Post on 31-Dec-2015
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W09D1:Sources of Magnetic Fields:
Ampere’s Law
Today’s Reading Assignment Course Notes:
Sections 9.3-9.4, 9.7, 9.10.2
Announcements
Math Review Week Nine Tuesday from 9-11 pm in 26-152
PS 7 due Tuesday April 9 at 9 pm in boxes outside 32-082 or 26-152
Next reading Assignment W09D2 Faraday’s Law Course Notes: Sections 10.1-10.4
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Today: Biot-Savart vs. Ampere
Biot-Savart Law
general current sourceex: finite wire
wire loop
Ampere’s law
symmetriccurrent source
ex: infinite wireinfinite current sheet
encId 0sB
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Last Time:Creating Magnetic Fields:
Biot-Savart
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The Biot-Savart Law
Current element of length carrying current I produces a magnetic field at the point P:
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The Biot-Savart Law: Infinite Wire
Magnetic Field of an Infinite Wire Carrying Current I from Biot-Savart:
See W06D3 Problem Solving http://web.mit.edu/8.02t/www/materials/ProblemSolving/solution05.pdf
More generally:
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3rd Maxwell Equation: Ampere’s Law
(we will eventually add one more term to this equation)
Open surface is bounded by closed path
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Ampere’s Law: The Idea
In order to have a B field around a loop, there must be current punching through the loop
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Concept Question: Line IntegralThe integral expression
1) is equal to the magnetic work done around a closed path.
2) is an infinite sum of the product of the tangent component of the magnetic field along a small element of the closed path with a small element of the path up to a choice of plus or minus sign.
3) is always zero.
4) is equal to the magnetic potential energy between two points.
5) None of the above.
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C.Q. Answer: Line Integral
2. A line integral by definition is the sum
We need to make a choice of integration direction (circulation) for the line integral. The small line element is tangent to the line and points in the direction of circulation. The dot product therefore is the product of the tangent component of the magnetic field in the direction of the line element. So the answer depends on which way we circulate around the path.
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Current Enclosed
Current density
Current enclosed is the flux of the current density through an open surface S bounded by the closed path. Because the unit normal to an open surface is not uniquely defined this expression is unique up to a plus or minus sign.
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Sign Conventions: Right Hand Rule
Integration direction clockwise for line integral requires that unit normal points into page for open surface integral
Current positive into page, negative out of page
:
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Sign Conventions: Right Hand Rule
Integration direction counterclockwise for line integral requires that unit normal points out of page for open surface integral
Current positive out of page, negative into page
:
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Concept Questions:Ampere’s Law
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Concept Question: Ampere’s Law
Integrating B around the loop shown gives us:
1. a positive number
2. a negative number
3. zero
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C.Q. Answer: Ampere’s Law
Answer: 3. Total enclosed current is zero, so
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Concept Question: Ampere’s Law
Integrating B around the loop in the clockwise direction shown gives us:
1. a positive number
2. a negative number
3. zero
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C.Q. Answer: Ampere’s Law
Answer: 2.
Net enclosed current is out of the page, so field is counter-clockwise (opposite to circulation direction)
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Applying Ampere’s Law1) Identify regions in which to calculate B field.
2) Choose Amperian closed path such that by symmetry B is zero or constant magnitude on the closed path!
3) Calculate
4) Calculate current enclosed:
5) Apply Ampere’s Law to solve for B: check signs
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Infinite Wire
I A cylindrical conductor has radius R and a uniform current density with total current I. we shall find the direction and magnitude of the magnetic field for the
two regions:
(1) outside wire (r ≥ R)
(2) inside wire (r < R)
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Worked Example: Ampere’s Law Infinite WireI
I
B
Amperian Closed Path:
B is Constant & Parallel
Current penetrates surface
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Example: Infinite WireRegion 1: Outside wire (r ≥ R)
0I
enc 0I
Cylindrical symmetry Amperian Circle
B-field counterclockwise
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Group Problem: Magnetic Field Inside Wire
I We just found B(r>R)
Now you find B(r<R)
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Infinite Wire: Plot of B vs. r
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2 R
IrBin
r
IBout
2
0
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Group Problem: Non-Uniform Cylindrical Wire
A cylindrical conductor has radius R and a non-uniform current density with total current:
Find B everywhere
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Other Geometries
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Two Loops
http://web.mit.edu/viz/EM/visualizations/magnetostatics/MagneticFieldConfigurations/tworings/tworings.htm
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Two Loops Moved Closer Together
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Multiple Wire Loops
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Demonstration:Long Solenoid
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=G%2018&show=0
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Magnetic Field of Solenoid
loosely wound tightly wound
For ideal solenoid, B is uniform inside & zero outside
Horiz.
comp.
cancel
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Magnetic Field of Ideal SolenoidUsing Ampere’s law: Think!
IencnlI n: # of turns per unit length
n N / L : # turns/unit length
Bl 0 0 0
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Group Problem: Current Sheet
A sheet of current (infinite in the y & z directions, of thickness d in the x direction) carries a uniform current density:
Find the direction and magnitude of B as a function of x.
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Ampere’s Law:Infinite Current Sheet
I
Amperian Loops:
B is Constant & Parallel OR Perpendicular OR Zero
I Penetrates
B
B
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Surface Current Density
A very thin sheet of current of width w carrying a current I in the positive z-direction has a surface current density
For sheet of thickness d , width w, and current I
K I / w
I Jdw Kw J K / d
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Solenoid is Two Current Sheets
Consider two sheets each of thickness d with current density J. Then surface current per unit length
Use either Ampere’s Law or superposition principle
K Jd nI
B 0K
0Jd
0nI
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Biot-Savart vs. Ampere
Biot-Savart Law
general current sourceex: finite wire
wire loop
Ampere’s law
symmetriccurrent source
ex: infinite wireinfinite current sheet
encId 0sB
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=
2 Current Sheets
Ampere’s Law:
IB
B
X XX
X
X
XX
X
XXX
X
X
XX
X
XXXXXXXXXXXX
B
Long
Circular
Symmetry(Infinite) Current Sheet
Solenoid
Torus