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!