1)A charged particle moves through a velocity selector at a constant speed in a

Question

1)A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 4.05 103 N/C, while the magnetic field is 0.360 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.60 cm. Find the charge-to-mass ratio of the particle. 2)The x, y, and z components of a magnetic field are Bx = 0.12 T, By = 0.16 T, and Bz = 0.17 T. A 25 cm wire is oriented along the z axis and carries a current of 3.6 A. What is the magnitude of the magnetic force that acts on this wire?

Explanation / Answer

First find velocity v, then solve for m/q. The velocity selector equation: Fmag = qvB = Felec = Eq ==> v = E/B. The cyclotron equation (ref.): r = mv/(qB) = mE/(qB^2) ==> q/m = E/(rB^2)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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