Physics for Scientists and Engineers Chapter 27: Sources of the Magnetic Field Copyright © 2004 by W. H. Freeman & Company Paul A. Tipler Gene Mosca Fifth.

Physics for Scientistsand Engineers

Chapter 27:Sources of the Magnetic Field

Copyright © 2004 by W. H. Freeman & Company

Paul A. Tipler • Gene Mosca

Fifth Edition

So far, we have learnt the effect of

B fields on charged particles,

on current carrying conductors,

Torque on a current loop

(magnetic moment)

Next: B field from a current

Lectures 13 & 14: The Magnetic Field Related to its Cause: ELECTROMAGNETISM

Learning Objectives: To Learn and Practice Methods used to Calculate the Magnetic Field of Steady Currents

Method 1: The Biot-Savart Law (Tipler 27-1&2)

Hans Christian OerstedDanish Physicist

(1777-1851)In 1820 Oersted demonstrated that a magnetic field exists near a current-carrying wire - first connection between electric and magnetic phenomena.

q

v

Pr

Magnetic Field of a Moving Charge

2

sin

r

qvB

φ∝

Perpendicular to the plane containing r and v

Taking into account direction: 2

ˆ

r

rvqB

∧∝

B

20 ˆ

4 r

rvqB

∧=

πμ

In SI units

-170 A m T 104 −×= πμ

the permeability of free space

r

rr =ˆ is a unit vector

20 ˆ

4 r

rvqB

∧=

πμ

Magnetic Field of a Current Element

I

l

r

P B into screen

2

0 sin

4 r

qvB

φπμ

=

20 sin

4 r

lIB

φπμ ×=

20 ˆ

4 r

rvqB

∧=

πμ

€

qv =qvδl

δl=

qδlδl

v

=qδl

δt= Iδl

Magnetic Field of a Current Element

In vector form

20 ˆ

4 r

rlIdBd

∧×=

πμ

The Magnetic Fields Established by Complete Circuits

Procedure

•Write down dB in terms of a single variable

•Integrate between the limits applicable to the problem

•Be careful about the directions of the vector quantities

•Use symmetry to simplify the problem

Magnetic Field of a Current Element

The Biot-Savart Law

In vector form

20 ˆ

4 r

rlIdBd

∧×=

πμ

€

dE =1

4πε0

λdy

r2ˆ r

dy

P

r

Directions of the fields!!!

Magnetic Field of a Current Element

I

l

r

P B into screen

20 ˆ

4 r

rlIdBd

∧×=

πμ

I

l rP

lI

lI

€

dB =μ0

4π×

Idl∧ˆ r

r2

€

dB =μ0

4π×

Idl

r2

€

B =μ0

4π×

I

r2dl

0

2πr

∫ =μ0

4π×

I

r2× 2πr =

μ0I

2r

The B Field Due to a Long Straight Wire (Tipler pg 865)

I

l

r

P

a

l

€

l = a × tanθ ××= 2secal

sec×= arφ cossin =

πμ

22

20

sec

cossec

4 a

IaB

××=

πμ ××= cos4

0

a

IB

B

20 sin

4 r

lIB

φπμ ×=

The B Field Due to a Long Straight Wire

I

l

r

P

a

l

πμ ××= cos4

0

a

IB

∫×=2

1

cos4

0

θ

θ

θθπ

μd

a

IB

( )120 sinsin

4θθ

π

μ−×=

a

IB

Exercise: Find the magnitude of the magnetic field at the centre of a square current loop of side l = 50 cm carrying a current I = 1.5 A.

ll/2 I

First “picture” the problem. The magnetic field is the sum of the contributions from the four sides of the loop

( )( ) T 6OO0 1039.345sin45sin

2

lI

44B −×=−−×

π

μ×=

The B Field Due to an Infinite Straight Wire

Ia

( )120 sinsin

4θθ

π

μ−×=

a

IB

2 and

2 12

ππ −==

a

IB

πμ2

0=

E Field of a line charge

∫ =×=0

2ε

λπ

lrlEAd.E

rE

02πελ

=a

IB

πμ2

0=

Lecture Questionnaire Summary

A: 100, B: 80, C: 60, D: 40, E: 20

F: 0

1. Could you clearly hear the lecturer? 89

2. Was the lecturer enthusiastic?97

3. How helpful were the lecturer’s responses to questions?

86

4. Did the Lecturer use board/OHP/PPT

in a clear way?93

5. Were the handouts useful? 93

6. How interesting was the module?78

7. Introduction to each lecture 88

8. How useful were the text books?79

9. Level of presentation 95

10. Speed of the lecture 94

11. Level of the problem sheets 87

7. How interesting was the module? 78

Coulomb

Gauss

Ampere

Faraday

78

78

78

78

78

Guo 97

Tesla

B at any Point on the Axis of a Single Current Loop(Tipler pg. 859)

dl parallel to z, B field at P has no z component. But it has both x and y components.dB perpendicular to r

