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) What radius does this expression predict for a proton traveling with a speed v = 0.93 c through a magnetic field B = 0.23 T if you use the nonrelativistic momentum (p = mv)? What radius does this expression predict for a proton traveling with a speed v = 0.93 c through a magnetic field B = 0.23 T if you use the relativistic momentum

Explanation / Answer

given

R =p/qB

where q =1.6*10^-19 C

a)

v=0.93c = 0.93*3*10^8 m/s

B =0.23 T

momentum p=mv

where m = mass of the proton = 1.67*10^-27 kg

radius R = p/qB = mv/qB = (1.67*10^-27*0.93*3*10^8 ) /(1.6*10^-19*0.23) =12.66 m

b)

v = 0.93 c m/s

B =0.23 T

relativistic momentum p = mv/sqrt(1-v^2/c^2) = (1.67*10^-27*0.93*3*10^8)/sqrt(1-0.93*0.93) =1.26*10^-18

radius R = p/qB = 1.26*10^-18/(1.6*10^-19*0.23) =34.24 m

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