181L_Lab_10-1

# Motion of the electrons is accelerated by the

This preview shows pages 1–3. Sign up to view the full content.

motion of the electrons is accelerated by the electric potential (V); applied between anode P and Cathode K [1]. Using the law of Conservation of Energy, we can determine the velocity of the electrons accelerated by V with the equation which results in ; where the mass of the electron (m) is measured in kg and the electrical charge of the electron (e) is measured in C. A Lorentz Force (F = evB) is established due to the electrons perpendicular motion with velocity (v) to the uniform magnetic field (B). This force gives the electrons their circular path in a plane perpendicular to that of the magnetic field’s. Establishing a magnetic field with magnetic flux density equal to B [b/m 2 ], we can obtain the equation where r is the radius and v is the velocity of the circular motion respectively. From the previous equation, we determine , which allows for the calculation of the charge to mass ratio, or e/m, being . The magnitude of the magnetic field created due to the Helmholtz coils (H), can be calculated using the coil’s radius (R) and passing current (I) using . Here we determine , where N denotes the turns of the wire in each coil. The e/m apparatus used for this laboratory

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
experiment is built of 130 turns per wire with a radius of 0.15m, establishing our final calculation of the magnetic field being . Procedure : (Broken into parts for convenience) A] Preparation 1. Connect the 6.3V power supply to the apparatus’ power terminal, then connect the 500V DC “B power supply” to the “B power” terminal of the apparatus (P = + [red terminal] and K = - [black terminal]).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern