7.3 Magnetic Fields and the Electron

p. 284

Electron Behaviour (part 1)J.J. Thomson (1856 – 1940) made meticulous measurements of cathode rays and measured an important property of these rays; the ratio of charge to their mass. Later on the cathode rays were found to be made up of electrons.

FB = BQv = Bev (where Q = e = charge on an electron)

J.J. Thomson used magnetic fields to bend the cathode rays (or electrons) after they had been accelerated by high voltage. The magnetic field pushes the electrons up or down. The accelerating voltage and magnetic field are adjusted until the electron beam moves along a straight path to the screen. At this point Fc = FB.

mv2

R= Bev = FBFc = (con’t)

7.3 Magnetic Fields and the Electron

p. 285

Electron Behaviour (part 2)mv2

R= Bev (Cancel out one velocity and re-arrange to solve for e/m ratio)

e

m=

v

BRThe charge to mass ratio as discovered by J.J. Thomas

Since both B and R are measurable and v can be determined from FB = FE.

FB = Bev = EQ = FE

E

Bv =And:

By combining the two equations the charge to mass ratio is:

e

m=

E

B2R= 1.76 x 1011 C/kg

7.3 Magnetic Fields and the Electron

p. 286

Electron Behaviour (part 3)

Afterwards another scientist Robert A. Millikan (1868 – 1953) successfully measured the charge on electrons: elementary charge = 1.6 x 10-19 C.

1.6 x 10-19 Cm

= 1.76 x 1011 C/kg

1.6 x 10-19 C

1.76 x 1011 C/kg= m = 9.1 x 10-31 kg = mass of an electron

The discovery of the both the charge and the mass of the smallest (at that time) sub-atomic particle was one of the great physics feats of the last century! Both J.J. Thomson (1906) and Robert A. Millikan (1923) received Nobel prizes for the achievements in Physics.

7.3 Magnetic Fields and the Electron

p. 286 - 287

Mass Spectrometer

A tool developed for measuring the masses of charged atoms. The mass spectrometer uses a combination of electric fields and magnetic fields to select out particles that have a certain velocity and charge.

E

B1

v =mv2

R= B2ev Fc = = FB

From circular motion, Magnetic force, and velocity:

B2QRv

=m =B2QR

E/B1

B2B1QR

E=

Since B1, B2, Q, R and E are all measureable quantities the mass, m, of the ion can be calculated.

The mass spectrometer can be used to measure mass of isotopes or to separate two closely related isotopes. It can also be used to measure the mass of charge molecules.

7.3 Magnetic Fields and the Electron

p. 288

Electric Motors (part 1)

A motor consists of a rotating coil made from a current carrying wire (armature) inside a magnetic field (either created by a permanent magnet or an electromagnet), and a split-ring commutator.

The external magnetic field exerts a force (a torque) on the armature causing it to rotate. The split-ring commutator changes the current direction just at the right time to keep the armature rotating.

The right hand rule gives the direction of the force on the wire in the magnetic field.

7.3 Magnetic Fields and the Electron

p. 289

Electric Motors (part 2)

The split-ting commutator in DC motors is a crucial element in motor design. The current must be revered in the coil to keep the armature rotating. Without the split-ring commutator the armature would simply rotate to one position and then rotate backwards.

When the coil is in the position of diagram (a) the current runs from the brush at A to the left half of the commutataor, then to the coil of wire on the armature. The current direction in the coil produces a S-pole on the right side and a N-pole on the left side. The coil then rotates due to the attraction of the poles of the coil and the external magnet.

The split-ring commutator as shown in diagram (b) causes he current to flow from left hand brush at B. The current now enters B and leaves A, opposite to diagram (a). The poles on the coil are switched and the external magnetic field causes a repulsion effect. Due to momentum the rotating coil the keeps on rotating. The process is repeated as along as there is current flowing in the armature.

7.3 Magnetic Field Strength

Key Questions

In this section, you should understand how to solve the following key questions.

Page 286 – Quick Check #2

Page 288 – Quick Check #1

Page 293 – 296 – Review 7.3 #3,5,6,7,8,10,13,15 & 18