Question

proton having an initial velocity of 17.2i Mm/s enters a uniform magnetic field of magnitude 0.290 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -17.2] Mm/s (a) Determine the direction of the magnetic field (b) Determine the radius of curvature of the proton's path while in the field (c) Determine the distance the proton traveled in the field (d) Determine the time interval for which the proton is in the field

plz solve this

Answer 1

a)

The direction of magnetic field is k

b)

Here centripertal force is balanced by magnetic force

mv²/r = qvB

=>r=mv/qB =(1.6727*10^-27)(17.2*10⁶)/(1.602*10^-19)(0.29)

r=0.6193 m

c)

Distance travelled by proton

D=2pir/4

=>D=2pi*0.6193/4 =0.97276 m

d)

Time period

T=2pim/Bq =2pi*(1.6727*10^-27)/(1.602*10^-19)(0.29)

T=2.262*10^-7 s

Time interval for the proton in the field is

t=T/4 =56.56 ns