PHYS2012
ELECTROMAGNETIC PROPOERTIES OF MATTER
WORKSHOP 5
Question 1
As blood is electrically conducting, it is possible to measure the blood velocity in an
artery electromagnetically by measuring the voltage produced across the artery when
a magnetic field is set up across it.
(a) Draw a diagram showing the artery, the magnetic field, the voltage across the
artery and its polarity and the electric field set up in the artery. Show the
connections of a voltmeter placed across the artery and the polarity of the meter
to give a positive reading.
(b) Carefully explain how the voltage is produced across the artery and how it is
related to the magnetic field B, the diameter of the artery d and the Hall voltage
V.
(c) An instrument generates a transverse magnetic field of 1.5010-3 T. When used
on an artery of internal diameter of 5.20 mm, the voltage produced across the
artery is 2.52 V. Assuming a uniform velocity profile across the artery,
calculate the velocity and volume flow rate dVol/dt of the blood.
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
1
Question 2
A schematic diagram of a copper Hall probe is shown.
Y
X
Z
w
I
t
B
(a)
Why is the bottom plate positive?
(b)
Derive the expression for the Hall Voltage
VHall 
(c)
iB
nqt
How can the number density of charge carriers be measured using a Hall
probe?
(d)
Calculate the magnitude of the magnetic field for a copper Hall probe of
thickness t = 125 m and width w = 20.0 mm, current in the strip of I = 25.0 A
and Hall voltage across the width of the strip of VHall = 11.0 V.
Density of copper  = 8.93103 kg.m-3 NA = 6.021023 mol-1
Molar mass copper M = 63.5 g.mol-1
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
2
Question 3
The figure shows a
figure from the
1934 patent
application of
Ernest Lawrence for
a cyclotron. Ions are
injected in the
centre of the device
and are accelerated
until they are
ejected once they
make it to the outer
regions. The two
semicircular halves of the devices are connected to a high-speed radio frequency (rf)
voltage generator with constant frequency.
(a)
Sketch the path of positive ion (H2+) traversing the magnetic field. Compare
this to the path of an electron travelling through the same magnetic field.
(b)
Explain how this device can accelerate charged particles.
(c)
Show that the radius of the orbit is given by
mv
R
qB
(d)
Show that the cyclotron frequency is given by
qB
fc 
2 m
(e)
Show that the kinetic energy of the ion is given by
1 q2 B 2 R 2
K
2
m
How is the magnetic flux B enclosed by the orbit with radius R related to the
angular momentum L of the ion?
(f)
(g)
A modern medical cyclotron has a diameter of d = 1.00 m and a magnetic field
of B = 2.00 T. It is used to accelerate H2+ ions.
(i)
Calculate the frequency of the rf source.
(ii)
Calculate the maximum speed of the ions.
(iii) Calculate the maximum kinetic energy of the ions in MeV.
(iv)
Are relativistic effects important for this cyclotron?
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
3
Solution: Question 1
(a)
Magnetic force on + charges
artery
+
+++++++
+
E
direction of blood flow
V
-
common
-------
Magnetic force on - charges
Magnetic field B
directed into page
(b)
Blood moving from left to right: magnetic force on positive charges is up and
magnetic force on negative charges is down (right hand rule).
Magnetic force on charge q is Fm = q v B
Charge separation gives rise to an electric field E across the artery opposing further
charge separation
Electric force on charge q is Fe = q E
A steady state situation is reached when Fe = Fm  v B = E
The electric field E is related to the potential difference across the artery (diameter d)
V=Ed=vBd
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
4
(c)
Therefore, the velocity of the blood is v = E / B = V / (d B)
The volume flow rate of the blood is dVol/dt = A v
B = 1.50010-3 T
d = 5.200 mm = 5.20010-3 m
V = 2.52 V = 2.5210-6 V
v = V / (d B) = (2.5210-6) / {(5.20010-3)( 1.50010-3)} m.s-1 = 0.323 m.s-1
A = (d/2)2
dVol/dt = A v = (5.20010-3 /2)2(0.323) m3.s-1 = 6.8610-6 m3.s-1
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
5
Solution: Question 2
Hall Effect: Charge carriers moving in a magnetic field experience a force, moving
them to one side of a conductor (Edwin Hall, U.S.A. 1879). This provides a way for
determining the sign of charge carriers in a current and a Hall probe can be used to
precisely measure the magnitude of a magnetic field or determine the number density
of the charge carriers.
(a) and (b)
Assume the current is in a positive X direction with an average drift velocity v and
with the magnetic field B in the Z direction. The width of the probe is w and the
thickness is t.
Y
X
+
Z
+
+
+
+
+
+
-
w
I
-
-
-
B
-
-
-
-
-
-
t
charge carriers are electrons for copper
Right hand rule  electrons are deflected down
 bottom of probe is negative
The magnetic force experienced by the charge carriers (negative electrons q = -e)
moving in the –X direction is
Fm  ( e )v  B  e v B j
Therefore negative charges move in the –Y direction producing a charge separation
across the conductor that produces an electric field transverse to the direction of the
current I. Hence, the bottom of the probe is negative.
Fe  e E j
Steady state:
Fe  Fm
 E vB
as with the blood moving through the artery.
The potential difference across the conductor of width w is
B
VB  VA    E  ds   E w  VHall
A
| VHall | = v w B
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
6
But I = n e v A  v = I / (n e A) = I / (n e w t) where A = w t
VHall 
IB
net
(c)
n
IB
t eVHall
From the measurements of I, B, t and VHall
 number density of charge carriers n
(d)
thickness, t = 125 m = 12510-6 m
Hall probe width, w = 20 mm = 2010-3 m
I = 25 A
VHall = 11 V = 11 10-6 V
e = 1.610-19 C
B=? T
n = ? charge carriers.m-3
 = 8.93103 kg.m-3
NA = 6.021023 mol-1
M = 63.5 g.mol-1 = 6.3510-2 kg.mol-1
Assume one conduction electron per copper atom
mass of one copper atom, m = M / NA
=Nm/V=nm
n =  / m =  NA / M = 8.4981028 charge carriers.m-3
B
n e tVHall
= 0.75 T
I
/physics/p2/em/wks5_09.doc
(Earth magnetic field ~ 510-5 T)
3/7/2016
7:08 AM
7
Solution: Question 3
(a)
Use right hand palm rule – a positive charged particle is bent (circular path) in the
opposite direction to a negatively charged particle. The H2+ ion has a much larger
radius of curvature compared with an electron. The H2+ ions will travel in a anticlockwise direction
B
I
+q
F
B
out of page
B
right hand
palm rule
I
right hand screw rule
Cyclotron
(b)
ac voltage accelerates charges across gap. Oscillator frequency independent of v and
R. Magnetic field can not increase the kinetic energy of the ions since the direction of
the magnetic force on the charges is always at right angles to the direction of motion
of the charged particle. Only the electric field between the dees produces an
acceleration.
(c)
Magnetic force = centripetal force
m v2
Fm  q v B sin   q v B
Fc 
R
(d)
period = circumference / speed
T
2 R 2  m v 2  m


