A beta^+ particle is moving at 19 500 m s^-1 parallel to the ground and due east

Question

A beta^+ particle is moving at 19 500 m s^-1 parallel to the ground and due east through a region of space in which there is a uniform 0.05 T magnetic field. If the magnetic field lines point upwards in the vertical direction, what is the direction and magnitude of the magnetic force on the beta^+ particle? In a region of space there is a uniform magnetic field of magnitude 0.25 m T pointing vertically straight down. If an electron is moving in the horizontal plane at a speed of 550 km s^-1, what will the radius of the resultant circular path be? Will the electron be moving clockwise or counter-clockwise when viewed from above? If a positron (same mass as an electron but with a charge of + q_e) is moving in the horizontal plane at speed of 550 km s^-1, what will the radius of the resultant circular path be? will the positron be moving clockwise or counter-clockwise when viewed from above? If a positron (same mass as an electron but with a charge of +qe) is moving in the horizontal plane at speed of 550 km s^-1, what will the radius of the resultant circular path be? Will the positron be moving clockwise or counter-clockwise when viewed from above? How fast would an alpha particle (4/2 alpha^2+) need 10 be traveling to have a path of the same radius as the electron in part (a)? An electron travelling parallel to the ground and due north enters a region of space in which there is uniform magnetic field of 0.5 mu T pointing straight up. The electron is travelling at a speed of 1000 km s^-1.

Explanation / Answer

39.2]

a) F =qvB = mv2/r

so, r = mv2/qvB = [9.11*10-31 * 550*103]/[1.6*10-19 * 0.25*10-3] = 1.2cm

b) the electron will be moving counter-clockwise

c) the radius of the circular path will be the same

d) the direction of motion of the positron will be opposite to the electron, so it will be clock-wise

e) charge on alpha particle = 2e

so, v = qBr/m = [2*1.6*10-19 * 0.25*10-3 * 0.012]/[6.64*10-24] = 0.145m/s

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