Question

A beam of protons traveling at 1.40 enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (the figure ). The beam travels a distance of 1.10 while in the field.

a). What is the magnitude of the magnetic field?

Answer 1

Using Force balance on the proton beam

Fc = Fm

Centripetal force = magnetic field

m*V^2/R = q*V*B

B = mV/qR

m = mass of proton

V = speed of proton = 1400 m/sec

q = charge on proton = +e

R = radius of circle

given that

(2*pi*R)/4 = 1.10 cm

R = 2*1.10*10^-2/pi

R = 0.007 m

Now Using these values

B = 1.67*10^-27*1400/(1.6*10^-19*0.007)

B = 2.09*10^-3 T

B = 2.09 mT