- #1
jayayo
- 13
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Homework Statement
The uniform 45.6 mT magnetic field in the picture below points in the positive z-direction. An electron enters the region of magnetic field with a speed of 5.29E+6 m/s and at an angle of θ = 30.1° above the xy-plane. Find the radius r and the pitch p of the electron's spiral trajectory. (Enter the radius r first and the pitch p second.)
Picture attached! Please let me know if you can't see it and I'll attempt to try attaching again :D
Where,
B= 45.8mT = 0.0456T
v=5.29*10^6m/s
Θ=30.1
m= 9.11*10^-32kg
q=1.60*10^-19C
2. The attempt at a solution
F=qvBsinΘ=mv^2/R
qBsinΘ=mv/R
R=mv/(qBsinΘ) = (9.11*10^-32kg)(5.29*10^6m/s)/[(1.60*10^-19C)( 0.0456T)*sin(90-30.1)]
=7.63*10-4m
For the second part of the question, could someone please explain to me the concept of pitch and possibly how the equation has been derived (d=vsinΘ*2Pi*m/[qB]), because I have searched google and I don't understand it much other than that it is the measurement of the height that the electron has traveled in one revolution~ Or is this equation something I should just memorize?
Thank you sooo much!