lecture 24 amperes law
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Lecture 24 amperes lawTRANSCRIPT
Lecture 24Ampère’s law
ACT: Video tape
Which kind of material would you use in a video tape?
A. Diamagnetic
B. Paramagnetic
C. Soft ferromagnetic
D. Hard ferromagnetic
Diamagnetism and paramagnetism are far too weak to be used for a video tape. Since we want the information to remain on the tape after recording it, we need a “hard” ferromagnet. These are the key to the information age—cassette tapes, hard drives, credit card strips,…
Circulation around wire
Draw an imaginary loop around a straight infinite wire and compute
B dl is perpendicular to B r
I
B
dl
r
B dl Brdrdθ
dr
20 0
002 2
I IB dl d d I
0
2
IB
r
0
2
Id
Ampère’s law
This result turns out to be true for ANY loop around ANY current.
We will not prove it in the general case. It is partially done in the book.
0 enclosedB dl I
Line integral
Current outside the loop do not make a contribution:
I
Here 0B dl
Exercise: Prove it for the infinite straight wire.
Calculating E and B fields
2
0
1 ˆ4
qE r
r
Coulomb Law
0
2
ˆ
4v r
B qr
Biot-Savart Law
Always true, can always use, but requires superposition:
enclosed
closed 0surf ace
q
E dA
Gauss Law
0 enclosedB dl I
Ampere’s Law
Always true. Useful to get E or B when charge, current distributions are symmetric
2
0
1 ˆ4
qE r
r
Coulomb Law
0
2
ˆ
4v r
B qr
Biot-Savart Law
Always true, can always use, but requires superposition:
enclosed
closed 0surf ace
q
E dA
Gauss Law
0 enclosedB dl I
Ampere’s Law
Always true. Useful to get E or B when charge, current distributions are symmetric
Direction of the Amperian loop
Same right-hand rule as in the B-field handy-trick:1) choose a direction for positive currents2) thumb in this direction, 3) fingers give direction of loop
dl
0 1 2( - )B dl I I
In this case
ACT: Four Amperian loops
Three parallel wires carry equal currents I as shown. Which of the four Amperian loops has the largest magnitude of ?
B dl
D
I I
I
A
B
C
A intercepts all three currentsB and D intercept twoC intercepts just one
Using Ampere’s law to find B
This is always true:
When there is enough symmetry, we can actually solve the integral!
Your Amperian loop should:
• contain the point where you want to find B
• respect the symmetry of the problem
• circulation in direction given by RHR (with respect to what we choose as positive current)
0 enclosedB dl I
Infinite straight wire
Symmetry: circle around the wire.
Close your eyes, let me rotate wire around the center of wireCould you tell that I rotated wire? NoCurrent distribution does not change if you rotate wire
B-field cannot change upon rotation
dl
B
I
B dl Bdl
B constant for all points on the loop
2B dl B r
r
02B r I
0
2
IB
r
Inside the infinite straight wire
A uniform current I runs through a very long wire of circular cross section with radius R as shown. What is the magnetic field at r < R?
Symmetry: as beforeAmperian loop: circle of radius r
B dl Bdl 2B dl B r
B
2 2
enclosed 2 2
r rI I I
R R
2
0 22
rB r I
R
022
I rB
R
R
B
r
Solenoid
A solenoid is a long, tightly wound helical coil of wire
B-field
DEMO: Solenoid B
lines
Solenoid symmetry
If solenoid is very long and tight, you can move solenoid back and forth; this leaves the distribution of current unchanged B-field will not change with back/forth translation B-field straight lines parallel to the solenoid axis
No field outside
Enclosed current is zero for any loop
Boutside = 0
B inside a solenoid
b c d a
a b c db
ab
a
B dl B dl B dl B dl B dl
Bdl
B dl Bh
B perpendicular to dl
B = 0 outside solenoid
0 0 0
I
Current in each turn: I Turns per length: n
enclosedI nI h
0Bh nI h
0 enclosedB dl I
I
0B nIDEMO:
Electromagnet
Uniform field
In-class example: Solenoid
A solenoid is made by winding 500 turns of wire evenly along a 20 cm long tube with radius 1 cm. What is the magnetic field at the central region of the solenoid (far from ends) if the current in the wire is 10 A?
A. 3.1410−4 T to the left
B. 3.1410−4 T to the right
C. 3.1410−2 T to the left
D. 3.1410−2 T to the right
E. None of the above
7
0
T m 500 turns4 10 10 A 0.0314 T
A 0.20 mB nI
Direction (RHR): to the right
Times when you cannot use Ampere’s law to find B
Symmetry is circular but…– no circular Amperian loop goes through center– Amperian loops that go through the center give beastly integrals
since B is not constant at all points on the Amperian loop
I
B
0 enclosed B dl I is true for the loop shown, but we cannot solve the integral!