A particle with a charge of q = -5.67nC is moving in a uniform magnetic field of

Question

A particle with a charge of q = -5.67nC is moving in a uniform magnetic field of B= ( 4.50T ) z^. The magnetic force on the particle is measured to be F=( 5.30microN )-y^.

Calculate the x component of the velocity of the particle.

Express your answer in meters per second.

What is the radius of the circular motion the particle will have in the magnetic field if the particle has a mass of 0.500g ?

What is the period of this circular motion?

What is the pitch of the motion if the velocity has a y component of 1.70m/s .

Explanation / Answer

Part 1)

Start with F = qvB

(5.3 X 10-6) = (5.67 X 10-9)(v)(4.50)

v = 207.7 m/s

Part 2)

Apply r = mv/qB

r = (5 X 10-4)(207.7)/(5.67 X 10-9)(4.5)

r = 4.07 X 106 m

Part C)

Apply d = vt where the d = the circumference

2pi(r) = vt

2pi(4.07 X 106) = (207.7)(t)

t = 1.23 X 105 sec

Part D)

d = vt

d = (1.70)(1.23 X 105)

d = 2.09 X 105 m

All of the solutions seem a little off, but they are correct based on the numbers and units you typed into the question. If anything is incorrect, please check your values and comment below. For example, is the B-field really 4.5 Tesla, or is it supposed to be 4.5 uT? and all others.

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