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Question 5
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 (Voltmeter terminals marked common (black) and V
(red)).
(b) Carefully explain how the voltage is produced across the artery.
(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. Assume a uniform velocity profile across the artery, calculate
the velocity and volume flow rate dVol/dt of the blood.
(d) How does the above application relate to the Hall Effect. Derive an expression
for the Hall Voltage and use this expression to calculate the magnitude of the
magnetic field for a copper Hall probe of thickness 125 m and width 20.0 mm,
current in the strip of 25.0 A and Hall voltage across the width of the strip of
11.0 V. How does this magnetic field compare with the Earth’s magnetic field?
Density of copper = 8.93103 kg.m-3 NA = 6.021023 mol-1
Molar mass copper = 63.5 g.mol-1
(Reference: Reece University Physics)
a04/p2/emproblems/em05.doc
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Solution
(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
(c)
The electric field E is related to the potential difference across the artery (diameter d)
V = E d or E = V / d
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
a04/p2/emproblems/em05.doc
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4:26 AM
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dVol/dt = A v = (5.20010-3 /2)2(0.323) m3.s-1 = 6.8610-6 m3.s-1
(d)
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.
Assume the current arises from positive charge carriers, q, moving in a positive X
direction with an average drift velocity v and with the magnetic field B in the Z
direction.
Y
X
thickness t
Z
width L
+q
A
+
B
The magnetic force experienced by the charge carriers is
Fm  q v  B  q v B j
Therefore positive 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
Fe  q E j
Steady state:
Fe  Fm
 E vB
as with the blood moving through the artery.
The potential difference across the conductor of width L is
B
VB  VA   E  ds  E L  VHall
A
| VHall | = v L B
a04/p2/emproblems/em05.doc
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4:26 AM
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But i = n q v A  v = i / (n q A) = i / (n q L t) where A = L t
VHall 
iB
nqt
Hall probe
width, L = 20 mm
thickness, t = 125 m = 1.2510-4 m
i = 25 A
q = 1.610-19 C
VHall = 11 V = 1.1 10-5 V
B=? T
n = ? charge carriers/m3
 = 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
mass of one copper atom, m = M/ NA
(assume 1 charge carrier / atom)
=Nm/V=nm
n =  / m =  NA / M = 8.4981028 charge carriers/m3
B
n q t VHall
= 0.75 T
i
a04/p2/emproblems/em05.doc
(Earth magnetic field ~ 510-5 T)
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