Certainly! Let's solve this problem step by step.
First, we start with the formula that relates the magnetic force to the charge, velocity, magnetic field strength, and the angle between the velocity and the magnetic field direction. The formula for the magnetic force \( F \) is given by:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge,
- \( v \) is the velocity,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity and the magnetic field.
We are given:
- \( q = 8.24 \times 10^{-7} \, \text{C} \) (Coulombs),
- \( F = 7.89 \times 10^{-8} \, \text{N} \) (Newtons),
- \( v = 2090 \, \text{m/s} \),
- \( \theta = 29.4^\circ \).
Firstly, let's convert the angle from degrees to radians because trigonometric functions in most scientific calculators and implementations use radians. The conversion is done using the formula:
[tex]\[ \text{angle in radians} = \theta \times \left(\frac{\pi}{180}\right) \][/tex]
So,
[tex]\[ 29.4^\circ = 29.4 \times \left(\frac{\pi}{180}\right) \approx 0.513 \, \text{radians} \][/tex]
Next, we rearrange the formula to solve for the magnetic field strength \( B \):
[tex]\[ B = \frac{F}{q \cdot v \cdot \sin(\theta)} \][/tex]
Now, substituting the given values:
- \( F = 7.89 \times 10^{-8} \, \text{N} \),
- \( q = 8.24 \times 10^{-7} \, \text{C} \),
- \( v = 2090 \, \text{m/s} \),
- \( \theta = 0.513 \, \text{radians} \).
Then, calculate the sine of the angle:
[tex]\[ \sin(0.513) \approx 0.490 \][/tex]
Now, substitute these values into the formula for \( B \):
[tex]\[ B = \frac{7.89 \times 10^{-8}}{8.24 \times 10^{-7} \cdot 2090 \cdot 0.490} \][/tex]
[tex]\[ B \approx \frac{7.89 \times 10^{-8}}{8.24 \times 10^{-7} \cdot 1024.1} \][/tex]
[tex]\[ B \approx \frac{7.89 \times 10^{-8}}{8.43 \times 10^{-4}} \][/tex]
[tex]\[ B \approx 9.33 \times 10^{-5} \, \text{T} \][/tex]
Therefore, the magnetic field strength is approximately:
[tex]\[ B \approx 9.33 \times 10^{-5} \, \text{T} \][/tex]
Expressing this in scientific notation as requested:
[tex]\[ B = 933.27 \times 10^{-7} \, \text{T} \][/tex]
Final result in the form requested would be:
[tex]\[ 933.27 \times 10^{-7} \, \text{T} \approx 933.27 \times 10^{-7} \, \text{T} \][/tex]
Thus, the strength of the magnetic field is approximately:
[tex]\[ 9.33 \times 10^{-5} \, \text{T} \][/tex] in the [tex]$\times 10^{-5}$[/tex] form.