v
vqB
qB
f 
R
mv
qB
1
qB

T 2 m
cyclotron frequency
(e)
1
1 q2 B 2 R 2
K  mv 2 
2
2
m
(f)
m  B A  B  R 2
m 
L  mv R
B
mv
qR
L
q
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
8
(g)
d = 1.00 m B = 2.00 T q = 2e m = 2 mp
e = 1.60210-19 C mp = 1.6710-27 kg
1 eV = 1.60210-19 J
1 MeV = 1.60210-13 J
Frequency of rf source = cyclotron frequency
f 
qB
 1.52  107 Hz  15.2 MHz
2 m
Max speed of ion
R = d/2
1 2 1 q2 B2 R2
mv 
2
2
m
qBR qBd
v

 4.8 107 m.s-1
m
2m
K
Max KE of ion
1 q2 B2 R2 1
K
 m v 2  24 MeV
2
m
2
v / c = 0.16
relativistic effects are not important
Cyclotrons can work because the cyclotron frequency does not depend on either the
radius or the velocity component perpendicular to the magnetic field
qB

.
m
Therefore, a set of identical particles have identical cyclotron frequency, irrespective
of the initial condition. The magnetic field in each of the semicircular parts takes the
particles through half a circle and the particles then get accelerated in the small gaps.
The field that is responsible for the acceleration needs to flip sign every half period,
and thus it has to be an ac voltage. Since the cyclotron frequency is constant, this
voltage is set equal to the cyclotron frequency. The particles are injected at low
energies in the centre and upon being accelerated, their orbits get larger, until they are
ejected when the radius approaches that of the semicircular parts.
/physics/p2/em/wks5_09.doc
3/7/2016
7:08 AM
9