Answer :
To determine the amount of the charge experiencing a magnetic force, we'll apply the Lorentz force equation. The Lorentz force equation for a charge moving through a magnetic field is given by:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the force experienced by the charge (in Newtons),
- [tex]\( q \)[/tex] is the charge (in Coulombs),
- [tex]\( v \)[/tex] is the velocity of the charge (in meters per second),
- [tex]\( B \)[/tex] is the magnetic field (in Tesla),
- [tex]\( \theta \)[/tex] is the angle between the velocity vector and the magnetic field (in degrees).
We are given:
- [tex]\( F = 7.89 \times 10^{-7} \text{ N} \)[/tex]
- [tex]\( v = 2090 \text{ m/s} \)[/tex]
- [tex]\( \theta = 29.4^{\circ} \)[/tex]
- [tex]\( B = 4.23 \times 10^{-3} \text{ T} \)[/tex]
Let's find the charge [tex]\( q \)[/tex].
1. Converting the angle from degrees to radians:
[tex]\[ \theta_{\text{radians}} = \text{radians}(\theta) = \theta \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{radians}} = 29.4 \times \frac{\pi}{180} \][/tex]
2. Rearranging the Lorentz force equation to solve for the charge [tex]\( q \)[/tex]:
[tex]\[ q = \frac{F}{v \cdot B \cdot \sin(\theta)} \][/tex]
3. Substituting the given values into the equation:
[tex]\[ q = \frac{7.89 \times 10^{-7}}{2090 \times 4.23 \times 10^{-3} \times \sin(29.4^{\circ})} \][/tex]
4. Calculating:
[tex]\[ \sin(29.4^{\circ}) \approx 0.49 \][/tex]
[tex]\[ B \cdot \sin(\theta) = 4.23 \times 10^{-3} \times 0.49 \approx 2.07 \times 10^{-3} \][/tex]
[tex]\[ q = \frac{7.89 \times 10^{-7}}{2090 \times 2.07 \times 10^{-3}} \][/tex]
5. Simplifying:
[tex]\[ 2090 \times 2.07 \times 10^{-3} \approx 4.32 \][/tex]
[tex]\[ q = \frac{7.89 \times 10^{-7}}{4.32} \approx 1.82 \times 10^{-7} \text{ C} \][/tex]
So, the amount of the charge is:
[tex]\[ q \approx 1.82 \times 10^{-7} \text{ C} \][/tex]
Therefore, the charge experiencing the magnetic force is [tex]\( \boxed{1.82 \times 10^{-7} \text{ C}} \)[/tex].
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the force experienced by the charge (in Newtons),
- [tex]\( q \)[/tex] is the charge (in Coulombs),
- [tex]\( v \)[/tex] is the velocity of the charge (in meters per second),
- [tex]\( B \)[/tex] is the magnetic field (in Tesla),
- [tex]\( \theta \)[/tex] is the angle between the velocity vector and the magnetic field (in degrees).
We are given:
- [tex]\( F = 7.89 \times 10^{-7} \text{ N} \)[/tex]
- [tex]\( v = 2090 \text{ m/s} \)[/tex]
- [tex]\( \theta = 29.4^{\circ} \)[/tex]
- [tex]\( B = 4.23 \times 10^{-3} \text{ T} \)[/tex]
Let's find the charge [tex]\( q \)[/tex].
1. Converting the angle from degrees to radians:
[tex]\[ \theta_{\text{radians}} = \text{radians}(\theta) = \theta \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{radians}} = 29.4 \times \frac{\pi}{180} \][/tex]
2. Rearranging the Lorentz force equation to solve for the charge [tex]\( q \)[/tex]:
[tex]\[ q = \frac{F}{v \cdot B \cdot \sin(\theta)} \][/tex]
3. Substituting the given values into the equation:
[tex]\[ q = \frac{7.89 \times 10^{-7}}{2090 \times 4.23 \times 10^{-3} \times \sin(29.4^{\circ})} \][/tex]
4. Calculating:
[tex]\[ \sin(29.4^{\circ}) \approx 0.49 \][/tex]
[tex]\[ B \cdot \sin(\theta) = 4.23 \times 10^{-3} \times 0.49 \approx 2.07 \times 10^{-3} \][/tex]
[tex]\[ q = \frac{7.89 \times 10^{-7}}{2090 \times 2.07 \times 10^{-3}} \][/tex]
5. Simplifying:
[tex]\[ 2090 \times 2.07 \times 10^{-3} \approx 4.32 \][/tex]
[tex]\[ q = \frac{7.89 \times 10^{-7}}{4.32} \approx 1.82 \times 10^{-7} \text{ C} \][/tex]
So, the amount of the charge is:
[tex]\[ q \approx 1.82 \times 10^{-7} \text{ C} \][/tex]
Therefore, the charge experiencing the magnetic force is [tex]\( \boxed{1.82 \times 10^{-7} \text{ C}} \)[/tex].
