VISUAL PHYSICS ONLINE

6 ELECTROMAGNETISM

P60 005

Two parallel metal plates were in a magnetic field B = 1.23´10^-3 T. The plates were separated by a distance d = 22.5 mm. An electron was accelerated by a 10.5 kV before entering the region between the plates.

(a) What was the speed of the electron upon entering the region between the plates?

(b) What was the magnitude and direction of the force due to the magnetic field?

(c) What was the magnitude and direction of the force due to the electric field if the electron continued on a straight path parallel to the plates?

(d) What was the magnitude and direction of the electric field between the plates?

(e) What was the voltage between the plates? Which plate was positively charged?

(f) How was this set-up used by J. J. Thompson used to measure the q/m ratio for an electron?

View solution below only after you have completed the answering the question.

Solution

B = 1.23´10^-3 T d = 22.5 m = 2.25´10^-3 m

accelerating voltage V_a = 10.5 kV = 10.5´10³ V

(a) v = ? m.s^-1

accelerating voltage does work on the electron increasing its kinetic energy

(b) right hand screw rule – magnetic force directed up the page

(c) magnetic force & electric force must cancel

– electric force directed down the page

(d) Electric field between the plates – same direction as force on a positive charge

– up the page

(e) voltage across the plates

– bottom plate is positive (electron attracted to bottom plate)

(f)

J.J Thompson’s e/m experiment

From measurement of E (or V and d), B and V_a the e/m ratio can be found.

Alternatively: Thomson adjusted the electric field strength and magnetic field strength to make the electrons pass straight through. This allowed him to calculate the velocity v of the electrons (cathode rays). He then passed the electrons through the magnetic field B alone, and from the radius R of their path calculated the q/m ratio