A point charge moving in a magnetic field of 1.24 Tesla experiences a force of 0

Question

A point charge moving in a magnetic field of 1.24 Tesla experiences a force of 0.615·10-11 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 select True or False for the charge Q.

1.

Q is positive, moving east.
Q is positive, moving west.
Q is negative, moving north.
Q is negative, moving west.
Q is negative, moving east.

2. The speed of the charge is 0.310·108 m/s. Calculate the magnitude of the charge.

3.The path of the charge in the magnetic field is a circle. Assume that the charge is positive and has a mass of 1.673·10-27 kg, the mass of a proton. What is the radius of the circle?

4. What is the orbital frequency of the charge in the previous problem?

Explanation / Answer

1) By using Flemins right hand rule

we can say that the charge Q is positive and moving east

2) F = q*v*B

q = F/(v*B) = 0.615*10^-11/(0.31*10^8*1.24) = 1.6*10^-19 C

3) radius is r = (m*v)/(q*B) = (1.673*10^-27*0.31*10^8)/(1.6*10^-19*1.24) = 0.261 m

4) orbital frequency is f = (q*B)/(2*pi*m) =(1.6*10^-19*1.24)/(2*3.142*1.673*10^-27) = 1.88*10^7 Hz

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