A proton moving at speed v = 1.13·106 m/s enters a region in space where a magne
ID: 1470015 • Letter: A
Question
A proton moving at speed v = 1.13·106 m/s enters a region in space where a magnetic field given B = (-0.531 T) $hat{z}$ exists. The velocity vector of the proton is at an angle = 62.9° with respect to the positive z-axis.
a) Calculate the radius, r, of the trajectory projected onto a plane perpendicular to the magnetic field (in the xy-plane).
b) Calculate the period, T, of the motion in that plane.
c) Calculate the frequency, f, of the motion in that plane.
d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).
Explanation / Answer
a)
magnetic force = centripetal force
Fb = Fc
q*v*B*sintheta = m*(v*sintheta)^2/r
r = (m*v*sintheta)/(qB)
r = (1.67*10^-27*1.13*10^6*sin62.9)/(1.6*10^-19*0.531)
r = 0.019 m = 1.9 cm
b)
T = (2*pi*r) / v*sintheta= (2*pi*0.019)/(1.13*10^6*sin62.9)
T = 1.2*10^-7 s
c)
f = 1/T = 8.33*10^6 Hz
d)
d = v*costheta*t
d = 1.13*10^6*cos62.9*1.2*10^-7
d = 0.062 m