Response 1 Response 1 An electron enters a uniform magnetic field B= 0.23 T at a

Question

Response 1 Response 1 An electron enters a uniform magnetic field B= 0.23 T at a 37 degree angle to vector B . Determine the radius r and pitch p (distance between loops) of the electron's helical path assuming its speed is 2.4 x 10^6 m/s. (See figure.) r = 4.7397e-5 p = 2.978e-4 Response 2 An electron enters a uniform magnetic field B= 0.23 T at a 37 degree angle to vector B . Determine the radius r and pitch p (distance between loops) of the electron's helical path assuming its speed is 2.4 x 10^6 m/s. (See figure.) r = .047397e-5 p = .2978 Response 3 An electron enters a uniform magnetic field B= 0.23 T at a 37 degree angle to vector B . Determine the radius r and pitch p (distance between loops) of the electron's helical path assuming its speed is 2.4 x 10^6 m/s. (See figure.) r = .047397 p = .2978

Explanation / Answer

the force due to magnetic field acting on the electronto keep it circular motion             F = BqV sin = mV^2 / r , centripetal force                                  r= mV / Bq sin      r = radius of the circular path        V = velocity of theelectron         B = magneticfield         q = 1.6*10^-19Col         = 37. pluggin these values we can solve for radisu ofthecircular path. b) time period of single rotation                T = 2m / Bq                V cos = horizatal component of velocity of electron            pitch = T * V cos.                         

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

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