1. W09D1: Sources of Magnetic Fields: Ampere’s Law
Today’s Reading Assignment Course Notes:
Sections , 9.7,
Class 18

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 or
Next reading Assignment W09D2 Faraday’s Law Course Notes: Sections

3. Today: Biot-Savart vs. Ampere
Biot-Savart Law
general
current source
ex: finite wire
wire loop
Ampere’s law
symmetric
ex: infinite wire
infinite current sheet
Class 20

4. Last Time: Creating Magnetic Fields: Biot-Savart
Class 20

5. The Biot-Savart Law
Current element of length carrying current I produces a magnetic field at the point P:
Class 20

6. The Biot-Savart Law: Infinite Wire
Magnetic Field of an Infinite Wire Carrying Current I from Biot-Savart:
See W06D3 Problem Solving
More generally:
Class 20

7. 3rd Maxwell Equation: Ampere’s Law (we will eventually add one more term to this equation)
Open surface is bounded by closed path
Class 20

8. Ampere’s Law: The Idea
In order to have a B field around a loop, there must be current punching through the loop
Class 20

9. Concept Question: Line Integral
The integral expression
is equal to the magnetic work done around a closed path.
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.
is always zero.
is equal to the magnetic potential energy between two points.
None of the above.
Class 20

10. 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.
Class 20

11. 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.
Class 12

12. 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 :
Class 20

13. 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 :
Class 20

14. Concept Questions: Ampere’s Law
Class 20

15. Concept Question: Ampere’s Law
Integrating B around the loop shown gives us:
a positive number
a negative number
zero
Class 20

16. C.Q. Answer: Ampere’s Law
Week 07, Day 2
C.Q. Answer: Ampere’s Law
Answer: 3. Total enclosed current is zero, so
Class 16

17. Concept Question: Ampere’s Law
Integrating B around the loop in the clockwise direction shown gives us:
a positive number
a negative number
zero
Class 20

18. C.Q. Answer: Ampere’s Law
Week 07, Day 2
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)
Class 16

19. Applying Ampere’s Law Identify regions in which to calculate B field.
Choose Amperian closed path such that by symmetry B is zero or constant magnitude on the closed path!
Calculate
Calculate current enclosed:
Apply Ampere’s Law to solve for B: check signs
Class 20

20. Week 06, Day 1
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)
Class 14

21. Worked Example: Ampere’s Law Infinite Wire
Week 06, Day 1
Worked Example: Ampere’s Law Infinite Wire I B I
Amperian Closed Path:
B is Constant & Parallel
Current penetrates surface
Class 14

22. Example: Infinite Wire
Week 06, Day 1
Example: Infinite Wire
Region 1: Outside wire (r ≥ R)
Cylindrical symmetry 
Amperian Circle
B-field counterclockwise
Class 14

23. Group Problem: Magnetic Field Inside Wire
Week 06, Day 1
Group Problem: Magnetic Field Inside Wire I
We just found B(r>R)
Now you find B(r<R)
Class 14 23

24. Infinite Wire: Plot of B vs. r
Week 06, Day 1
Infinite Wire: Plot of B vs. r
Class 14

25. Group Problem: Non-Uniform Cylindrical Wire
A cylindrical conductor has radius R and a non-uniform current density with total current:
Find B everywhere
Class 20

26. Other Geometries
Class 20

27. Two Loops
Class 20

28. Two Loops Moved Closer Together
Class 20

29. Multiple Wire Loops
Class 20

30. Multiple Wire Loops – Solenoid
Class 20

31. Demonstration: Long Solenoid
Class 20

32. Magnetic Field of Solenoid
Horiz.
comp.
cancel
loosely wound
tightly wound
For ideal solenoid, B is uniform inside & zero outside
Class 20

33. Magnetic Field of Ideal Solenoid
Using Ampere’s law: Think!
Class 20

34. 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.
Class 20

35. Ampere’s Law: Infinite Current SheetB I B
Amperian Loops:
B is Constant & Parallel OR Perpendicular OR Zero
I Penetrates
Class 20

36. 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
Class 20

37. 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
Class 20

38. infinite current sheet
Biot-Savart vs. Ampere
Biot-Savart Law
general
current source
ex: finite wire
wire loop
Ampere’s law
symmetric
ex: infinite wire
infinite current sheet
Class 20

39. (Infinite) Current Sheet
Ampere’s Law: I B
Long
Circular
Symmetry
(Infinite) Current Sheet X B X =
2 Current Sheets
Solenoid
Torus
Class 20