I

P

dl1

a

dB1

x

dl2

dB2

r

x

y

dB1

dB2

dBtotalP

•By symmetry, components of dB perpendicular to the axis sum to zero

Each dB has a component parallel to the x-axis of magnitude

cos×= dBdBx

( )220

4 ax

dlIdB

+×=

π

μWhere

and

( ) 21

22cos

ax

a

+=θ

( )( ) 21

2222

0

4 ax

a

ax

dlIdBx

++×=

π

μ

Integrating:

( ) ∫∫+

== dlax

IadBB xx

23

22

0

4π

μ

( ) 23

22

20

2 ax

IaBx

+=∴

μ

B at any Point on the Axis of a Single Current Loop(Tipler pg. 859)

( ) 23

22

20

2 ax

IaBx

+=∴

μ

A “small” loop of current is called a magnetic dipole

Exercise:

Obtain an expression for the magnetic field

(i) At the centre of the loop

a2

IB 0

0

μ=

(ii) At x >> a

( ) 23

22

20

2 ax

IaBx

+=

μConsider x >> a

3

20

2x

IaBx

μ= Magnetic dipole moment:

[ ] [ ]loopofareacurrent ×=μ

30 2

4 xBx

μπμ

=

Compare with electric dipole

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ +

−

⎟⎠⎞

⎜⎝⎛ −

= 220

22

4

1

ar

q

ar

qE

πε

The E-field due to an Electric Dipole - Calculation

To simplify the calculation, we will only compute the field along the axis

r

E

-q +q

a

a/2

304

2

r

qaE

πε≈

For r >> a

This applies to any points along the line of the dipole,

and only for points along the line of the dipole.

Along this particular direction, the E field from the

positive charge is in opposite direction to that from

the negative charge.

Electric Dipole

Magnetic Dipole

30

2

4

1

x

pEx πε

=

30 2

4 xBx

μπμ

=

B-field lines encircle the current that acts as their source. B-field lines are continuous loops (lecture 11 - Gauss’s Flux Law for Magnetism)

ReviewThe magnetic field set up by a current-carrying conductor can be found from the Biot-Savart law. This law asserts that the contribution dB to the field set up by a current element Idl at a point P, a distance r from the current element, is

20 ˆ

4 r

rlIdBd

∧×=

πμ

REMEMBER THIS

Calculating Magnetic Field Strengths Method II Ampere’s Law (Tipler 27-4)

r

IB

πμ2

0=

( ) IrB 02 μπ =×

€

B.dl = B.2πr = μ0I∫Thus it appears that

The B Field Due to a Long Straight Wire

We know that

€

B.dl = μ0I∫

Right hand rule and the

direction of the B field from a current

In general, the line integral of B around any closed mathematical path equals μ0 times the current intercepted by the area spanning the path.

∫ = IldB 0. μ Ampere’s Law

•Left hand side is a line integral round a closed loop•I is the current enclosed by the loop

B-Field Inside a Long Solid Cylindrical Conductor Carrying Uniformly Distributed Current

(Example 27-9 – Tipler)

First Example

€

B.dl = 2πrB∫

I

R

r

€

Iencl = Iπr2

πR2=

Ir2

R2

Enclosed current:

20

2 R

rIB

πμ

=

2

202R

IrrB

μπ = r

0 200 400 600 800 1000 12000

r=R

B(r)

r

B field inside a conductor, OKE field inside a conductor, Not OK.

r

IB

πμ2

0=2

0

2 R

rIB

πμ

=

B-Field Inside a Long Solenoid

Second Example

∫ =BLldB.

nLIBL 0μ=

n = number of turns per unit length

nIB 0μ=

Field of a Toroidal Solenoid(Tipler pg. 872-873)

Third Example

Path 2Path 1 - no current

enclosed B = 0

Path 2 - no net current enclosed B = 0

Path 3 - net current enclosed

∫ = rBldB π2. NI0μ=

N = total number of turns

r

NIB

πμ20=

Path 1

Ampere’s law is useful for calculating the magnetic field only when there is a high degree of symmetry.

Need B to be constant in magnitude along the enclosed path and tangent to any such path.

(a)zero

(b)into the screen

(c)out of the screen

(d)toward the top or bottom of the screen

(e)toward one of the wires.

Short Exercise 1:Two wires lie in the plane of the screen and carry equal currents in opposite directions. At a point midway between the wires, the magnetic field is

Short Exercise 1:Two wires lie in the plane of the screen and carry equal currents in opposite directions. At a point midway between the wires, the magnetic field is

(a)zero

(b)into the screen

(c)out of the screen

(d)toward the top or bottom of the screen

(e)toward one of the wires.

The Force Between Two Long Parallel Currents

r

IB

πμ2

0=

The B Field Due to a Long Straight Wire

r

IB

πμ2

0=

Force F on a length L of the upper conductor is:

r

LIILBIF

πμ20 ′

=′=

I

I

FB

B r

IB

πμ20 ′

=′

r

LIIBILF

πμ20 ′

=′=′

Force F on a length L of the lower conductor is:

F

F = F Attraction!

