Question

ituricioIl Of Held and a) East b) Up c) Down d) South 2. A proton moving at a speed of 3.8 x 106 m/s cuts across the lines of a magnetic field at an angle of 60 degrees. The strength of the field is 0.25 x 10-4 T. What is the magnitude of the force acting on the proton? (ap- 1.6x 10-19 C) are held in a 3. At the Fermilab accelerator, singly-charged ions with momentum 4.8 x 10-16 kg m/s circular orbit of radius 1 km by an upward magnetic field. What B-field must be used to maintain the ions in this orbit ? (qion = 1.6 x 10-19 C) a) 3.0 T b) 4.8 x 10-6T c) 3.8 T d) None of these

#2!!

Answer 1

The magnetic field force acting on the charged particle is, F = q_p vB sin theta = (l.6 times 10^-19C) (3.8 times 10^6 m/s) (0.25 times 10^-4 T) sin (60 degree) approximate 1.32 times 10^-17 N (3) If the charged particle is normally enter to the magnetic field, the magnetic field force provides the centripetal force Bvq = mv^2/r r = mv/Bq Here, momentum of the particle is, P = mv Therefore, r = p/Bq B = p/rq = (4.8 times 10^-16 kg middot m/s)/(10^3 m) (1.6 times 10^-19 C) = 3.0 T(option a is correct.)