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!!!!
Question
Given Information:
v = 3.5×10⁵ m/s
θ = 60°
Magnetic field = 0.60 T
no. of electrons = 6.242×10¹⁸
radius = 0.750 m
Required Information:
Magnetic force = F = ?
Magnetic field = B = ?
Answer:
Part A) F = 2.909×10⁻¹⁴ N
Part B) B = 5.34×10⁻⁷ T
Explanation:
Part A:
The magnetic force can be found using
F = qvBsin(θ)
Where q = 1.60×10⁻¹⁹ C and v is the speed of electron, B is the magnetic field and θ is the angle between v and B.
F = 1.60×10⁻¹⁹*3.5×10⁵*0.60sin(60°)
F = 2.909×10⁻¹⁴ 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 6.242×10¹⁸ electrons per second so
I = 1.25×10¹⁹/6.242×10¹⁸
I = 2.0032 A
B = 4πx10⁻⁷*2.0032/2π*0.750
B = 5.34×10⁻⁷ T