Question #282993

An electron traveling at relativistic speed moves perpendicular to a magnetic field of 0.20 𝑇. Its path is circular, with a radius of 15 𝑚. Find: (a) the momentum, (b) the speed, and (c) the kinetic energy of the electron. Recall that, in nonrelativistic situations, the magnetic force 𝑞𝑣𝐵 furnishes the centripetal force 𝑚𝑣 2⁄𝑟. Thus, since 𝑝 = 𝑚𝑣 it follows that 𝑝 = 𝑞𝐵𝑟 and this relation holds even when relativistic effects are important. 


1
Expert's answer
2022-02-13T12:12:43-0500

a)

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

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

p=qBr=4.81019 kgms,p=qBr=4.8\cdot 10^{-19}~\frac{kg\cdot m}s,

b)

v=pm=5.31011 ms,v=\frac pm=5.3\cdot 10^{11}~\frac ms,

c)

E=mv22=2.5109 J.E=\frac {mv^2}2=2.5\cdot 10^{-9}~J.


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