Question #199978

What will be the radius of curvature of the path of a 3.0 KeV proton in a

perpendicular magnetic field of magnitude 0.8 T ?


1
Expert's answer
2021-05-30T13:30:35-0400

Larmor radius is given by a formula:

R=mvqBR = \frac {mv}{qB}

The energy of electron:

E=mc21(vc)2E = \frac{mc^2}{1-(\frac v c)^2}

In this case we can express a speed of electron:

v=cEmc2E=534976msv=c\sqrt{\frac{E-mc^2}{E}} = 534976 \frac m s

mq=103\frac m q = 10^{-3}

R=668mR = 668 m






Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS