Question #282993

An electron traveling at relativistic speed moves perpendicular to a magnetic field of 0.20 š‘‡. Its path is circular, with a radius of 15 š‘š. Find: (a) the momentum, (b) the speed, and (c) the kinetic energy of the electron. Recall that, in nonrelativistic situations, the magnetic force š‘žš‘£šµ furnishes the centripetal force š‘šš‘£ 2ā„š‘Ÿ. Thus, since š‘ = š‘šš‘£ it follows that š‘ = š‘žšµš‘Ÿ and this relation holds even when relativistic effects are important. 


Expert's answer

a)

qvB=mv2r,qvB=\frac{mv^2}r,

qB=mvr,qB=\frac{mv}r,

p=qBr=4.8ā‹…10āˆ’19 kgā‹…ms,p=qBr=4.8\cdot 10^{-19}~\frac{kg\cdot m}s,

b)

v=pm=5.3ā‹…1011 ms,v=\frac pm=5.3\cdot 10^{11}~\frac ms,

c)

E=mv22=2.5ā‹…10āˆ’9 J.E=\frac {mv^2}2=2.5\cdot 10^{-9}~J.


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!

LATEST TUTORIALS
APPROVED BY CLIENTS