Lecture 24 amperes law

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!