A long, straight wire carries a current of 8.60 A. An electron is traveling in t

Question

A long, straight wire carries a current of 8.60 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and traveling with a speed of 5.90 Times 10^4 m/s directly toward the wire, what is the magnitude of the force that the magnetic field of the current exerts on the electron? Express your answer with the appropriate units. What is the direction of this force? the same as the direction of the current perpendicular to the direction of the current opposite to the direction of the current

Explanation / Answer

Break it down to what is given first:
0 = 4(pi) x 10^-7 (permeability constant in a vacuum)
I = 8.60 A
r = 4.50 cm = 0.045 m
v = 5.90 x 10^4 (I think that's what you wrote as the velocity but i'm not sure because your formatting was confusing. If this isn't what the velocity was, then use my same method but just plug in your own value for v)
= 90 degrees (because the electron is moving directly towards the wire, which would imply a perpendicular motion to the current)
q= charge of electron = 1.6x10^-19

B= (0)(I) / (2)(pi)(r)

Plug in your values:

B = (4pi x 10^-7)(8.60) / (2)(pi)(0.045)
= 3.82222... x 10^-5 (I stored this answer in my calculator as A)

F= Bqv(sin)

F = (A)(1.6x10^-19)(5.9x10^4)(1)

F = 3.607 x 10^-19 N

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