2T=(2*pi*massproton)/(chargeproton*B)
So
a) B=(pi*1.67*10^-27)/(1.6*10^-19*155*10^-9)
b) Doubling the kinetic energy changes the velocity. Period T does not depend on velocity so the answer is 155 ns.
A charged particle moves into a region of uniform magnetic field B, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 155 ns in the region.
a) What is the magnitude of B?
b)If the particle is sent back through the magnetic field (along the same initial path) but with 5.00 times its previous kinetic energy, how much time does it spend in the field during this trip?
1 answer