Question

A beam of protons traveling at 1.40 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field in a direction perpendicular to its original direction (Figure 1). The beam travels a distance of 1.70 cm while in the field.

What is the magnitude of the magnetic field? Express your answer in teslas to three significant figures.

Answer 1

since field is perpendicula to velocity
so, motion is circular motion

also, it comes out of field with velocity perpendicular to its intial velocity
so, it travel quarter circle
so,
distance travel = (3.14*r)/2
1.70 = (3.14*r)/2
r = 1.08 cm
= 0.0108 m

given,
v = 1.40 km/s
= 1400 m/s
m = mass of proton
= (1.67*10^-27) Kg

use,
B = (m*v)/(q*r)
= (1.67*10^-27*1400)/(1.6*10^-19*0.0108)
= (2338*10^-27)/(0.01728*10^-19)
= (1.35*10^-3) tesla

Answer: (1.35*10^-3) tesla