A point charge moving in a magnetic field of 1.30 Tesla experiences a force of 5

Question

A point charge moving in a magnetic field of 1.30 Tesla experiences a force of 5.158E-11 N. The velocity of the charge is perpendicular to the magnetic field.

1A

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 west and the force points out of the page, then give ALL possible correct answers for the charge Q

A) Q is positive, moving south

B) Q is negative, moving south

C) Q is negative, moving east

D) Q is positive, moving north

E) Q is negative, moving north

1B) The speed of the charge is 0.310E8 m/s. Calculate the magnitude of the charge.

1C) The path of the charge in the magnetic field is a circle. Assume that the charge is positive and has a mass of 26.77E-27kg. What is the radius of the circle?

1D) What is the orbital frequency of the charge in the previous problem?

Explanation / Answer

B = magnetic field                              towards west

F = force                                 out of the page

using right hand rule , the direction of motion of charge comes out to be north if the charge is positive

using right hand rule , the direction of motion of charge comes out to be south if the charge is positive

B) Q is negative, moving south

D) Q is positive, moving north

1 B)

v = speed of charge = 0.310 x 108 m/s

q = magnitude of charge

F = magnetic force = 5.158 x 10-11 N

B = magnetic field = 1.30 T

using the formula

F = q v B Sin90

5.158 x 10-11 = q (0.310 x 108) (1.30)

q = 1.28 x 10-18 C

1c)

m = mass = 26.77 x 10-27 kg

r = radius of circle

magnetic force provides the necessary centripetal force to move in circle

Fb = Fc

q v B Sin90 = m v2/r

r = mv/(qB)

r = (26.77 x 10-27) (0.310 x 108)/((1.28 x 10-18) (1.30))

r = 0.498 m

1D)

orbital frequency is given as

f = qB/(2pim) = (1.28 x 10-18 ) (1.30)/(2 x 3.14 (26.77 x 10-27)) = 9.898 x 106 Hz

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