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Today’s lecture is brought to you by…
…the Right Hand Rule.
Announcements
Exam 2 is two weeks from yesterday.
Contact me by the end of next Wednesdays lecture if you
have special circumstances different than for exam 1.
 Exam 2 will cover chapters 24.3-27.
 Physics 2135 Test Rooms, Spring 2016:
Instructor
Dr. Hale
Dr. Kurter
Dr. Madison
Mr. Noble
Dr. Parris
Mr. Upshaw
Sections
G, J
D, F
H, N
B, E
K, M
A, C, L
Special Accommodations
Exam is from
5:00-6:00 pm!
Room
125 BCH
104 Physics
199 Toomey
B-10 Bert.
G-31 EECH
G-3 Schrenk
Testing Center
Know the exam time!
Find your room ahead of time!
If at 5:00 on test day you are lost, go to 104 Physics and check the exam
room schedule, then go to the appropriate room and take the exam there.
Today’s agenda:
Review and some interesting consequences of
F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
Reminder: signs
F = qv  B
Include the sign on q, properly account for the directions of any
two of the vectors, and the direction of the third vector is
calculated “automatically.”
F = q vB sinθ
If you determine the direction “by hand,” use the magnitude of
the charge.
Everything in this equation is a magnitude.
The sign of r had better be +!
mv
r=
qB
Reminder: left- and right-hand axes
This is a right-handed
coordinate system:
y
z
This is not:
z
x
y
For the magnetism part of physics 2135, you MUST use righthand axes.
And you’d better use your
right hand when applying the
right-hand rule!
x
Handy way to “see” if you have drawn right-hand axes:
y
Z
?
y
x
z
x
z
?
y
x
I personally find the three-fingered axis system to often (but not
always) be the most useful way to apply the right-hand rule.


“In F = IL  B and F = qv  B does it matter
which finger I use for what?”

You’ll learn about F = IL x B
later in today’s lecture.
F = IL  B
F = qv  B
 
 


No, as long as
you keep the
right order. All
three of these
will work:
 












This works:


This doesn’t:

Switching only two is wrong!

“The right-hand rule is unfair! Physics is discriminating against
left-handers!”
No, you can get the same results with left-hand axes and lefthand rules. See this web page.
But Physics 24 does discriminate against left-handers!
This is Captain Jack Crossproduct.
He visits our classes occasionally
(see the physics on the blackboard
behind him).
You don’t want to see what he
does with his scimitar when he
sees a left hand used for the right
hand rule!
The right hand rule is just a way of
determining vector directions in a cross
product without having to do math.
Magnetic and Electric Forces
The electric force acts in the direction of the electric field.
FE = qE
The electric force is nonzero even if v=0.
The magnetic force acts perpendicular to the magnetic field.
FB = qv  B
The magnetic force is zero if v=0.

FB  v = 0  =  q 0  B = 0
Magnetic and Electric Forces
The electric force does work in displacing a charged particle.
FE = qE
E
+
F
D
WF =F  D =FD = qED
The magnetic force does no work in displacing a charged
particle!
B v
FB = qv  B
WF =F  ds =0
+ds
Amazing!
F
not really
Today’s agenda:
Review and some interesting consequences of F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying
wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
Magnetic Forces on Currents
So far, I’ve lectured about magnetic forces on moving charged
particles.
F = qv  B
Actually, magnetic forces were observed on current-carrying
wires long before we discovered what the fundamental charged
particles are.
Experiment, followed by theoretical understanding, gives
F = IL  B.
For reading clarity, I’ll use L
instead of the l your text uses.
If you know about charged particles, you can derive this from the equation for the
force on a moving charged particle. It is valid for a straight wire in a uniform
magnetic field.
Here is a picture to help you visualize. It came from
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html.
Homework Hint
F = ILB sin F max = ILB
For fixed L, B: this is the biggest force
you can get from a given current I.
For bigger F, you need bigger I.
F
       
       
B
I L
       
       
F = IL  B
Valid for straight wire, length L inside region of magnetic field,
constant magnetic field, constant current I, direction of L is
direction of conventional current I.
You could apply this equation to a beam of charged particles moving through
space, even if the charged particles are not confined to a wire.
What if the wire is not straight?
B


I
 





















 
ds
 





dF
 






dF = I ds  B
F =  dF

F = I ds  B

Integrate over the part
of the wire that is in the
magnetic field region.
Homework Hint: if you have a tiny piece of a wire, just calculate dF; no need to integrate.
Note: I generally use ds for an infinitesimal piece of wire,
instead of dl.
Font choice may make “l” look like “1.”
It’s a pain to search the fonts for a script lowercase l: l.
Example: a wire carrying current I consists of a semicircle of
radius R and two horizontal straight portions each of length L.
It is in a region of constant magnetic field as shown. What is
the net magnetic force on the wire?
I
       
