When a particle of charge q and momentum p enters a uniform magnetic field at ri

Question

When a particle of charge q and momentum p enters a uniform magnetic field at right angles it follows a circular path of radius R=p/qB, as shown in the figure.(Figure 1)

Part A

What radius does this expression predict for a proton traveling with a speed v = 0.96 c through a magnetic field B = 0.23 T if you use the nonrelativistic momentum (p=mv)?

Part B

What radius does this expression predict for a proton traveling with a speed v = 0.96 c through a magnetic field B = 0.23 T if you use the relativistic momentum (p=mv/1?v2/c2?????????)?

Explanation / Answer

a)if you use non relativistic equations,

radius=p/(q*B)

=mass*speed/(charge*magnetic field)

=1.67*10^(-27)*0.96*3*10^8/(1.6*10^(-19)*0.23)=13.07 m

part b:

relativistic momentum is given by

p=m*v/sqrt(1-(v/c)^2)

=1.67*10^(-27)*0.96*3*10^8/sqrt(1-0.96^2)
=1.7177*10^(-18) kg.m/s

then radius=p/(q*B)

=1.7177*10^(-18)/(1.6*10^(-19)*0.23)

=46.667 m

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