An a-particle has a charge of +2e and a mass of 6.64 10-27 kg. It is accelerated

Question

An a-particle has a charge of +2e and a mass of 6.64 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.06 106 V and then enters a uniform magnetic field whose magnitude is 3.40 T. The a-particle moves perpendicular to the magnetic field at all times.
(a) What is the speed of the a-particle?
= 1.01e7 m/s

(b) What is the magnitude of the magnetic force on it?
= 1.1e-11 N

(c) What is the radius of its circular path?
for part C I keep getting the answer to be .713 m but it keeps telling me I am wrong. CAN SOMEONE PLEASE TELL ME WHAT I AM DOING WRONG FOR PART C AND WHAT THE CORRECT ANSWER SHOULD BE?

Explanation / Answer

Charge q = +2 e = 3.2*10^-19 C mass m = 6.64*10^-27 Kg Speed v = 1.01*10^7 m/s B = 3.4 T Radius of the orbit r = mv/qB                                 = (6.64*10^-27*1.01*10^7)/(3.2*10^-19*3.4)                                 = (6.7064*10^-20)/(108.8*10^-20)                                 = 0.0616 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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