An electron is moving through a magnetic field whose magnitude is8.70e -4 T. The

Question

An electron is moving through a magnetic field whose magnitude is8.70e-4 T. The electron experiences only a magneticforce and has an acceleration of magnitude 3.50e14m/s2. At a certain instant, it has a speed of6.80e6 m/s. Determine the angle (less than 90)between the electron's velocity and the magnetic field.

Explanation / Answer

We know that            F = B q v sin ==> sin = F / B q v                              = m a / B q v                                  = 9.1 x 10-31 * 3.50 x 1014 / 8.70 x10-4 * 1.6 x 10-19 * 6.80 x106                                   = 0.33648 Therefore              = 19.66o

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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