Question

A 0.32 μc particle moves with a speed of 18 m/s through a region where the magnetic field has a strength of 0.95 t . you may want to review (pages 773 - 777) . part a at what angle to the field is the particle moving if the force exerted on it is 4.8×10−6n? express your answer using two significant figures. θ = ∘ request answer part b at what angle to the field is the particle moving if the force exerted on it is 3.0×10−6n? express your answer using two significant figures.

Answer 1

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.

Answer 2

Its all about Astrophysics bro.