The document summarizes Ampere's law and its applications. It begins by stating Ampere's law as an integral form: the line integral of the magnetic field B around any closed loop is equal to the permeability times the current enclosed by the loop. It then provides examples of using Ampere's law to calculate B for an infinite straight wire and inside a long solenoid. The document notes that B can be found using Ampere's law when there is sufficient symmetry but that integrals may be impossible to solve in other cases lacking symmetry.

1. Lecture 24
Ampère’s law

2. 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,…

3. Circulation around wire
r r
Draw an imaginary loop around a straight infinite wire and compute Ñ ×
∫ B dl
→
B
→
dl
rdθ
I
r
r
B is perpendicular to r
r r
B × = Brd θ
dl
µ0 I
=
dθ
2π
dr
→
r
B =
r r
Ñ ×dl =
∫B
µ0I
µ0 I
Ñ2π d θ = 2π
∫
∫
2π
0
d θ = µ 0I
µ0 I
2π r

4. 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.
r r
Ñ ×dl = µ0I enclosed
∫B
Line integral
Current outside the loop do not make a contribution:
I
r r
Here Ñ × = 0
∫ B dl
Exercise: Prove it for the infinite straight wire.

5. Calculating E and B fields
Always true, can always use, but requires superposition:
r
E =
1 q
ˆ
r
2
4πε 0 r
Coulomb Law
r µ vr × r
ˆ
0
B =
q 2
4π
r
Biot-Savart Law
Always true. Useful to get E or B when charge, current
distributions are symmetric
∫
closed
surface
r
r q
E × = enclosed
dA
ε0
Gauss Law
r r
Ñ ×dl =µ0I enclosed
∫B
Ampere’s Law

6. Direction of the Amperian loop
Same right-hand rule as in the B-field handy-trick:
1) choose a direction for positive currents
2) thumb in this direction,
3) fingers give direction of loop
dl
In this case
r r
Ñ ×dl =µ0 (I1 - I2 )
∫B

7. ACT: Four Amperian loops
Three parallel wires carry equal currents I as shown.r Which of the four
r
Amperian loops has the largest magnitude of Ñ ×
∫ B dl ?
I
I
A
B
C
D
I
A intercepts all three currents
B and D intercept two
C intercepts just one

8. Using Ampere’s law to find B
r r
This is always true: Ñ ×
∫ B dl =µ0I enclosed
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)

9. Infinite straight wire
Symmetry: circle around the wire.
Close your eyes, let me rotate wire around the center of wire
Could you tell that I rotated wire? No
Current distribution does not change if you rotate wire
B-field cannot change upon rotation
r
B
I
→
r
dl
r r
Ñ ×dl = Ñ = B Ñ = B 2π r
∫B
∫ Bdl ∫ dl
B constant for all
points on the loop
B 2π r = µ0I
µ0 I
B=
2π r

10. 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?
r
B
Symmetry: as before
Amperian loop: circle of radius r
r r
Ñ ×dl = Ñ = B Ñ = B 2π r
∫B
∫ Bdl ∫ dl
I enclosed
r2
B 2π r = µ0 2 I
R
πr 2
r2
=
I = 2I
2
πR
R
B=
B
µ0Ir
2π R 2
R
r

11. Solenoid
A solenoid is a long, tightly wound helical coil of wire
DEMO:
Solenoid B
lines
B-field

12. 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

13. No field outside
Enclosed current is zero for any loop
Boutside = 0

14. B inside a solenoid
I
B = 0 outside solenoid
B perpendicular to dl
0
r r b r r c r r d r 0r a r 0v
Ñ ×dl = ∫ B ×dl + ∫ B ×dl + ∫ B ×dl + ∫ B ×dl
∫B
b
a
b
= ∫ Bdl
a
b
= B ∫ dl = Bh
a
c
d

15. I
Current in each turn: I
Turns per length: n
I enclosed = nIh
r r
Ñ ×dl = µ0I enclosed
∫B
Bh = µ0nIh
B = µ0nI
Uniform field
DEMO:
Electromagnet

16. 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. 3.14×10−4 T to the left
B. 3.14×10−4 T to the right
C. 3.14×10−2 T to the left
D. 3.14×10−2 T to the right
E. None of the above

T m  500 turns
B = µ0nI =  4π × 10 −7
( 10 A) = 0.0314 T
÷
A  0.20 m

Direction (RHR): to the right

17. Times when you cannot use Ampere’s law to find B
B
I
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
r r
Ñ ×dl =µ0I enclosed is true for the loop shown, but we cannot
∫B
solve the integral!