To find the magnetic field strength, 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 electric charge,
- [tex]\( v \)[/tex] is the velocity of the charge,
- [tex]\( B \)[/tex] is the magnetic field strength, and
- [tex]\( \theta \)[/tex] is the angle between the velocity and the magnetic field.
Given the problem data:
- [tex]\( F = 1.5 \times 10^2 \, \text{N} \)[/tex]
- [tex]\( q = 1.4 \times 10^{-7} \, \text{C} \)[/tex]
- [tex]\( v = 1.3 \times 10^6 \, \text{m/s} \)[/tex]
- [tex]\( \theta = 75^\circ \)[/tex]
We need to find [tex]\( B \)[/tex], the magnetic field strength. First, we rearrange the formula to solve for [tex]\( B \)[/tex]:
[tex]\[ B = \frac{F}{q \cdot v \cdot \sin(\theta)} \][/tex]
Before substituting the values, we need to convert the angle from degrees to radians because the sine function in the formula operates in radians:
[tex]\[ \theta \, (\text{in radians}) = 75^\circ \times \left( \frac{\pi}{180^\circ} \right) \approx 1.309 \, \text{radians} \][/tex]
Now substitute all the given values into the formula:
[tex]\[ B = \frac{1.5 \times 10^2 \, \text{N}}{1.4 \times 10^{-7} \, \text{C} \cdot 1.3 \times 10^6 \, \text{m/s} \cdot \sin(1.309)} \][/tex]
We can calculate the sine of 75 degrees:
[tex]\[ \sin(75^\circ) \approx 0.9659 \][/tex]
So,
[tex]\[ B = \frac{1.5 \times 10^2}{1.4 \times 10^{-7} \cdot 1.3 \times 10^6 \cdot 0.9659} \][/tex]
[tex]\[ B \approx \frac{1.5 \times 10^2}{1.75722 \times 10^{-1}} \][/tex]
[tex]\[ B \approx 853.25 \, \text{T} \][/tex]
Therefore, the magnetic field strength is approximately [tex]\( 853.25 \, \text{T} \)[/tex], which matches closely to [tex]\( 8.5 \times 10^2 \, \text{T} \)[/tex].
So the correct answer is:
[tex]\[ \boxed{8.5 \times 10^2 \, \text{T}} \][/tex]
