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Forces & Magnetic Dipoles
B
x
q
F
F
q
.
m
m  AI
  
  mB

U  m  B

Today...
• Application of equation for trajectory of charged
particle in a constant magnetic field: Mass
Spectrometer
• Magnetic Force on a current-carrying wire
• Current Loops
• Magnetic Dipole Moment
• Torque (when in constant B field)  Motors
• Potential Energy (when in constant B field)
• Nuclear Magnetic Resonance Imaging
Text Reference: Chapter 28.1 and 28.3
Examples: 28.2, 28.3, and 28.6 through 28.10
Last Time…
• Moving charged particles are deflected in magnetic
fields F  q v  B
mv
• Circular orbits R 
qB
• If we use a known voltage V to accelerate a particle
q
2V
2
1
 2 2
qV  2 mv
m R B
• Several applications of this
• Thomson (1897) measures q/m ratio for “cathode rays”
•
• All have same q/m ratio, for any material source
• Electrons are a fundamental constituent of all matter!
Accelerators for particle physics
• One can easily show that the time to make an orbit does not
depend on the size of the orbit, or the velocity of the particle
→ Cyclotron
Mass Spectrometer
• Measure m/q to identify substances
m R2 B2

q
2V
1. Electrostatically accelerated electrons knock
electron(s) off the atom  positive ion (q=|e|)
2. Accelerate the ion in a known potential U=qV
3. Pass the ions through a known B field
–
Deflection depends on mass: Lighter deflects more, heavier less
Mass Spectrometer, cont.
4. Electrically detect the ions which “made it through”
5. Change B (or V) and try again:
Applications:
Paleoceanography: Determine relative abundances of isotopes
(they decay at different rates  geological age)
Space exploration: Determine what’s on the moon, Mars, etc.
Check for spacecraft leaks.
Detect chemical and biol. weapons (nerve gas, anthrax, etc.).
See http://www.colby.edu/chemistry/OChem/DEMOS/MassSpec.html
Yet another example
• Measuring curvature of
charged particle in
magnetic field is usual
method for determining
momentum of particle in
modern experiments:
e.g.
e+
mv
2mK
R

qB
qB
- charged
particle
+ charged
particle
B
e-
End view: B into screen
Magnetic Force on a Current
• Consider a current-carrying wire in the
N
presence of a magnetic field B.
• There will be a force on each of the charges
moving in the wire. What will be the total force
dF on a length dl of the wire?
• Suppose current is made up of n charges/volume
each carrying charge q and moving with velocity
v through a wire of cross-section A.
 
• Force on each charge = qv  B

 
• Total force = dF  nA(dl )qv  B
• Current =
dq nAv(dt )q
I

 nAvq 
dt
dt
Yikes! Simpler: For a straight length of wire L
carrying a current I, the force on it is:
 

dF  Idl  B

 
F  IL  B
S
Magnetic Force on a Current Loop
• Consider loop in magnetic field as
on right: If field is ^ to plane of
loop, the net force on loop is 0!
– Force on top path cancels force
on bottom path (F = IBL)
x
x
Fx
x
x
x
x
x
x
x
x
x
x
x
x
– Force on right path cancels
force on left path. (F = IBL)
• If plane of loop is not ^ to field, there
will be a non-zero torque on the loop!
x
x
x
x
x
B
x
x
x
x F
x
B
x
F
1
F
x x
x x
x x
x x
x xI
F
F
.
Preflight 13:
A square loop of wire is carrying current
in the counterclockwise direction. There
is a horizontal uniform magnetic field
pointing to the right.
2) What is the force on section a-b of the loop?
a) zero
b) out of the page
c) into the page
3) What is the force on section b-c of the loop?
a) zero
b) out of the page
c) into the page
4) What is the net force on the loop?
a) zero
b) out of the page
c) into the page

