Since the charge to mass ratio of the electron is known very accurately (use the
ID: 2280853 • Letter: S
Question
Since the charge to mass ratio of the electron is known very accurately (use the value given in the lab manual), the experiment you are about to do can be used to measure the magnitude of a uniform magnetic field. Electrons are accelerated from rest through a potential difference of 350 V. Then these electrons enter a uniform magnetic field that is perpendicular to their initial directon. Due to the interaction with the magnetic field, they move in a circular path of radius 0.752 cm. What is the magnitude of the magnetic field?
mT
Explanation / Answer
KE=qV
(1/2)mv^2=qV
v=sqrt[2*V*q/m]
v=sqrt[2*350*1.602*10^-19/9.11*10^-31]
v=1.11*10^7 m/s
since
r=mv/qB
=>B =mv/qr =(9.11*10^-31)*1.11*10^7/(1.6*10^-19)*(0.752*10^-2)
B=8.39 mT=8.4 mT (approx)