(a) What is the speed of a proton that has total energy1000GeV ? (b) What is the angular frequency v of a proton with the speed calculated in part (a) in a magnetic field of4.00T ? Use both the nonrelativistic and the correct relativistic expression, and compare the results.

Short Answer

Expert verified

The speed of a proton is v=0.99999956cand angular frequency in case of nonrelativistic is ωnr=3.83×108rad/sand in case of relativistic is ωr=3.59×105rad/s.

Step by step solution

01

Definition of Speed

The term speed may be defined as the ratio of distance and time

02

Determine the speed of a proton is and angular frequency in case of nonrelativistic and in case of relativistic 

The relativistic total energy of a particle is E=γmc2

So

1000GeV=γ(0.938GeV)γ=1000GeV/0.938GeVγ=1066

and the speed of the proton is
1/(1-v2/c2)=1066v2=1-8.798×10(-7)v2=0.999999c2v=0.99999956c

Hence, the speed of a proton is v=0.99999956c

Now the relation between angular frequency and magnetic field is

ω=|q|B/m

For nonrelativistic angular frequency

ωnr=(1.60×10(-19)C)(4T)/(1.67×10(-27)Kg)ωnr=3.83×108rad/s

For relativistic angular frequency

ωr=(3.83×108rad/s)/1066ωr=3.59×105rad/s

Hence, angular frequency in case of nonrelativistic is ωnr=3.83×108rad/sand in case of relativistic isωr=3.59×105rad/s. .



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