Question

A point charge with a mass of 3.62 ng and a charge of +2.44 uC moves in the x-y plane with a velocity of 3.00 x 104 m/s in a direction 15° above the +x-axis. At time t=0, the point charge enters a uniform magnetic field of strength 1.25 T that points in the +y-direction. Assume that the point charge remains immersed in the uniform magnetic field after time t=0. a. (5 points) What is the magnitude and direction of the magnetic force that the magnetic field exerts on the point charge at time t=0? b. (5 points) How does the x-component of the charge's initial velocity affect the motion of this point charge as it moves through the uniform magnetic field? Does its magnitude change? Does its direction change? Explain your reasoning. c. (5 points) How does the y-component of the charge's initial velocity affect the motion of this point charge as it moves through the uniform magnetic field? Does its magnitude change? Does its direction change? Explain your reasoning. d. (5 points) Use your answers from parts 1b & lc to explain why the path of this point charge is helical (corkscrew-shaped). Explain your reasoning. e. (2.5 points) Determine the radius of the circular part of the point charge's helical path.

Answer 1

a) F = 9,838 10⁻² N + z axis , e) r = 2.318 10⁻² m

Step-by-step explanation:

Before starting let's reduce the magnitudes to the SI system

m = 2.62 ng = 2.62 10-9 g = 2.62 10 -12 kg

q = 2.44 mC = 2.44 10-6 C

a) to find the magnetic force we use the equation

F = q v x B

We use this equation in the scalar form and the direction is found by the rule of the right hand

The angle is between the velocity vector and the magnetic field

θ = 90 - 15 = 75 °

F = q v B sin θ

F = 2.44 10⁻⁶ 3 10⁴ 1.25 sin (75)

F = 9,838 10⁻² N

In the rule of the right hand the thumb goes in the direction of the speed, the fingers extended in the direction of the field and the palm points in the direction of the force, if the charge is positive

The thumb on the x axis, the fingers on the + y axis and the palm on the + z axis

b and c) The velocity has two components, one on the axis and which, because it is parallel to the field, does not feel any force and one component the x-axis that is perpendicular to the field and feels the force, which is perpendicular to the velocity, whereby its magnitude remains constant and its direction changes.

The component y axis of velocity no force to be parallel to the magnetic field.

d) The trajectory of the particle is circular on the x-axis, the direction of the force and linear on the y-axis, for this reason total movement is a helix with a circular base

e) To calculate the radius let's use Newton's second law

F = m a

q vₓ B = m vₓ² / r

r = m vₓ / q B

r = 2.62 10⁻¹² 3 10⁴ cos 15 / (2.62 10⁻⁶ 1.25)

r = 2.318 10⁻² m