Answer :
To determine the velocity of the charge moving in a magnetic field, we can use the formula for the magnetic force on a moving charge:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- \( F \) is the magnetic force on the charge.
- \( q \) is the charge of the particle.
- \( v \) is the velocity of the charge.
- \( B \) is the magnetic field strength.
- \( \theta \) is the angle between the velocity vector of the charge and the magnetic field.
Given:
- The charge, \( q = 2.99 \times 10^{-6} \, \text{C} \)
- The angle, \( \theta = 10.0^{\circ} \)
- The magnetic field strength, \( B = 5.00 \times 10^{-5} \, \text{T} \)
- The force, \( F = 2.14 \times 10^{-8} \, \text{N} \)
First, we need to convert the angle from degrees to radians because the sine function in the formula uses radians. The conversion is:
[tex]\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \][/tex]
Now, solve for velocity \( v \):
[tex]\[ v = \frac{F}{q \cdot B \cdot \sin(\theta)} \][/tex]
Substitute the given values:
[tex]\[ \theta_{\text{radians}} = 10.0 \times \frac{\pi}{180} \approx 0.1745 \, \text{radians} \][/tex]
The sine of the angle in radians:
[tex]\[ \sin(0.1745) \approx 0.1736 \][/tex]
Now plug in all the values:
[tex]\[ v = \frac{2.14 \times 10^{-8} \, \text{N}}{2.99 \times 10^{-6} \, \text{C} \times 5.00 \times 10^{-5} \, \text{T} \times 0.1736} \][/tex]
[tex]\[ v \approx \frac{2.14 \times 10^{-8}}{2.99 \times 10^{-6} \times 5.00 \times 10^{-5} \times 0.1736} \][/tex]
[tex]\[ v \approx \frac{2.14 \times 10^{-8}}{2.594 \times 10^{-10}} \][/tex]
[tex]\[ v \approx 824.332 \, \text{m/s} \][/tex]
Thus, the velocity of the charge is approximately:
[tex]\[ \boxed{824.332 \, \text{m/s}} \][/tex]
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- \( F \) is the magnetic force on the charge.
- \( q \) is the charge of the particle.
- \( v \) is the velocity of the charge.
- \( B \) is the magnetic field strength.
- \( \theta \) is the angle between the velocity vector of the charge and the magnetic field.
Given:
- The charge, \( q = 2.99 \times 10^{-6} \, \text{C} \)
- The angle, \( \theta = 10.0^{\circ} \)
- The magnetic field strength, \( B = 5.00 \times 10^{-5} \, \text{T} \)
- The force, \( F = 2.14 \times 10^{-8} \, \text{N} \)
First, we need to convert the angle from degrees to radians because the sine function in the formula uses radians. The conversion is:
[tex]\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \][/tex]
Now, solve for velocity \( v \):
[tex]\[ v = \frac{F}{q \cdot B \cdot \sin(\theta)} \][/tex]
Substitute the given values:
[tex]\[ \theta_{\text{radians}} = 10.0 \times \frac{\pi}{180} \approx 0.1745 \, \text{radians} \][/tex]
The sine of the angle in radians:
[tex]\[ \sin(0.1745) \approx 0.1736 \][/tex]
Now plug in all the values:
[tex]\[ v = \frac{2.14 \times 10^{-8} \, \text{N}}{2.99 \times 10^{-6} \, \text{C} \times 5.00 \times 10^{-5} \, \text{T} \times 0.1736} \][/tex]
[tex]\[ v \approx \frac{2.14 \times 10^{-8}}{2.99 \times 10^{-6} \times 5.00 \times 10^{-5} \times 0.1736} \][/tex]
[tex]\[ v \approx \frac{2.14 \times 10^{-8}}{2.594 \times 10^{-10}} \][/tex]
[tex]\[ v \approx 824.332 \, \text{m/s} \][/tex]
Thus, the velocity of the charge is approximately:
[tex]\[ \boxed{824.332 \, \text{m/s}} \][/tex]
