To find the magnitude of the magnetic force acting on the charge, we can use the formula for the magnetic force on a moving charge, which is given by:
[tex]\[ F = q \cdot v \cdot B \cdot \sin(\theta) \][/tex]
where:
- [tex]\( F \)[/tex] is the magnetic force,
- [tex]\( q \)[/tex] is the charge,
- [tex]\( v \)[/tex] is the velocity,
- [tex]\( B \)[/tex] is the magnetic field strength,
- [tex]\( \theta \)[/tex] is the angle between the velocity vector and the magnetic field direction.
The given data is:
- The charge [tex]\( q = 6.8 \mu C \)[/tex] (which is [tex]\( 6.8 \times 10^{-6} \)[/tex] Coulombs),
- The velocity [tex]\( v = 6.5 \times 10^4 \)[/tex] meters per second,
- The angle [tex]\( \theta = 15^\circ \)[/tex],
- The magnetic field strength [tex]\( B = 1.4 \)[/tex] Tesla.
First, we need to convert the angle in degrees to radians because the sine function in our formula requires the angle to be in radians.
[tex]\[ \theta = 15^\circ \][/tex]
[tex]\[ \theta_{rad} = \frac{15 \times \pi}{180} \][/tex]
[tex]\[ \theta_{rad} \approx 0.2618 \text{ radians} \][/tex]
Now we can plug the values into the formula:
[tex]\[ F = 6.8 \times 10^{-6} \, \text{C} \times 6.5 \times 10^4 \, \text{m/s} \times 1.4 \, \text{T} \times \sin(0.2618) \][/tex]
Using the sine value [tex]\( \sin(0.2618) \approx 0.2588 \)[/tex],
[tex]\[ F \approx 6.8 \times 10^{-6} \times 6.5 \times 10^4 \times 1.4 \times 0.2588 \][/tex]
[tex]\[ F \approx 0.16015722510943983 \][/tex]
Rounding this result to two significant figures, we get:
[tex]\[ F \approx 0.16 \, \text{N} \][/tex]
Therefore, the magnitude of the magnetic force acting on the charge is closest to:
[tex]\[ 1.6 \times 10^{-1} \, \text{N} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{1.6 \times 10^{-1} \, \text{N}} \][/tex]
