Question

A proton, moving with velocity v = -234 +6662 m/s encounters a magnetic field B = 3.252 UT. What is the radius of the circular part of the proton's trajectory?

Answer 1

r=mv/(qB), where r is radius of the circular part of the charged particle, m is its mass,v is magnitude of velocity of the charged particle perpendicular to the magnetic field, q is its charge and B is magnetic field.

Here,m=mass of proton=1.6726*10^-27 kg, v=234 m/s(only X component of velocity is perpendicular to the magnetic field),q=charge on proton=1.6*10^-19 C, B=3.25 micro-tesla=3.25*10^-6 T.

So,required radius=1.6726*10^-27*234/(1.6*10^-19*3.25*10^-6)=0.75267 m.