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Physics 202, Lecture 13Today’s Topics

Sources of the Magnetic Field (Ch. 30) Calculating the B field due to currents

Biot-Savart Law Examples: ring, straight wire Force between parallel wires

Ampere’s Law: infinite wire, solenoid, toroid Displacement Current: Ampere-Maxwell Magnetism in Matter

On WebAssign Tonight: Homework #6: due 10/22 ,10 PM. Optional reading quiz: due 10/19, 7 PM

Magnetic Fields of Current Distributions

Two ways to calculate the electric field directly:

"Brute force"

– Coulomb’s Law

"High symmetry"

– Gauss’ Law

d!E =

1

4!"0

dq

r 2r̂

!E id!A! =

qin

"0

First, review: back to electrostatics:

ε0 = 8.85x10-12 C2/(N m2): permittivity of free space

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Magnetic Fields of Current Distributions

× I

– Biot-Savart Law (“Brute force”)

–AMPERIAN LOOP

Two Ways to calculate the magnetic field:

– Ampere’s Law (“high symmetry”)

d!B =

µ0I

4!

d!l " r̂

r2

!Bid!s"! = µ

0Ienclosed

µ0 = 4πx10-7 T⋅m/A: permeability of free space

Biot-Savart Law...

I B field “circulates” around the wireUse right-hand rule: thumb along I,fingers curl in direction of B.

XdB

dl

rθ

…add up the pieces

d!B =

µ0I

4!

d!l ! r̂

r2

=µ

0I

4!

d!l !!r

r3

3

B Field of Circular Current Loop on Axis (Text example 30.3) B field on axis of current loop is:

2/322

2

0

)(2 Rx

IRBx

+=

µ

Bcenter

=µ0I

2R

See also: center of arc (text example 30.2)

B of Circular Current Loop: Field LinesB

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B Field of Straight Wire, length L (Text example 30.1) B field at point P is:

When the length of thewire is infinity:

)cos(cos4

210

!!"

µ#=

a

IB

θ1 θ2

a

IB

!

µ

2

0=

example Ch. 30, #3(return to this later with Ampere’s Law)

Superposition:

Which figure represents the B field generated by thecurrent I.

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

Quick Question 1: Magnetic Field and Current

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Quick Question 2: Magnetic Field and Current

Which figure represents the B field generated by thecurrent I.

Forces between Current-Carrying WiresA current-carrying wire experiences a force in an external

B-field, but also produces a B-field of its own.

dIa

Ib

!F

!F

dIa

Ib

!F

!F

Parallel currents: attract

Opposite currents: repel

Magnetic Force:

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Example: Two Parallel Currents• Force on length L of wire b due to field of wire a:

Ba=µ0Ia

2!d

Fb= I

bd!l !!Ba="µ0IbIaL

2#d

dIa

Ib

!F

!F

x

• Force on length L of wire a due to field of wire b:

Bb=µ0Ib

2!d

Fa= I

ad!l !!Bb="µ0IaIbL

2#d

Ampere’s Law Ampere’s Law:

applies to any closed path, any static B field useful for practical purposes

only for situations with high symmetry

Ampere’s Law can be derived from Biot-Savart Law Generalized form: Ampere-Maxwell (later in lecture)

!Bid!s = µ

0Ienclosed

anyclosed path

"!

I

ds

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Use symmetry (and Ampere’s Law)

Ampere’s Law: B-field of ∞ Straight Wire

× I

!B id"s = µ

0#! I

!B id"s = Brd! = 2"rB =## µ

0## I

B =µ

0I

2!r

Quicker and easier method for B field of infinite wire (due to symmetry -- compare Gauss’ Law)

Choose loop to be circle of radius R centered on thewire in a plane ⊥ to wire.Why?

Magnitude of B is constant (function of R only)Direction of B is parallel to the path.

B Field Inside a Long Wire

Total current I flows through wire ofradius a into the screen as shown.

x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x xa

r

⇒

What is the B field inside the wire?

By symmetry -- take the path to be a circle of radius r:

!Bid!s = B2!r""

•Current passing through circle: Ienclosed

=r2

a2I

Therefore: Ampere’s Law gives

!Bid!s = B2!r = µ

oIenclosed"" B =

µ0Ir

2!a2

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B Field of a Long Wire

• Outside the wire:( r > a )

B =µ

0I

2! r

Inside the wire: (r < a)

B =µ

0I

2!

r

a2

r

B

a

Ampere’s Law: Toroid (Text example 30.5) Using Ampere’s Law,

B field inside a toroid is found to be:

r

NIB

!

µ

2

0=

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Ampere’s Law: ToroidToroid: N turns with current I.

Bφ=0 outside toroid!

Bφ inside: consider circle of radius r,centered at the center of the toroid.

x

x

x

xx

xx

x

x x

xx x

x

x

x

•

••

•

• •

•

••

•

• •

•

•

••

r

B

(Consider integrating B on circle outside toroid: net current zero)

!Bid!s = B2!r = µ

oIenclosed""

B =µ0NI

2!r

B Field of a Solenoid

Solenoid: source of uniform B field (inside of it)

If a << L, the B field is to first order contained within the solenoid,in the axial direction, and of constant magnitude.

• Solenoid: current I flows through a wirewrapped n turns per unit length on acylinder of radius a and length L. L

aTo calculate the B-field, use Biot-Savart

and add up the field from the different loops.

In this limit, can calculate the field using Ampere's Law!

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Ampere’s Law: Solenoid The B field inside an ideal solenoid is:

(see board)nIB0

µ=

ideal solenoid

n=N/L

segment 3 at ∞

Right-hand rule: direction of field lines.

In this example, since the field lines leave the left end,the left end is the north pole.

Solenoid: Field Lines

Like a bar magnet! (except it can be turned on and off)