Question

1.) A solenoid used to produce magnetic fields for research purposes is 1.6 m long, with an inner radius of 30 cm and 1200 turns of wire. When running, the solenoid produced a field of 1.5 T in the center. Given this, how large a current does it carry? Express your answer using two significant figures.

2.) An electron moves with a speed of 4.0×10⁷ m/s in the directions shown in the figure. A 0.50 T magnetic field points in the positive x-direction.(Figure 1)

What is the magnitude of the magnetic force in figure (a)? What is the magnitude of the magnetic force in figure (b)?

Figure can be seen at:

http://www.chegg.com/homework-help/questions-and-answers/electron-moves-speed-40-10-7-m-s-directions-shown-figure-080-t-magnetic-field-points-posit-q2976124

Answer 1

@1)

Lenght of the solenoid = L = 1.6 m

No.of turns of solenoid = N = 1200

magnetic field produced at its center = B = 1.5 T

Let , current passing through the solenoid is = I  

It is known by the formula,

            magnetic field due to solenoid at its center

                             is B = μ_o  N I / L

                   where   μ_o   = 4π x 10^-7 A /m

                        thus, 1.0 T = (4π x 10^-7 A /m )(1200)(I) / (1.6)

                                    1.061x10^-4 x10⁷= I  

                                            thus, I = 1.061x10³ A

                                                  or current = I = 1.061 K A________________answer