted on a moving charged particle:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
Where:
- [tex]\( F \)[/tex] is the magnetic force (in Newtons, [tex]\( N \)[/tex])
- [tex]\( q \)[/tex] is the charge of the particle (in Coulombs, [tex]\( C \)[/tex])
- [tex]\( v \)[/tex] is the velocity of the particle (in meters per second, [tex]\( m/s \)[/tex])
- [tex]\( B \)[/tex] is the magnetic field strength (in Tesla, [tex]\( T \)[/tex])
- [tex]\( \theta \)[/tex] is the angle between the velocity and the magnetic field (in degrees, [tex]\( ^\circ \)[/tex])
We need to solve for [tex]\( q \)[/tex]. Rearranging the formula to solve for [tex]\( q \)[/tex], we get:
[tex]\[ q = \frac{F}{v \cdot B \cdot \sin(\theta)} \][/tex]
Given values:
- [tex]\( v = 5.2 \times 10^4 \, m/s \)[/tex]
- [tex]\( \theta = 35^\circ \)[/tex]
- [tex]\( B = 0.0045 \, T \)[/tex]
- [tex]\( F = 0.003087 \, N \)[/tex]
First, convert the angle from degrees to radians:
[tex]\[ \theta_{rad} = \frac{35 \times \pi}{180} \][/tex]
[tex]\[ \theta_{rad} \approx 0.610865238 \][/tex]
Now we substitute all values into the formula:
[tex]\[ q = \frac{0.003087 \, N}{(5.2 \times 10^4 \, m/s) \cdot 0.0045 \, T \cdot \sin(0.610865238)} \][/tex]
From the earlier calculation, we know that:
[tex]\[ \sin(0.610865238) \approx 0.573576436 \][/tex]
Now perform the calculation:
[tex]\[ q = \frac{0.003087}{5.2 \times 10^4 \times 0.0045 \times 0.573576436} \][/tex]
[tex]\[ q \approx 2.300008657 \times 10^{-5} \, C \][/tex]
The magnitude of the charge is approximately [tex]\( 2.300008657 \times 10^{-5} \, C \)[/tex], which can be converted to microcoulombs ([tex]\( \mu C \)[/tex]):
[tex]\[ q \approx 2.3 \times 10^{-5} \, C \][/tex]
[tex]\[ q = 23 \, \mu C \][/tex]
Thus, the magnitude of the charge is [tex]\( 23 \, \mu C \)[/tex], corresponding to the second option provided:
[tex]\[ \boxed{23 \, \mu C} \][/tex]
