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. (Eigure 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)? Express your answer using two significant figures. 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 (p = mv/square squareroot 1 - v^2/c^2)? Express your answer using two significant figures.

Explanation / Answer

Hi,

This problem requires to evaluate an equation for two different cases:

v = 0.93 c ; c = 3*108 m/s ; B = 0.23 T ; q = 1.602*10-19 C ; m = 1.673*10-27 kg

In this case we are assuming that the particle is a proton

Part A (non-relativistic case)

R = mv/qB = (0.93)(1.673*10-27 kg)(3*108 m/s) / [(0.23 T)(1.602*10-19 C)] = 12.67 m

Part B (relativistic case)

R = y (mv/qB) ; where y = 1/[1 - (v2/c2)]1/2 = 1/[1 - (0.93)2]1/2 = 2.72

R = 2.72 (0.93)(1.673*10-27 kg)(3*108 m/s) / [(0.23 T)(1.602*10-19 C)] = (2.72)*(12.67 m) = 34.46 m

I hope it helps.

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