1820 Hans Christian Oersted - Physics and Astronomy...

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Transcript of 1820 Hans Christian Oersted - Physics and Astronomy...

1820 Hans Christian Oersted

Hans Christian Ørsted

Magnetic Field Produced by a Moving Charge

Magnetic Field Produced by a Moving Charge

03

( )4

q v rBr

μπ

×=

r rr

Most convenient view for determining the direction of the magnetic field

produced by a moving charge.

Warning: don’t confuse unit vector

||/ˆˆ4

12

00

rrrrq

qFE rrr

r===

πε

rr rr vector with||/ˆˆ rrr =

Magnetic Field Produced by a Moving Charge

03

( )4

q v rBr

μπ

×=

r rr

Most convenient view for determining the direction of the magnetic field

produced by a moving charge.

Battery A

Units

2

0 0

1cμ ε

=

70

7 2 2

7 2

4 10 /

4 10 /4 10 /

T m A

N s CN A

μ π

π

π

= × ⋅

= × ⋅

= ×

7 2 2 2

0

1 (10 / )4

N s C cπε

−= ⋅

permeability of the vacuum

sec)//(][ meterCoulombNewtonsB ⋅=

mANsmCNteslaT ⋅=⋅= /1//1)(1

03

( )4

q v rBr

μπ

×=

r rr

BvqFrrr

×=

Forces Produced by Moving Charges

Force generated by magnetic interaction for oppositely moving equal-sign charges has repulsive sign.

22

0 0 2B

E

F vvF c

μ ε= =

Congratulation: we arrive to the idea of relativity!

Biot-Savart Law

Infinitesimally small element of a current carrying wire produces an infinitesimally small magnetic field

0μ is called permeability of free space

0 03 3

( ) ( )4 4

q v r I d l rB dBr r

μ μπ π

× ×= ⇒ =

r rr rr r

dQ nqSdl vdQ vnqSdl Idl= ⇒ = =r r uur

70

7 2 2

7 2

4 10 /

4 10 /4 10 /

T m A

N s CN A

μ π

π

π

= × ⋅

= × ⋅

= ×

once again vector’s direction is delegated from v to dl

( )

( )

03

03/ 20 2 2

/0

3/ 20 2

max 2 2

02 2

0max

( )4

24

24 1

/ sin

24

1 tan2

a

z

a x

z

I d l rdBr

I xB dyx y

I dzx z

az y x tgx a

I aBx x a

IR

μπμπ

μπ

α α

μπ

μ απ

×=

= − =+

= −+

= = =+

= −+

= −

r rr

Magnetic Field Produced by a Current

y x z× = −

( )

( )

2 2 3/ 20

3/ 20 2 20

/

3/ 20 20

max 2 2

2 2 2 20 0

4 ( )2

4

24 1

sin

1 2 14 4

x

a

x

a x

x

xdyd Ex y

xE dyx y

dzz

az tgx a

a QEx x a x x a

λπελπε

λπε

α α

λπε πε

=+

= =+

=+

= =+

= =+ +

ur

Electric Field Produced by a Line of Charge, Chapter 21,Example 21.11

Magnetic Field Produced by a Linear Current:again Flatlandia

0| |2

IBr

μπ

=02rEr

λπε

=

E versus Bsimilarity: axial symmetry in both casesdifferences: 1) E is radial; B is circular2) instead of Gauss’s law -- Ampere’s law

Flatlandia: when dependence on one of the coordinates drops out; cylinders or a world confined/reduced to a plane

Magnetic Field Produced by a Straight Current Carrying Wire:

0| |2

IBr

μπ

=

70

7 2 2

7 2

4 10 /

4 10 /4 10 /

T m A

N s CN A

μ π

π

π

= × ⋅

= × ⋅

= ×

70 ( ) 4 10| |

2 ( ) 2I I Amper T mBr r meter Amper

μ ππ π

−× ⋅= = ×

Magnetic Field of Two Parallel Wires with Oppositely Directed Currents

Principle of superposition:Magnetic field created by few current-carrying wires should be summed according to the vector summation rules.

Forces between Two Parallel Wires with Oppositely Directed Currents :

Force per unit length:

0 1 2

2F I IL r

μπ

=

Currents are not like charges:Oppositely directed currents repel each other!Same directed currents attract each other.

7 22 74 10 // (1 ) 2 10 /

2 1N AF L A N mm

ππ

−−×

= = ××

Example 28.5

Forces between Two Parallel Wires

0 1 2

7 22

3

4

2

4.5 15,0004 10 // (15,000 )2 4.5 10

10 /

F I IL r

r mm I AN AF L A

mN m

μπ

ππ

=

= =

×=

× ×=

Pulse currents break magnets!

Electromagnets (most of magnets used in industry are the electromagnets)

An electromagnet contains a current carrying wire with numerous turns

Magnetic Field along the Axis of a Current Loop

0 03 2 2 2 2 1/ 2

20

2 2 3/ 2

2 27

2 2 3/ 2 3

( )4 4 ( ) ( )

2 ( )( )2 10 ( ) /

( ) ( )

xI d l r I d l adB dB

r x a x aIaB

x aIa mampers T m amper

x a m

μ μπ π

μ

π −

×= ⇒ =

+ +

= =+

= × ⋅+

r rrr

Charge Q

Current I

Electro-magnet: Magnetic Field created by an N-loops coil

20

2 2 3/ 2

0 magnet02 2 3/ 2 2 2 3/ 2

2 ( )

2 ( ) 2 ( )

NIaBx aINS

x a x a

μ

μ μμπ π

= =+

= =+ +

magnetitem item magnet

potential energy ( )

cos

magnetic

U B Bμ μ φ= − ⋅ = −ur ur

magnetic field created by the current magnetizes iron items

Gauss’s law for Magnetism

magnetic charges do not exist

magnetic lines have no ends

∫ =⋅S

enclosedQSdE0ε

rr

0∫ =⋅ SdBrr

Some magnetic field lines remind electric dipole. Where is the difference?.

Do we have something instead of the Gauss’s law? Ampere’s law

∫contour

0B dl Iμ=ur r

orientation of the contour !

Caution: Sign of the currents enclosed by the contour are determined by the orientation of the contour.

For this orientation of the contour (anticlockwise), currents I1 and I3 are positive while I3 is negative.

Contour may be not in plain, and currents may be not perpendicular to the contour plane.

Ampere’s law (Example 28.8)

Most effective when it is applied for symmetrical configurations:

cylindrical conductor ∫contour

0B dl Iμ=ur r

Ampere’s law (Example 28.9)

Most effective when it is applied for symmetrical configurations:

solenoid

Find similarity between solenoid and capacitor

∫contour

0B dl Iμ=ur r

∫contour

0

0

B dl BL nLI

B nI

μ

μ

= =

=

ur r

Ampere’s law (Example 28.10)

Most effective when it is applied for symmetrical configurations:

Toroidal solenoid∫

contour0B dl Iμ=

ur r

∫contour

0

0

2

2

B dl rB NI

NIBr

π μ

μπ

= =

=

ur r

What information (may be questions for the Exam) produces this picture?

Magnetic field produced by a clock-like current in the coils (look on the picture from the left side) is directed from left to right. A positive charge responding to the magnetic field rotates counter-clock wise ( i.e., opposite to the current which induces this magnetic field). Magnetic field produced by the rotating charge tries to compensate the original magnetic field.

Current-loops with oppositely directed, i.e., anti-parallel, currents repulse. (Current loops with parallel currents attract.) This is why rotating charge repels from coil 1 and goes toward the middle of the magnetic bottle.Check that you fully understand all these statements.