Pre-Lab Charge to Mass Ratio of an Electron Students will calculate a magnetic f

Question

Pre-Lab Charge to Mass Ratio of an Electron

Students will calculate a magnetic field magnitude from a current measurement this week in lab. This measurement of the magnetic field is taken in the presence of the Earth’s magnetic field. Knowing this one may ask if the Earth’s magnetic field could influence data results.

Following the equation for the magnetic field in the laboratory write up and assuming that the current for the Helmholtz coils (low voltage power supply) was set at 2Amps, what is the resulting magnetic field produced by the coils? (Be sure to include units.)

The Earth’s magnetic field ranges from approximately 25-65T. Discuss whether you believe this value could potentially influence your data collection and the final calculations determining e/m.

Explanation / Answer

B= 7.80*10^-4 I

F=qvB, Newtons law of circular motion: a=V^2/R and F=ma, therefore by substituting one equation as F=ma=me[v^2/R]=qevB... The potential energy imparted to the electrons by accelerating the voltage +V can be written as PE(form potential)=qe(+V)=KEe=1/2mev^2... If we solve for the velocity in the force equation (2) we find v=qeB*r/me^2.
substituting the velocity in the energy equation we find: qe(+V)=1/2me(qeB*R/me)^2... simplifying this equation we find: qe/me=2V/B^2R^2

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.