Question

A) Equally charged chlorine ions of mass 37 amu now enter the spectrometer. How close to the detector slit will they impact?

B) A point charge moving in a magnetic field of 1.18 Tesla experiences a force of 0.774E-11 N. The velocity of the charge is perpendicular to the magnetic field.In this problem, we use the points of the compass and `into' and `out of' to indicate directions with respect to the page. If the magnetic field points south and the force points out of the page, then give ALL possible correct answers for the charge Q (i.e., B, AC, BCD...).

a) Q is positive, moving west
b) Q is negative, moving west
c) Q is negative, moving east
d) Q is positive, moving east
e) Q is negative, moving north

C) The speed of the charge is 0.410E8 m/s. Calculate the magnitude of the charge.

D) The path of the charge in the magnetic field is a circle. Assume that the charge is positive and has a mass of 1.673E-27kg, the mass of a proton. What is the radius of the circle?

Answer 1

(A)

d = 2 R ( here R is the radius)

since the magnetic force serves as the centripetal force, we have:

m v^2 / R = q v B

B = m v / (R q) = 35*1.66e-27*(8.60e5)/0.75/1.6e-19 = 0.417 T

another way to see (1) is that R = m v/q B, i.e., the radius R is proportional to mass m, so it the diameter.

d' / d = m' / m

d' = (m' / m) d

distance is d' - d = (m'/m - 1) d = 0.0857 m (or 8.57 cm)

(B)

Ans:

a) Q is positive, moving west

c) Q is negative, moving east

(determined using right palm rule)

(C)

F = qvB

=> q = F/vB = (0.774E-11)/( 0.410E8 * 1.18)

= 1.6 x 10^-19 C

(D)

F = qvB

Also: F = mv²/r

Equating both:

qvB = mv²/r

=> r = mv/qB

= (1.673E-27 * 0.410E8 )/(1.6 x 10^-19 *1.18)

= 0.3627m