F = F

I

I

FB

BF

Attraction!

What happens when the currents are in opposite directions?

force per unit length r

II

L

F

πμ20 ′

=

This fundamental magnetic effect was first studied by Ampere (1822)

(Ans. Repulsion)

Short Exercise 2:Two parallel wires carry currents I1 and I2 (= 2I1) in the same direction. The forces F1 and F2 on the wires are related by:

•F1 = F2

•F1 = 2F2

•2F1 = F2

•F1 = 4F2

•4F1 = F2

Short Exercise 2:Two parallel wires carry currents I1 and I2 (= 2I1) in the same direction. The forces F1 and F2 on the wires are related by:

•F1 = F2

•F1 = 2F2

•2F1 = F2

•F1 = 4F2

•4F1 = F2

Definition of the Ampere

The ampere is that steady current which, flowing in two infinitely long straight parallel conductors of negligible cross-sectional area placed 1 m apart in a vacuum, causes each wire to exert a force of 2 x10-7 N on each metre of the other wire.

€

F

L=

μ0I ′ I

2πr

4π ×10−7 ×1×1

2π ×1= 2 ×10−7

A current of one ampere carries a charge of one coulomb per second

Definition of the Coulomb

Review and SummaryAmpere’s Law

For current distributions involving a high degree of symmetry, Ampere’s law

∫ = IldB 0. μ

can be used (instead of the Biot-Savart law) to calculate the magnetic field where

(i) is the line integral round a closed loop, and

(ii) I is the current enclosed by the loop

∫ ldB.

Maxwell’s Equations – the story so far!

Laws of Magnetostatics

∫ =

=∫

enclosed

surfaceclosed

IldB

AdB

0

.

0.

μ

Laws of Electrostatics

∫ =

=∫

0.

.0

ldE

qAdE enclosed

surfaceclosed ε

For E and B fields that do not vary with time

Exercise: The diagram shows two currents associated with infinitely long wires, one current of 8 A into the screen, the other current is 8 A out of the screen. Find

∫ μ= Ild.B 0 for each path indicated.

μ0 8 A-μ0 8 A (field is opposite to the line of integration)

0

€

B.dl = μ0I∫ = μ0 × 8C1

Can we obtain B from this?

C2:

€

B.dl = μ0I∫ = 0Is B=0 everywhere along the path?

Exercise: Two straight rods 50 cm long and 1.5 mm apart carry a current of 15 A in opposite directions. One rod lies vertically above the other. What mass must be placed on the upper rod to balance the magnetic force of repulsion.

kg 3

0

1053.1m

r2

LIImg

−×=π′μ

=

i.e. the magnetic force between two current-carrying wires is relatively small, even for currents as large as 15 A separated by only 1.5 mm

An extended,but brief, summary

20 ˆ

4 r

rvqB

∧=

πμ

20 ˆ

4 r

rlIdBd

∧×=

πμ

r

IB

πμ2

0=∫ = IldB 0. μ

€

rF m = q

r v ×

r B

€

φB = B.d Asurface

∫ = 0

BlIF ∧=

BAI ∧=τB∧=μτBU .μ−=

m 1

A 1

N 102 7

==

=′=

×= −

Lr

II

FL

r

IIF

πμ20 ′

=

-1-2270 m C s J 104 −×= πμGiving

A long straight cylindrical conductor has an outer radius b and a hollow core of radius a. The current density varies with radius r for a r b as j = j0r/a where j0 is a constant.

Find an expression for the total current flowing in the conductor. Find expressions for the strength of the magnetic field in the three regions (i) r < a, (ii) a r b and (iii) r > b.

( )

[ ] 033

0

0

3

22 jab

ardr

a

rjI

a

rjrj

b

a

−π

=π=

=

∫

For r < a, B = 0 as no current linked

 a r b

( ) [ ]

( ) [ ]3300

30000

1

3

3

222

arra

jrB

ra

jrdr

a

rjrrB

r

a

−μ

=

πμ=πμ=π ∫

r > b

[ ]

[ ]3300

33000

1

3

3

2

22

abra

j

abar

j

r

IB

−=

−==

μ

π

π

μ

π

μ

The Relationship Between μ0 and ε0

-1-2270 m C s J 104 −×= πμ

-12-1120 m C J 10854178.8 −×=ε

(by definition)

(by experiment)

2-216

00

s m 10987551788.81

×=εμ

1-8

00

s m 1099792458.21

×=εμ

00

1

εμ=∴c

Next Section of Course:

Dealing with time varying electric and magnetic fields

Ampere’s Law

What about ?∫ ldB.

∫ =0. ldE

For electrostatic electric fields

In fact 0. ≠∫ ldBBut B.dl does not represent work done!

∫ =surfaceclosed

AdB

0.From Gauss’s Flux law for magnetism

Static E fields are conservative

James Clerk Maxwell Sir Isaac Newton

Jean-Baptiste Biot André Marie Ampère

Ludwig Boltzmann