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TRANSCRIPT
Contact me by the end of today’s lecture if you have special circumstances different than for exam 1.
Exam 2 is next week.
Announcements
Exam 2 will cover chapters 24.3-27.
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.
Exam is from 5:00-6:00 pm!
Physics 2135 Test Rooms, Fall 2015:
Instructor Sections RoomDr. Hagen J, R G-3 SchrenkDr. Hale K, P 120 Butler-CarltonDr. Hor B, D G-3 SchrenkDr. Kurter A, C 104 PhysicsDr. Madison G, L 204 McNuttDr. Parris M, Q 125 Butler-CarltonMr. Upshaw N, H G-5 Humanities/Soc. Sci.Dr. Vojta E, F 199 Toomey
Special Accommodations Testing Center
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.”
θF= q vB sin
If you determine the direction “by hand,” use the magnitude of the charge.
mv
r=qB
Everything in this equation is a magnitude. The sign of r had better be +!
Reminder: left- and right-hand axes
This is a right-handed coordinate system:
For the magnetism part of physics 2135, you MUST use right-hand axes.
And you’d better use your right hand when applying the right-hand rule!
x
y
z
This is not:
x
z
y
Handy way to “see” if you have drawn right-hand axes:
x
y
Z
x
z
y
?
x
y
z
?
I personally find the three-fingered axis system to often (but not always) be the most useful way to apply the right-hand rule.
No, as long as you keep the right order. All three of these will work:
“In and does it matter which finger I use for what?”
F=IL B
F=qv B
F=IL B
F=qv B
No, as long as you keep the right order.
You’ll learn about F = IL x B later in today’s lecture.
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 left-hand rules. See this web page.
This is Captain Jack Crossproduct.
But Physics 24 does discriminate against left-handers!
He visits our classes occasionally(see the physics on the blackboardbehind 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.
The electric force acts in the direction of the electric field.
Magnetic and Electric Forces
The magnetic force acts perpendicular to the magnetic field.
EF =qE
BF =qv B
The electric force is nonzero even if v=0.
The magnetic force is zero if v=0.
BF v=0 = q 0 B=0
The electric force does work in displacing a charged particle.
Magnetic and Electric Forces
EF =qE
BF =qv B
The magnetic force does no work in displacing a charged particle!
EF
+
D
FW =F D=FD=qED
+
v B
F
ds
FW =F ds=0
Amazing! 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.
So far, I’ve lectured about magnetic forces on moving charged particles.
Magnetic Forces on Currents
Actually, magnetic forces were observed on current-carrying wires long before we discovered what the fundamental charged particles are.
F=qv B
Experiment, followed by theoretical understanding, gives
F=IL B.
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.
For reading clarity, I’ll use
L instead of the l your text uses.
Here is a picture to help you visualize. It came from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html.
Homework Hint
max
F =ILB sin F =ILBFor fixed L, B: this is the biggest force you can get from a given current I.For bigger F, you need bigger I.
LI
F
B
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?
I
B
dsdF
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?
B
L L
R
I
x
y
There is no magnetic force on the portions of the wire outside the magnetic field region.
B
L L
R
I
x
y
First look at the two straight sections.
F=IL B
1 2F =F =ILB
L B, so
F1 F2
B
L L
R
I
x
y
Next look at the semicircular section.
dF=I ds B.
ds B, so
F1 F2
Calculate the infinitesimal force dF on an infinitesimal ds of current-carrying wire.
dF
dds
ds subtends the angle from to +d.The infinitesimal force is
ds=R d .Arc length
dF=I ds B.
dF=I R d B.Finally,
Why did I call that angle instead of
?Because we usually use for the angle in the cross product.
B
L L
R
I
x
y
Calculate the y- component of F. F1 F2dF
ddsdFy
ydF =I R d B sin
y y0F = dF
y 0
F = I R d B sin
y 0
F =I R B sin d
0
yF = -I R B cos
yF =2 I R BInteresting—just the force on a straight horizontal wire of length 2R.
