An electron orbits a proton at a radius of 5.29x10^-11 m. The speed of the elect

Question

An electron orbits a proton at a radius of 5.29x10^-11 m. The speed of the electron in its orbit is 5.1x10^4 m/s. The charge on a proton is +1.6x10^-19 C, electron charge is-1.6x10^-19 C. Electron mass is 9.1x10^-31 kg and proton mass is 1.7x10^-27 kg. What is the magnetic field at the location of the proton due to the motion of the electron? HINTS: #1 Think of the orbiting electron as constituing an electric current. With what current (in Amps) is that electron equivalent? #2 Not all of the quanitities listed above are relevant to this problem

Explanation / Answer

The magnetic field will be found by the formula

B= oI/2R, where I = Q/t

The time for one revolution = 2r/v

Thus the equation becomes

B = oqv/4r2

B = (4 X 10-7)(1.6 X 10-19)(5.1 X 104)/(4)(5.29 X 10-11)2

B = .292 T

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.