Question

(9-19)Constants A long, straight wire carries a current of 8.60 A. An electron is traveling in the vicinity of the wire

Part A

At the instant when the electron is 4.40 cm from the wire and traveling with a speed of 5.40×10⁴ 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.

Answer 1

given data:

I = 8.60 A

r = 0.044 m

v = 5.40 x 10⁴ m/s

= 90 degrees

q = 1.6 x 10^-19 C

B= [(₀)*(I)] / [(2)*(pi)*(r)]
B = [(4 x 10^-7)*(8.60)] / [(2)()(0.044)]
B = 3.909090 x 10^-5

F= B*q*v*(sin)
F = (3.909090 x 10^-5)*(1.6 x 10^-19)*(5.4 x 10⁴)*(1)
F = 3.377 x 10^-19 N