EXAMPLE SOLUTIONS – Topic 6: MAGNETIC FORCES AND FIELDS
1) Explain how a mass spectrograph (spectrometer) works. Draw a figure (include velocity selector
portion). Assume you will be using positive ions. Indicate what is measured, and what characteristics of
the ions are obtained from these measurements.
Solution: The basic structure of a mass
spectrograph (or mass spectrometer) is indicated region 2
in the figure. Ions are produced from a source at ion source region 1
'O', they are accelerated through some potential + + + + + +
difference, and enter region 1 which comprises
the 'velocity selector' We have selected a
magnetic field B into the paper in this region. O
qvB = qE
Since the ions are traveling to the right, v x B is
upward. Since the ions are + ions, then Fmag is
also up. To balance this force we apply a
- - - - - -
potential difference across the parallel plates 'a' E field
magnetic field
& 'b', giving us an electric field downward, and
hence, an electric force downward.
Hence, the only ions which will pass through hole and into region 2 are those for which the 2 forces cancel.
That is:
Fmag = q v B = q E = Felec or v = E/B.
Since we set up the magnetic & electric fields we can control them (adjust them) to obtain any ratio E/B we
want. Hence, we can select ions with any desired velocity.
The + ions now enter region 2 where they are acted on by the magnetic force only. They are initially
deflected upward, and will travel a circular path under the influence of the magnetic force. Hence, they
undergo UCM and:
m v2/r = q v B v/r = q B/m .
They travel with constant angular velocity. Since we are interested in the properties of the ions, we solve for
q/m.
q/m = v/(rB) & substituting v = E/B q/m = E/B2r .
Hence: Measurement of r q/m (charge to mass ratio of ion)
If q is known (singly or doubly ionized) have determination of m .
Assuming we have an element with, say, 3 isotopes present, and we produce all singly ionized ions, then we
will have 3 different radii which will be directly proportional to the masses. Exposure of film to these ions
can permit density studies of the exposed areas, and hence, a determination of relative abundances of the 3
ions.
2) Protons are used in a Bainbridge mass spectrograph. The fields in the velocity selector portion are set
at: E = 4000 V/cm; B = 0.2 T. After emerging from the velocity selector, the protons are travelling at
right angles to the magnetic induction of 0.2 T. Determine the radius of the proton's path.
Solution: The figure is as shown. Region 1 is
called a 'velocity selector' since the only region 2
particles which pass through undeflected must ion source region 1
have a precise velocity given by: + + + + + +
Felec = q E = Fmag = q v B
O
qvB = qE
Setting these equal we have: v = E/B .
Thus the particles which enter region 2 all
- - - - - -
have a velocity: E field
magnetic field
v = E/B = (4 x 105)/(0.2) = 2 x 106 m/sec.
In region II, the particle moves under the influence of the magnetic force only, and we have Uniform
Circular Motion. Hence,
m v2/R = q v B .
Solving for R we have:
R = mv/qB = (1.6 x 10-27)(2 x 106)/(1.6 x 10-19)(.2) = (2 x 10-21)/(2 x 10-20)
= 10-1 m = 10 cm.
3) A charged particle travelling with a speed of 106 m/sec enters a
magnetic field of 0.05 T. The particle is deflected as shown, and vo
travels in a circular path of radius 20 cm. What properties of this
particle can you deduce from the given information? (Show all
work!)
Solution: The first property of the particle to consider is the sign of the
charge. From the figure we see that v B is upward, but the particle
is deflected downward. Hence, the sign of the charge is negative. vo
The motion of a charged particle in a uniform magnetic field is
Uniform Circular Motion. Hence we have:
m v2/R = q v B .
Solving for q/m we have:
q/m = v/RB = (106)/(.05)(.2) = 108 C/kg .
If we assume a charge of q = 1.6 x 10-19 Coulombs, then we can calculate the particle's mass. We would
have:
m = q/108 = 1.6 x 10-27 kg.
