To find the magnitude of the magnetic force acting on a moving charge in a magnetic field, we use the formula:
[tex]\[ F = qvB\sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnitude of the magnetic force,
- [tex]\( q \)[/tex] is the charge,
- [tex]\( v \)[/tex] is the velocity of the charge,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity and the magnetic field.
Let's break this down step by step:
1. Convert the charge [tex]\( q \)[/tex] from microCoulombs [tex]\(\mu C\)[/tex] to Coulombs [tex]\( C \)[/tex]:
[tex]\[ q = 6.8 \mu C = 6.8 \times 10^{-6} C \][/tex]
2. Identify the velocity [tex]\( v \)[/tex]:
[tex]\[ v = 6.5 \times 10^4 \, \text{m/s} \][/tex]
3. Identify the angle [tex]\( \theta \)[/tex] and convert it from degrees to radians. Note that [tex]\( \sin(\theta) \)[/tex] in the formula requires the angle in radians:
[tex]\[ \theta = 15^\circ \][/tex]
[tex]\[ \theta \, (\text{in radians}) = \theta \times \frac{\pi}{180} = 15 \times \frac{\pi}{180} = \frac{\pi}{12} \, \text{radians} \][/tex]
4. Identify the magnetic field strength [tex]\( B \)[/tex]:
[tex]\[ B = 1.4 \, \text{T} \][/tex]
5. Calculate the magnetic force:
[tex]\[ F = qvB\sin(\theta) \][/tex]
Since:
[tex]\[ \sin\left(\frac{\pi}{12} \right) = \sin(15^\circ) \][/tex]
Plugging in all the given values:
[tex]\[
F = (6.8 \times 10^{-6} \, \text{C}) \cdot (6.5 \times 10^4 \, \text{m/s}) \cdot (1.4 \, \text{T}) \cdot \sin(15^\circ)
\][/tex]
The value of [tex]\(\sin(15^\circ)\)[/tex] is approximately [tex]\(0.2588\)[/tex].
Therefore:
[tex]\[
F \approx (6.8 \times 10^{-6}) \cdot (6.5 \times 10^4) \cdot (1.4) \cdot (0.2588)
\][/tex]
After performing the calculation, the magnitude of the magnetic force [tex]\( F \)[/tex] is approximately:
[tex]\[
F \approx 0.1601572251 \, \text{N}
\][/tex]
So, the correct answer from the given choices is:
[tex]\[ 1.6 \times 10^{-1} \, \text{N} \][/tex]