 
F  IL  B
ab: Fab = 0 = Fcd since the wire is parallel to B.
bc: Fbc = ILB RHR: I is up, B is to the right, so F
points into the screen.
By symmetry:
Fda   Fbc

Fnet  Fab  Fbc  Fcd  Fda  0
Lecture 13, Act 1
• A current I flows in a wire which is
formed in the shape of an isosceles
right triangle as shown. A constant
magnetic field exists in the -z direction.
– What is Fy, net force on the wire in
the y-direction?
(a) Fy < 0
(b) Fy = 0
y
B
(c) Fy > 0
x
x
x
x
x
x
x
x
x x
x x
x x
x Lx
x x
x2L x
x x
x x
x
x
x
Lx
x
x
x
x
x
x
x
x
x
x
x
x
x
Lecture 13, Act 1
• A current I flows in a wire which is
formed in the shape of an isosceles
right triangle as shown. A constant
magnetic field exists in the -z direction.
– What is Fy, net force on the wire in
the y-direction?
(a) Fy < 0
(b) Fy = 0
y
B
x
x
x
x
x
x
x
x
x x
x x
x x
x Lx
x x
x2L x
x x
x x
x
x
x
Lx
x
x
x
x
x
x
x
x
x
x
x
x
x
(c) Fy > 0
• The forces on each segment are determined by:

 
F  IL  B
• From symmetry, Fx = 0
BIL
• For the y-component: F1 y  F2 y 
45˚
F1
F2
F3 y   BI ( 2 L)
• Therefore: Fy  F1 y  F2 y  F3 y  0
F3
There is never a net force on a
loop in a uniform field!
2
Calculation of Torque
• Suppose the coil has width w (the
side we see) and length L (into the
screen). The torque is given by:
  
τ  r F

w

τ  2 F sin q 
2

F = IBL 
where
  AIB sinq
B
x
q
w
F
F
.
r
A = wL = area of loop
F
• Note: if loop ^ B, sinq = 0   = 0
maximum  occurs when loop parallel to B
r ×F
Applications: Galvanometers
(≡Dial Meters)
We have seen that a magnet can exert a torque on a loop of
current – aligns the loop’s “dipole moment” with the field.
– In this picture the loop (and hence the
needle) wants to rotate clockwise
– The spring produces a torque in the
opposite direction
– The needle will sit at its equilibrium
position
Current increased
 μ = I • Area increases
 Torque from B increases
 Angle of needle increases
Current decreased
 μ decreases
 Torque from B decreases
 Angle of needle decreases
This is how almost all dial meters work—voltmeters, ammeters,
speedometers, RPMs, etc.
Motors
Free rotation
of spindle
B
Slightly tip the loop
Restoring force from the magnetic torque
Oscillations
Now turn the current off, just as the loop’s μ is
aligned with B
Loop “coasts” around until its μ is ~antialigned with B
Turn current back on
Magnetic torque gives another kick to the loop
Continuous rotation in steady state
Motors, continued
Even better
Have the current change directions every half rotation
Torque acts the entire time
Two ways to change current in loop:
1. Use a fixed voltage, but change the circuit (e.g., break
connection every half cycle
 DC motors
2. Keep the current fixed, oscillate the source voltage
AC motors
VS I
t
2
Lecture 13, Act 2
• What should we do to
increase the speed (rpm)
2A
of a DC motor?
(a) Increase I
(b) Increase B
B
(c) Increase number of loops
2B • What should we do to increase the speed of an AC motor?
(a) Increase I
(b) Increase B
(c) Increase number of loops
Lecture 13, Act 2
• What should we do to
increase the speed (rpm)
2A
of a DC motor?
(a) Increase I
(b) Increase B
B
(c) Increase number of loops
All of the above, though for variable speed
motors, it is always the current that is adjusted.
Lecture 13, Act 2
• What should we do to
increase the speed (rpm)
2B
of an AC motor?
(a) Increase I
(b) Increase B
B
(c) Increase number of loops
Again, all of the above!
BUT, while the answers above are still true, the
main change is to vary the drive frequency—this
must match the rotation rate.
Magnetic Dipole Moment
• We can define the magnetic dipole moment of a
current loop as follows:
magnitude:
m  AI
q
direction: ^ to plane of the
loop in the direction the thumb
of right hand points if fingers
curl in the direction of current.
B
x
F
q
F
m
• Torque on loop can then be rewritten as:
  