B
L L
R
I
x
y
F1 F2dF
dds
Or, you can calculate the x component of F.
Does symmetry give you Fx immediately?
xdF =-I R d B cos
x 0
F =- I R d B cos
x 0
F =-I R B cos d
0
xF =- I R B sin
xF =0
dFx
Sometimes-Useful Homework Hint
Symmetry is your friend.
B
L L
R
I
x
y
Total force:F1 F2dF
ds
Fy
1 2 yF=F + F + F
F=ILB + ILB + 2IRB
F=2IB L + R
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.
ˆF=2IB L + R j We probably should
write the force in vector form.
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?
B
R
I
x
y
We calculated the force on the semicircular part in the previous example (current is flowing in the same direction there as before).
CF =2 I R B
FC
B
R
I
x
y
FC
Next look at the straight section.
SF =IL B
SF =2IRB
L B, and L=2R so
Fs is directed in the –y direction (right hand rule).
FS
ˆ ˆnet S cF =F +F =-2IRB j+2IRB j=0
The net force on the closed loop is zero!
This is true in general for closed loopsin 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 LoopsWe 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.No net force means no motion NOT.
Example: a rectangular current loop of area A is placed in a uniform magnetic field. Calculate the torque on the loop.
BI
Let the loop carry a counterclockwise current I and have length L and width W.
W
L
The drawing is not meant to imply that the top and bottom parts are outside the magnetic field region.
B
I
W
L
There is no force on the “horizontal” segments because the current and magnetic field are in the “same” direction.
The vertical segment on the left “feels” a force “out of the page.”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.
FL
FR
There is no force on the “horizontal” segments because the current and magnetic field are in the “same” direction. Homework hint (know why).
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.
FL
FR
B
IL IRW2
W2
=r F .In general, torque is
R R
W 1= F WILB
2 2
L L
W 1= F WILB
2 2
net R L= =WILB=IAB
area of loop = WL
When the magnetic field is not parallel to the plane of the loop…
FL
FR
B
IL
IR
W2
W
sin 2
R R
W 1= F sin WILB sin
2 2
L L
W 1= F sin WILB sin
2 2
net R L= =WILB sin =IAB sin
A
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 .
FL
FR
B
IL
IR
W2
W
sin 2
A
= IA B
Magnetic Moment of a Current Loop
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.
= IA
= B
Your starting equation sheet has:
= N I A (N=1 for a single loop)
Homework Hint
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:
= B p
= E
p
U= - E
U= - B
Electric Dipole
Magnetic Dipole
Homework Hint
Homework Hint
Example: a magnetic dipole of moment is in a uniform magnetic field . 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?
B
Example: a magnetic dipole of moment is in a uniform magnetic field . 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?
B
cos
U= - IA B = - IAB Potential energy is zero whencos = 0, or when = 90 or 270.
IR
IL
A
B
IL
IR
A
B = 90
Example: a magnetic dipole of moment is in a uniform magnetic field . 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?
B
Potential energy is minimum when cos = 1, or when = 0 or 180.
IL
IR
A B
IR
IL
A
B
= 0
cos
U= - IA B = - IAB
Example: a magnetic dipole of moment is in a uniform magnetic field . 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?
B
sin
= IA B = IAB Torque magnitude is minimum when sin = 0, or when = 0 or 180.
IL
IR
A B
IR
IL
A
B
= 0
Example: a magnetic dipole of moment is in a uniform magnetic field . 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?
B
sin
= IA B = IAB Torque magnitude is maximum when sin = 1, or when = 90 or 270.
IR
IL
A
B
IL
IR
A
B = 90
Example: a magnetic dipole of moment is in a uniform magnetic field . Under what conditions is the dipole’s potential energy maximum?
B
IR
IL
A
B
cos
U= - IA B = - IAB 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…
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1
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.
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 10 meter rail gun at the University of Texas.
…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.
Today’s lecture was brought to you by…
…the Right Hand Rule.
A final word from our sponsors.