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 mak

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!!!!

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Acacia 4 years 2021-09-01T01:03:25+00:00 1 Answers 100 views 0

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  1. lethu
    0
    2021-09-01T01:05:07+00:00
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    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

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