τ  μ B
  AIB sinq 
• Note: if loop consists of N turns, m = NAI
.
Bar Magnet Analogy
• You can think of a magnetic dipole moment as a bar
magnet:
μ
N
=
– In a magnetic field they both experience a torque trying to line
them up with the field
– As you increase I of the loop  stronger bar magnet
– N loops  N bar magnets
• We will see next lecture that such a current loop does produce
magnetic fields, similar to a bar magnet. In fact, atomic scale
current loops were once thought to completely explain magnetic
materials (in some sense they still are!).
Electric Dipole Analogy
+q
F
E
q
p
F
  
τ  r F


F  qE
B
x
.
-q
F
q
F
.
m
  
τ  r F
  
F  IL  B (per turn)


p  2qa
μ  NAI
  
τ  p E
  
τ  μ B
Preflight 13:
A square loop of wire is carrying current
in the counterclockwise direction. There
is a horizontal uniform magnetic field
pointing to the right.
6) What is the net torque on the loop?
a) zero
b) up
c) down
d) out of the page
e) into the page
   The torque due to Fcb is up, since r
τ  r F
(from the center of the loop) is to the
right, and Fcb is into the page. Same
goes for Fda.
   m points out of the page (curl your fingers in the direction of
τ  μ B
the current around the loop, and your thumb gives the direction
of m). Use the RHR to find the direction of  to be up.
Potential Energy of Dipole
• Work must be done to change the
orientation of a dipole (current loop)
in the presence of a magnetic field.
q
F
• Define a potential energy U (with zero at
position of max torque) corresponding to
this work.
θ
U
 τdθ
90
B
x
F
q
.
m
θ

Therefore,
U  μB cos θ 
θ
90
U
 μB sin θdθ
90
 
 U   μB cos θ  U   μ  B
Potential Energy of Dipole
m
B
m
x
B
x
m
x
=0
 = mB X
=0
U = -mB
U=0
U = mB
negative work
3
positive work
B
Preflight 13:
Two current carrying loops are oriented
in a uniform magnetic field. The loops
are nearly identical, except the direction
of current is reversed.
8) What direction is the torque on loop 1?
a) clockwise
b) counter-clockwise
c) zero
9) How does the torque on the two loops compare?
a) τ1 > τ2
b) τ1 = τ2
c) τ1 < τ2
10) Which loop occupies a potential energy minimum,
and is therefore stable?
a) loop 1
b) loop 2
c) the same
Loop 1: m points to the left, so the angle
between m and B is equal to 180º, hence  = 0.
Loop 2: m points to the right, so the angle between m and B is equal to
0º, hence  = 0.
 
U  μ  B
Loop 1: U1 = m B
Loop 2: U2 = m B
 U2 is a minimum.
Lecture 13, Act 3
• A circular loop of radius R carries current I
as shown in the diagram. A constant
3A
magnetic field B exists in the +x direction.
Initially the loop is in the x-y plane.
– The coil will rotate to which of the
following positions?
y
y
w
a
z
3B
R
a
I b
x
(c) It will not rotate
a
b
B
w
(b)
(a)
y
b
z
What is the potential energy U0 of the loop in its initial position?
(a) U0 is minimum
(b) U0 is maximum
(c) neither
Lecture 13, Act 3
• A circular loop of radius R carries current I
as shown in the diagram. A constant
3A
magnetic field B exists in the +x direction.
Initially the loop is in the x-y plane.
– The coil will rotate to which of the
following positions?
y
y
w
a
z
•
•
•
•
R
a
I b
x
(c) It will not rotate
a
b
B
w
(b)
(a)
y
b
z
  
