Question

Electrons are accelerated from rest through a potential difference of 350 V and then enter a uniform magnetic field that is perpendicular to the velocity of the electrons. The electrons travel along a curved path with the radius of 7.5 cm, because of a magnetic force.

a. What is the magnitude of the magnetic field?

b. Estimate the period of this circular path?

c. What is the angular speed of the electrons?

d. If protons are used instead of electrons, what would be the radius of the path?

e. If the potential difference is increased to 500 V, what would be the radius of electrons' path?

Answer 1

Part A)

Start with PE = KE

qV = .5mv²

(1.6 X 10^-19)(350) = (.5)(9.11 X 10^-31)(v)²

v = 1.109 X 10⁷ m/s

Then r = mv/qB

(.075) = (9.11 X 10^-31)(1.109 X 10⁷)/(1.6 X 10^-19)(B)

B = 8.42 X 10-4 T which is .842 mT

Part B)

Apply d = vt where d is the circumference

2(pi)(.075)/(1.109 X 10⁷) = T

T = 4.25 X 10^-8 sec (which is 42.5 ns)

Part C)

w = v/r

w = 1.109 X 10⁷/.075

w = 1.48 X 10⁸ rad/s

Part D)

Again PE = KE

qV = .5mv²

(1.6 X 10^-19)(350) = (.5)(1.67 X 10^-27)(v)²

v = 2.59 X 10⁵ m/s

Then r = mv/qB

(r) = (1.67 X 10^-27)(2.59 X 10⁵)/(1.6 X 10^-19)(8.42 X 10^-4)

r = 3.21 m

Part E

Start with PE = KE

qV = .5mv²

(1.6 X 10^-19)(500) = (.5)(9.11 X 10^-31)(v)²

v = 1.33 X 10⁷ m/s

Then r = mv/qB

(r) = (9.11 X 10^-31)(1.33 X 10⁷)/(1.6 X 10^-19)(8.42 X 10^-4)

r = .0896 m (which is 8.96 cm)