1.) A proton moving with a speed of 4.0 x 10^+6 m/s through a magnetic field of
ID: 2116200 • Letter: 1
Question
1.) A proton moving with a speed of 4.0 x 10^+6 m/s through a magnetic field of 1.7 T experiences a magnetic force of 8.2 x 10^-13 N. What is the angle between the proton's velocity and the field?
2.) An electron moving along the positive x-axis perpendicular to a magnetic field experiences a magnetic deflection in the negative y-direction. What is the direction of the magnetic field over this region?
3.) A proton moves with a speed of 8.0 x 10^6 m/s along the x-axis. It enters a region where there is a field of magnitude 2.5 T directed along the y-axis. Calculate the magnetic force and acceleration of the proton.
4.) An alpha particle, which is the nucleus of a Helium atom, is moving northward with a velocity of 3.8 x 10^5 m/s in a region where the magnetic field is 1.9 T and points horizontally to the east. What are the magnitude and direction of the magnetic force on this alpha particle?
5.) A proton is moving in a circular orbit of radius 0.14 m in a uniform magnetic field of magnitude 0.35 T directed perpendicular to the velocity of the proton. Find the orbital speed of the proton.
6.) If the proton in question 5 were replaced with an electron moving at the same orbital velocity, what is the radius of the electron's circular orbit?
THANK YOU!!!!!
Explanation / Answer
1.) F=qvBsin(theta), for magnitude.
8.2*10^-13=1.6*10^-19*(4*10^6)*1.7*sin(theta)
Theta=arcsin(all the mess that you throw in the calculator, but the 10^6 cancel :) )
Theta= 48.9 degrees
2.) By the right hand rule (and then reversing the direction since it is an electron), x^ cross z^ is -y^. But Reverse the direction because it's an electron, so the field is in the -z^ direction.
3.) F=qvB
F= 1.6*10^-19*8*10^6*2.5 = 3.2*10^-12 N
F=ma=1.627*10^-27*a, so a= 1.966*10^15 m/s/s
4.) If your paper is oriented North-is-up and East-is-right, then the directino of force is INTO the paper/table.
F=qvB=(2*1.6*10^-19)*3.8*10^5*1.9 = 2.31*10^-13 N
5.) F_centripetal= Fmagnetic.
mv^2/r=qvB
v=qBr/m, where q/m is the charge-to-mass ratio of the proton, 0.983*10^8.
v=0.983*10^8*0.35*0.14= 4.8*10^6 m/s
6.) Now to keep v the same, but changing to (q/m)', of the electron (which is 1.7588*10^11), note that the radius must decrease. (q/m)' is 1789 times greater, so the radius must be 1/1789 times smaller. That is, 0.14/1789 = 78 micrometers.
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