The coil will rotate if the torque on it is non-zero: τ  μ  B
The magnetic moment m is in +z direction.
Therefore the torque  is in the +y direction.
Therefore the loop will rotate as shown in (b).
Lecture 13, Act 3
• A circular loop of radius R carries current I
as shown in the diagram. A constant
magnetic field B exists in the +x direction.
3B
Initially the loop is in the x-y plane.
– What is the potential energy U0 of the
loop in its initial position?
y
B
R
I b
a
x
(a) U0 is minimum
(b) U0 is maximum
(c) neither
 
– The potential energy of the loop is given by: U   μ  B
– In its initial position, the loop’s magnetic moment vector points in
the +z direction, so initial potential energy is ZERO
– This does NOT mean that the potential energy is a minimum!!!
– When the loop is in the y-z plane and its magnetic moment points
in the same direction as the field, its potential energy is NEGATIVE
and is in fact the minimum.
– Since U0 is not minimum, the coil will rotate, converting potential
energy to kinetic energy!
Example: Loop in a B-Field
A circular loop has radius R = 5 cm and carries current I = 2 A in the
counterclockwise direction. A magnetic field B =0.5 T exists in the
negative z-direction. The loop is at an angle q = 45 to the xy-plane.
B
y
x
x
x
x
x
x
x
z
x x
x x
x x
xI x
x x
x x
x x
What is the magnetic moment m of the loop?
x
x
x
x
x
x
x
m = p r2 I = .0157 Am2
The direction of m is perpendicular to the
plane of the loop as in the figure.
Find the x and z components of m :
x
z
m
X
mx = –m sin45 = –.0111 Am2
B
mz = m cos45 = .0111 Am2
q
yX
x
Summary
• Mass Spectrometer
• Force due to B on I dF  I dl  B
• Magnetic dipole m  AIN
  
– torque
  mB
– potential energy U   m  B
zero defined at q  90
• Applications: dials, motors, NMR, …
• Next time: calculating B-fields from
currents
Reading assignment: Chapter 29.1 through 29.3
MRI (Magnetic Resonance Imaging) ≡
NMR (Nuclear Magnetic Resonance)
A single proton (like the one in
every hydrogen atom) has a charge
(+|e|) and an intrinsic angular
momentum (“spin”). If we (naively)
imagine the charge circulating in a
loop  magnetic dipole moment μ.
In an external B-field:
– Classically: there will be torques unless m is aligned along B or
against it.
– QM: The spin is always ~aligned along B or against it
Aligned: U1  m B
Anti-aligned: U 2  m B
Energy Difference: U  U 2  U1  2m B
MRI / NMR Example
Aligned: U1  m B
Anti-aligned: U 2  m B
Energy Difference: U  U 2  U1  2m B
μproton = 1.36•10-26 Am2
B = 1 Tesla (=104 Gauss)
U  2m B  2.7 10 J
26
(note: this is a big field!)
In QM, you will learn that photon
energy = frequency • Planck’s constant
h ≡ 6.6•10-34 J s
2.7  1026 J
f 
 41 MHz
-34
6.6  10 J s
What does this have to do with
?
MRI / NMR continued
If we “bathe” the protons in radio waves at this frequency, the
protons can flip back and forth.
If we detect this flipping  hydrogen!
The presence of other molecules can partially shield the applied B,
thus changing the resonant frequency (“chemical shift”).
Looking at what the resonant frequency is  what molecules are
nearby.
Finally, because f  U  B, if we put a strong magnetic field
gradient across the sample, we can look at individual slices, with
~millimeter spatial resolution.
B
Small B
low freq.
Bigger B
high freq.
Signal at the right frequency only from this slice!
See it in action!
Thanks to