B
       
R
       
 L     L 
y
x
       
There is no magnetic force on the portions of the wire outside
the magnetic field region.
First look at the two
straight sections.
F = IL  B
L  B, so
F1 =F2 = ILB
I
y
     
F1
     
R
     
 L    
 
F2
B
 
 
L 
       
x
Next look at the
semicircular section.
Calculate the
infinitesimal force dF
on an infinitesimal ds
of current-carrying
wire.
I
y
    
F1 dF ds d
    
R
    
 L   
  
F2
B
  
  
 L 
       
x
ds subtends the angle from  to +d.
Why did I call that
angle  instead of ?
The infinitesimal force is dF =I ds  B.
Because we usually
use  for the angle
in the cross product.
ds  B, so dF = I ds B.
Arc length ds =R d.
Finally, dF =I R d B.
Calculate the ycomponent of F.
dFy = I R d B sin
I

Fy =  dFy
y
0

Fy =  I R d B sin
0
dFy 
 
d
F1 dF
ds
 
R

 
 L   
  
 L 
       
x

Fy = I R B  sind
0

Fy =  -I R B cos  0
Fy = 2 I R B
  
F2
B
  
Interesting—just the force on a
straight horizontal wire of length 2R.
Does symmetry give
you Fx immediately?
Or, you can calculate
the x component of F.
I
y
dFx = -I R d B cos

Fx = - I R d B cos
 

F1 dF ds d
 
dFx
R

 
 L   
  
F2
B
  
  
 L 
       
x
0

Fx = -I R B  cosd
0

Fx = -  I R B sin 0
Fx = 0
Sometimes-Useful Homework Hint
Symmetry is your friend.
Total force:
F = F1 + F2 + Fy
F = ILB + ILB + 2IRB
I
y
Fy
     
F1 dF ds
     
R
     
 L    
 
F2
B
 
 
L 
       
F = 2IB L + R 
F = 2IB L + R  ˆj
x
We probably should write
the force in vector form.
Possible homework hint: how would the result differ if the magnetic field were directed along
the +x direction? If you have difficulty visualizing the direction of the force using the right
hand rule, pick a ds along each different segment of the wire, express it in unit vector
notation, and calculate the cross product.
Example: a semicircular closed loop of radius R carries current
I. It is in a region of constant magnetic field as shown. What
is the net magnetic force on the loop of wire?
FC
       
B
       
R
       
y
x
   I    
       
We calculated the force on the semicircular part in the previous
example (current is flowing in the same direction there as
before).
F =2 I R B
C
Next look at the
straight section.

FS = IL  B

L  B, and L=2R so

y
FS = 2IRB

FC
      
B
      
R
      
  I    
       
FS
x
Fs is directed in the –y direction (right hand rule).
Fnet =FS +Fc = -2IRB ˆj+2IRB ˆj = 0
The net force on the closed loop is zero!
This is true in general for closed loops
in a uniform magnetic field.
Today’s agenda:
Review and some interesting consequences of F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
Magnetic Forces and Torques on Current Loops
We showed (not in general, but illustrated the technique) that
the net force on a current loop in a uniform magnetic field is
zero.
No net force means no motion
motion.NOT.
Example: a rectangular current loop of area A is placed in a
uniform magnetic field. Calculate the torque on the loop.
L
I
B
W
Let the loop carry a
counterclockwise
current I and have
length L and width W.
The drawing is not meant to imply
that the top and bottom parts are
outside the magnetic field region.
There is no force on the “horizontal”
segments because the current and
magnetic field are in the “same”
direction. Homework hint (know why).
The vertical segment on the left
“feels” a force “out of the page.”
L
FL
FR B
I
W
The vertical segment on the right
“feels” a force “into the page.”
The two forces have the same magnitude: FL = FR = I L B.
Because FL and FR are in opposite directions, there is no net
force on the current loop, but there is a net torque.
Let’s look “down” along the “green axis” I just drew.
Top view of current loop, looking
“down,” at the instant the plane of the
loop is parallel to the magnetic field.
In general, torque is  = r  F .
W
1
R = FR  WILB
2
2
L =
FR
IL
FL W
2
W
1
FL  WILB
2
2
net = R  L = WILB = IAB
area of loop = WL
 IR
W
2
B
When the magnetic field is not parallel
to the plane of the loop…
W
1
R = FR sin   WILB sin 
2
2
L =
W
1
FL sin   WILB sin 
2
2
FR
W
sin 
2

