Question

(a) At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.00 ✕ 10^6 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.05 ✕ 10^−5 T? (b) What would the radius (in m) of the path be if the proton had the same speed as the electron? (c) What would the radius (in m) be if the proton had the same kinetic energy as the electron? (d) What would the radius (in m) be if the proton had the same momentum as the electron?

Answer 1

The radius of the electron can be found as

re = me*ve/(B*q)

= 9.11*10^-31*7.00*10^6/(1.05*10^-5*1.6*10^-19)

= 3.79 m

now the radius of the charged particle in the uniform magnetic field, r = m*v/(B*q)

rp = re

mp*vp/(B*q) = me*ve/(B*q)

vp = ve*(me/mp)

= 7.00*10^6*(9.11*10^-31/(1.67*10^-27))

= 3818.56 m/s

2) rp = m*v/(B*q)

= 1.67*10^-27*7.00*10^6/(1.05*10^-5*1.6*10^-19)

= 6958.33 m

3) r = m*v/(B*q)

= P/(B*q)

= sqrt(2*m*KE)/(B*q)

Now when KE is constant

rp/re = sqrt(mp/me)

rp = re*sqrt(mp/me)

= 3.79*sqrt(1.67*10^-27/(9.1*10^-31))

= 162.36m

4)

r = m*v/(B*q)

= P/(B*q)

so, when P is constant

rp = re

= 3.79 m