We note that this is the same as the mass of a proton. There is a particle which is exactly like a proton in
all aspects, but it has a negative charge. The particle is called an anti-proton.
4) An electron at point 'A' in the figure has a speed of 107 m/sec.
Find (a) the magnitude & direction of the magnetic field that will
vo
cause the electron to follow the semicircular path from A to B,
and (b) the time required for the e- to move from A to B. (The
distance from A to B is 10 cm).
A B
Solution: First consider the required direction of the magnetic field. The electron is initially deflected to the
right. Since it is charged negatively, then the direction of v B must be to the left. Hence a field into the
paper is required. The field may have other components (e.g. upward), but these will not effect the motion.
Thus assuming the total field is perpendicular to the paper, we then have:
m v2/r = q v B B = m v/q r = (9.1 x 10-31)(107)/(1.6 x 10-19)(5 x 10-2) = 1.14 x 10-3 T.
The path length from A to B is r . Hence the time to travel is:
t = r/v = (3.1416)(5 x 10-2)/(107) = 15.7 x 10-9 sec. or 1.57 x 10-8 sec.
5) You wish to magnetically suspend a 1 m wire having a mass of 10 g in a horizontal magnetic field of
field strength 5 Tesla. What current in the wire is required? Draw a figure showing the wire's orientation
and required current direction.
Solution: Since the field is horizontal and the weight (mg) is
downward, we need a magnetic force upward. The force on a wire F
carrying a current in a magnetic field is given by: mag
F wire = I L B where the vector L is in the direction of the
current. To produce and upward force we need the current to be I # B
coming out of the paper. We will then have:
mg
I L B sin 90 = m g or I = m g/L B
= (10 x 10-3)(9.8)/(1)(5) = 1.96 x 10-2 = 19.6 mA .
6) What is the maximum torque on a coil 5 x 12 cm, of 600 turns, when carrying a current of 10-5 A in a
uniform field of magnitude 0.1 T?
Solution: A loop of wire carrying a current acts as a magnetic dipole. Hence when placed in a uniform field
the net force on the loop will be zero. There can, however, be a non-zero torque on the loop. This torque is
given (for a single loop) by:
net = I A B
where A is the area vector to the loop. The magnitude is net = I A B sin where is the angle between the
area vector and the magnetic field. To obtain a maximum torque we align the loop so that A & B are at right
angles (sin 90 = 1). Since we have a number of turns (N), then we multiply the torque on one turn by the total
number:
net = N I A B = (600)(10-5)(.1)(5 x 10-2)(12 x 10-2) = 3.6 x 10-6 N-m (radians)
7) Explain how a cyclotron works and what it is used for.
A cyclotron is a device to accelerate elementary particles (usually
protons, or deuterons) to high energies. It consists of two metal half cans
(cyclotron 'Dees') under a strong magnetic field. An ion source (we
assume protons) is located in one of the Dees. When the particle is
emitted, it will travel in a circle under the influence of the magnetic field.
+
The protons will travel in circular paths according to:
mv2/r = Fmag = qvB v/r = qB/m . B
Hence, the angular velocity (v/r) is constant, and hence the period, and
frequency is fixed. For protons & a field of 2 Teslas we would have: Cyclotron 'Dees'
= v/r = (1.6 x 10-19)(2)/(1.6 x 10-27) = 2 x 108 .
The 'cyclotron frequency' for protons is 2 x 108 rad/sec, or = 33 MHz.
The protons are accelerated by applying an ac-voltage of 33 MHz across the 'Dees'. If a proton enters the gap
between the Dees at the moment Vab is a '+' maximum, then it will remain in phase with the ac voltage and
will be accelerated each time it enters the gap. After making many revolutions which build up its energy the
proton is then kicked out of the Dees. Typically, proton energies of up to 6-10 Mev are possible with the
cyclotron.