To find the speed of the charge moving through the magnetic field, we can use the formula for the magnetic force on a moving charge:
[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 (speed) of the charge.
- [tex]\( B \)[/tex] is the magnetic field strength.
- [tex]\( \theta \)[/tex] is the angle between the velocity vector of the charge and the magnetic field.
We need to solve for the speed [tex]\( v \)[/tex]. Rearrange the formula to get:
[tex]\[ v = \frac{F}{q \cdot B \cdot \sin(\theta)} \][/tex]
Let's list the given values:
- [tex]\( F = 3.5 \times 10^{-2} \)[/tex] N
- [tex]\( q = 8.4 \times 10^{-4} \)[/tex] C
- [tex]\( B = 6.7 \times 10^{-3} \)[/tex] T
- [tex]\( \theta = 35^\circ \)[/tex]
First, we need to convert the angle from degrees to radians because the trigonometric functions in most calculations use radians. We have:
[tex]\[ \theta = 35^\circ \][/tex]
The conversion factor between degrees and radians is:
[tex]\[ 1^\circ = \frac{\pi}{180} \][/tex]
So,
[tex]\[ \theta \text{ (in radians)} = 35 \times \frac{\pi}{180} = 35 \times \frac{3.141592653589793}{180} \approx 0.610865 \text{ radians} \][/tex]
Next, we calculate the sine of the angle:
[tex]\[ \sin(\theta) = \sin(0.610865) \approx 0.573576 \][/tex]
Now, substitute all the known values into the formula for [tex]\( v \)[/tex]:
[tex]\[ v = \frac{3.5 \times 10^{-2}}{8.4 \times 10^{-4} \cdot 6.7 \times 10^{-3} \cdot 0.573576} \][/tex]
Calculate the denominator:
[tex]\[ q \cdot B \cdot \sin(\theta) = 8.4 \times 10^{-4} \cdot 6.7 \times 10^{-3} \cdot 0.573576 \approx 3.23692288 \times 10^{-6} \][/tex]
Now, divide the force by this value:
[tex]\[ v = \frac{3.5 \times 10^{-2}}{3.23692288 \times 10^{-6}} \approx 10842.330818539169 \, \text{m/s} \][/tex]
Therefore, the speed of the charge moving through the magnetic field is approximately:
[tex]\[ \boxed{1.1 \times 10^4 \, \text{m/s}} \][/tex]
