Final answer:
The force on a particle in a magnetic field can be determined using F = qvBsin(\theta). By rearranging the equation to solve for \theta, we can find the angle the particle's velocity makes with the magnetic field for any given force, charge, velocity, and field strength.
Step-by-step explanation:
The situation described is a classic example of Lorenz force experienced by a charged particle moving through a magnetic field. The force exerted on a charged particle when it moves through a magnetic field can be calculated with the following equation:
F = qvBsin(\theta)
where:
- F is the magnetic force,
- q is the charge of the particle,
- v is the velocity of the particle,
- B is the magnetic field strength, and
- \theta is the angle between the velocity vector and the magnetic field.
To find the angle \(\theta\), we can rearrange the equation:
\theta = arcsin(\frac{F}{q v B})
For part A:
Using the information given, where F = 4.8 \times 10^{-6} N, q = 0.32 \times 10^{-6} C, v = 18 m/s, and B = 0.95 T, we can calculate the angle.
\(\theta = arcsin(\frac{4.8 \times 10^{-6}}{0.32 \times 10^{-6} \times 18 \times 0.95}) = arcsin(\frac{4.8}{0.32 \times 18 \times 0.95})\)
This returns an angle \(\theta\), which will be in degrees once evaluated using a calculator.
For part B, you would apply the same approach but with F = 3.0 \times 10^{-6} N, to find the different angle.