Question

A) An electron is moving with a speed of 3.5 x105 m/s when it encounters a magnetic field of 0.60T. The direction of the magnetic field makes an angle of60.0° with respect to the velocity of the electron. Whatis the magnitude of the magnetic force on the electron?B) Each second, 1.25 x 1019 electrons ina narrow beam pass through a small hole in a wall. The beam isperpendicular to the wall. Using Ampere’s law, determinethe magnitude of the magnetic field in the wall at a radius of0.750 m from the center of the beam.I know the answer to A is 2.9 x 10^-14 N and the answer to Bis 5.33 x 10^-7 T. Can anyone show me the steps? Ithink I've tried everything!! I will rate immediately!!!!

Answer 1

Given Information:  

v = 3.5x10⁵ m/s

θ = 60°

Magnetic field = 0.60 T

no. of electrons = 6.242x10¹⁸

radius = 0.750 m

Required Information:  

Magnetic force = F = ?

Magnetic field = B = ?

Answer:

Part A) F = 2.909x10⁻¹⁴ N

Part B) B = 5.34x10⁻⁷ T

Explanation:

Part A:

The magnetic force can be found using

F = qvBsin(θ)

Where q = 1.60x10⁻¹⁹ C and v is the speed of electron, B is the magnetic field and  θ is the angle between v and B.

F = 1.60x10⁻¹⁹*3.5x10⁵*0.60sin(60°)

F = 2.909x10⁻¹⁴ N

Part B:

From the Ampere's law, the magnetic field can be found using

B = μ₀I/2πr

Where μ₀ = 4πx10⁻⁷ is the permeability of free space, I is the current, and B is the magnitude of the magnetic field produced.

1 ampere of current has 6.242x10¹⁸ electrons per second so

I = 1.25x10¹⁹/6.242x10¹⁸

I = 2.0032 A

B = 4πx10⁻⁷*2.0032/2π*0.750

B = 5.34x10⁻⁷ T