Answer to Question #282993 in Mechanics | Relativity for BIGIRIMANA Elie

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 π‘š. Find: (a) the momentum, (b) the speed, and (c) the kinetic energy of the electron. Recall that, in nonrelativistic situations, the magnetic force π‘žπ‘£π΅ furnishes the centripetal force π‘šπ‘£ 2β„π‘Ÿ. Thus, since 𝑝 = π‘šπ‘£ it follows that 𝑝 = π‘žπ΅π‘Ÿ and this relation holds even when relativistic effects are important. 


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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=4.8β‹…10βˆ’19 kgβ‹…ms,p=qBr=4.8\cdot 10^{-19}~\frac{kg\cdot m}s,

b)

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

c)

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


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