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

Question

When a particle of charge g 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. What radius does this expression predict for a proton traveling with a speed upsilon = 0.94 c through a magnetic field B = 0.19 T if you use the nonrelativistic momentum (p = m upsilon)? What radius does this expression predict for a proton traveling with a speed upsilon = 0.94 c through a magnetic field B 0.19 T if you use the relativistic momentum (p = m upsilon/Squareroot 1 - upsilon^2/c^2)?

Explanation / Answer

given equation

R = p / q B

Part A )

v = 0.94 c

B = 0.19 T

p = m v

p = 9.1 X 10-31 X 0.94 X C

p = 9.1 X 10-31 X 0.94 X 3 X 108

p = 2.566 X 10-22 kg-m/s

R = p / q B

R = 2.566 X 10-22 / 1.6 X 10-19 X 0.19

R = 8.421 X 10-3 m

R = 8.421 mm

Part B )

p = m v / ( 1 - v2 / c2 )1/2

p = 9.1 X 10-31 X 0.94 X 3 X 108 / ( 1 - 0.942 )1/2

p = 7.521 X 10-22 kg-m/s

R = p / q B

R = 7.521 X 10-22 / 1.6 X 10-19 X 0.19

R = 0.02474 m

R = 24.74 mm

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