IL
FL
 IR
B

W
2
A
net = R  L = WILB sin  = IAB sin 
Define A to be a vector whose magnitude is the area of the
loop and whose direction is given by the right hand rule (cross
A into B to get ). Then
 = IA  B .
Magnetic Moment of a Current Loop
W
sin 
2
 = IA  B
Alternative way to get direction of A:
curl your fingers (right hand) around
the loop in the direction of the current;
thumb points in direction of A.
IA is defined to be the magnetic
moment of the current loop.
FR

IL
FL
 IR
B

W
2
 = IA
 =  B
Homework Hint
Your starting equation sheet has:
=N I A
(N =1 for a single loop)
A
Energy of a magnetic dipole in a magnetic field
You don’t realize it yet, but we have been talking about
magnetic dipoles for the last 5 slides.
A current loop, or any other body that experiences a magnetic
torque as given above, is called a magnetic dipole.
Energy of a magnetic dipole?
You already know this:
Today:
Electric Dipole
Magnetic Dipole
 = p E
Homework Hint
 =  B
U = - p E
Homework Hint
U = -  B
Example: a magnetic dipole of moment  is in a uniform
magnetic field B . Under what conditions is the dipole’s potential
energy zero? Minimum? Under what conditions is the
magnitude of the torque on the dipole minimum? Maximum?
Example: a magnetic dipole of moment  is in a uniform
magnetic field B . Under what conditions is the dipole’s potential
energy zero? Minimum? Under what conditions is the
magnitude of the torque on the dipole minimum? Maximum?
IR

IL

IL

B
A
U= - IA  B = - IAB cos 
IR

 = 90
B
A
Potential energy is zero when
cos  = 0, or when  = 90 or 270.
Example: a magnetic dipole of moment  is in a uniform
magnetic field B . Under what conditions is the dipole’s potential
energy zero? Minimum? Under what conditions is the
magnitude of the torque on the dipole minimum? Maximum?
 IR
IR

IL

B
A
U= - IA  B = - IAB cos 
 = 0 A
B
IL
Potential energy is minimum when
cos  = 1, or when  = 0 or 180.
Example: a magnetic dipole of moment  is in a uniform
magnetic field B . Under what conditions is the dipole’s potential
energy zero? Minimum? Under what conditions is the
magnitude of the torque on the dipole minimum? Maximum?
 IR
IR

IL

B
A
 = IA  B = IAB sin 
 = 0 A
B
IL
Torque magnitude is minimum when
sin  = 0, or when  = 0 or 180.
Example: a magnetic dipole of moment  is in a uniform
magnetic field B . Under what conditions is the dipole’s potential
energy zero? Minimum? Under what conditions is the
magnitude of the torque on the dipole minimum? Maximum?
IR

IL

IL

B
A
 = IA  B = IAB sin 
IR

 = 90
B
A
Torque magnitude is maximum when
sin  = 1, or when  = 90 or 270.
Example: a magnetic dipole of moment  is in a uniform
magnetic field B . Under what conditions is the dipole’s potential
energy maximum?
IR

IL

B
A
U= - IA  B = - IAB cos 
I left this for you to figure out.
Today’s agenda:
Review and some interesting consequences of F=qvxB.
You must understand the similarities and differences between electric forces and magnetic
forces on charged particles.
Magnetic forces on currents and current-carrying wires.
You must be able to calculate the magnetic force on currents.
Magnetic forces and torques on current loops.
You must be able to calculate the torque and magnetic moment for a current-carrying wire
in a uniform magnetic field.
Applications: galvanometers, electric motors, rail
guns.
You must be able to use your understanding of magnetic forces and magnetic fields to
describe how electromagnetic devices operate.
The Galvanometer
Now you can understand how a galvanometer works…
When a current is passed through a coil connected to a needle,
the coil experiences a torque and deflects. See the link below
for more details.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1
Electric Motors
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/mothow.html#c1
Hyperphysics has nice interactive graphics showing how dc and
ac motors work.
No lecture on magnetic forces would be complete without...
…the rail gun!
Current in the rails (perhaps millions of
amps) gives rise to magnetic field (we
will study this after exam 2). Projectile
is a conductor making contact with both
rails. Magnetic field of rails exerts force
on current-carrying projectile.
The 10 meter rail gun at the University of Texas.
Today’s
lecture
brought
to you by…
A final
wordwas
from
our sponsors.
…the Right Hand Rule.