Page 438 - Physics Coursebook 2015 (A level)

426
Cambridge International A Level Physics
WORKED EXAMPLE
You can also write this equation in terms of the momentum p of the particle, that is:
p = Ber
The equation r = mv shows that:
■■ faster-moving particles move in bigger circles (r ∝ v)
■■ particles with greater masses also move in bigger circles
(they have more inertia: r ∝ m)
■■ a stronger field makes the particles move in tighter circles
( r ∝ B1 ) .
This is made use of in a variety of scientific applications, such as particle accelerators and mass spectrometers.
It can also be used to find the charge-to-mass ratio e
of an electron. me
The charge-to-mass ratio of an electron
Experiments to find the mass of an electron first involve
finding the charge-to-mass ratio me . This is known as e
the specific charge on the electron – the word ‘specific’ here means ‘per unit mass’.
1
An electron is travelling at right angles to a uniform magnetic field of flux density 1.2 mT. The speed of the electron is 8.0 × 106 m s−1. Calculate the radius of circle described by this electron. (For an electron, charge e = 1.60 × 10−19 C and mass me = 9.11 ×
10−31 kg.)
Step1 Calculatethemagneticforceontheelectron. F=Bev=1.2×10−3×1.60×10−19 ×8.0×106 F=1.536×10−15N
Step2 Useyourknowledgeofmotioninacircleto determine the radius r.
mev2 F=r
Therefore:
r = mev2 = 9.11×10−31 ×(8.0×106)2
F 1.536 × 10−15 r≈3.8×10−2m (3.8cm)
Note: The same result could have been obtained simply by using the equation:
mev r= Be
Be
Using the equation for an electron travelling in a circle in a magnetic field, we have e = v . Clearly,
meBr e measurements of v, B and r are needed to measure me .
There are difficulties in measuring B and r. For example, it is difficult to measure r with a rule outside the tube
in Figure 27.8 because of parallax error. Also, v must
be measured, and you need to know how this is done. One way is to use the cathode–anode voltage Vca. This p.d. causes each electron to accelerate as it moves from the cathode to the anode. If an individual electron has charge −e then an amount of work e × Vca is done on each electron. This is its kinetic energy as it leaves the anode:
QUESTIONS
e V c a = 12 m e v 2
where m is electron mass and v is the speed of the
4
5
6
Look at the photograph of the electron beam
in the fine-beam tube (Figure 27.8). In which direction is the magnetic field (into or out of the plane of the photograph)?
The particles in the circular beam shown in Figure 27.8 all travel round in the same orbit. What can you deduce about their mass, charge and speed?
An electron beam in a vacuum tube is directed at right angles to a magnetic field, so that it travels along a circular path. Predict the effect on the size and shape of the path that would be produced (separately) by each of the following changes:
a increasing the magnetic flux density
b reversing the direction of the magnetic field
c slowing down the electrons
d tilting the beam, so that the electrons have a component of velocity along the magnetic field.
e
electron. 1
2
Eliminating v from the two equations eVca = 2 mev r = mev gives:
and
Be
e = 2Vca
me r2B2
Hence, if we measure Vca, r and B, we can calculate me .
As we shall see shortly, the electron charge e can be e
measured more directly, and this allows us to calculate the
electron mass me from the value of me . e