To solve this problem, we'll use the formula for the magnetic force on a moving charge. The formula is:
[tex]\[ F = qvB \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnetic force on the charge,
- [tex]\( q \)[/tex] is the charge, given as [tex]\( 8.4 \times 10^{-4} \, \text{C} \)[/tex],
- [tex]\( v \)[/tex] is the velocity of the charge (which we'll assume to be 1 m/s for this problem as it's not provided),
- [tex]\( B \)[/tex] is the magnetic field strength, given as [tex]\( 6.7 \times 10^{-3} \, \text{T} \)[/tex],
- [tex]\( \theta \)[/tex] is the angle between the velocity of the charge and the magnetic field, given as [tex]\( 35^\circ \)[/tex].
Let's go through the solution step-by-step:
1. Convert the Angle to Radians:
The angle is given in degrees, but for trigonometric functions, we need to convert it to radians. The conversion factor is [tex]\( \pi \)[/tex] radians = [tex]\( 180^\circ \)[/tex].
Therefore, the angle in radians is:
[tex]\[
\theta_{\text{rad}} = 35^\circ \times \left(\frac{\pi}{180}\right)
\][/tex]
This simplifies to:
[tex]\[
\theta_{\text{rad}} \approx 0.6108652381980153 \, \text{radians}
\][/tex]
2. Calculate the Sine of the Angle:
Next, we need to calculate [tex]\( \sin(\theta_{\text{rad}}) \)[/tex]. For [tex]\( \theta_{\text{rad}} \approx 0.6108652381980153 \)[/tex]:
[tex]\[
\sin(0.6108652381980153) \approx 0.5736
\][/tex]
3. Calculate the Magnetic Force:
Using the formula [tex]\( F = qvB \sin(\theta) \)[/tex]:
[tex]\[
F = (8.4 \times 10^{-4} \, \text{C}) \times (1 \, \text{m/s}) \times (6.7 \times 10^{-3} \, \text{T}) \times \sin(0.6108652381980153)
\][/tex]
[tex]\[
F \approx 8.4 \times 10^{-4} \times 6.7 \times 10^{-3} \times 0.5736
\][/tex]
Simplifying this:
[tex]\[
F \approx 3.2280881837836872 \times 10^{-6} \, \text{N}
\][/tex]
Therefore, the angle in radians is approximately [tex]\( 0.6108652381980153 \)[/tex] radians, and the magnetic force on the charge is approximately [tex]\( 3.2280881837836872 \times 10^{-6} \)[/tex] Newtons.
