A point charge of [tex]$6.8 \mu C$[/tex] moves at [tex]$6.5 \times 10^4 \, \text{m/s}$[/tex] at an angle of [tex][tex]$15^{\circ}$[/tex][/tex] to a magnetic field that has a field strength of [tex]1.4 \, \text{T}$[/tex].

What is the magnitude of the magnetic force acting on the charge?

A. [tex]$1.6 \times 10^{-1} \, \text{N}$[/tex]
B. [tex]$6.0 \times 10^{-1} \, \text{N}[tex]$[/tex]
C. [tex]$[/tex]1.6 \times 10^5 \, \text{N}$[/tex]
D. [tex]$6.0 \times 10^5 \, \text{N}$[/tex]



Answer :

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]