PS 250: Lecture 19 Magnetic Fields due to Currents

J. B. SnivelyOctober 19th, 2015

Today’s Class

Magnetic Field of a Moving ChargeMagnetic Field of a CurrentMagnetic Force on Parallel ConductorsSummary

B-Field of Moving Charge

“Mu-sub-zero” is the permeability of free spacer-hat is the unit vector pointing along vector rDecreases in amplitude with 1/r2Field lines curve around moving charge

⇥B =µo

4�

q⇥v ⇥ r̂

r2⇥B =

µo

4�

q⇥v ⇥ ⇥r

r3or

B-Field of Moving Charge

Permeability of Free Spacek =

1

4⇥�o

= 10�7[N · s2/C2]c2[m/s]

µo

=1

�o

c2= 4⇥ � 10�7[T ·m/A]

Since 1983, the meter is defined in terms of speed of light, thus speed of light can be specified exactly!

c =1

p�o

µo

= 2.99792458⇥ 108 ' 3⇥ 108[m/s]

Today’s Class

Magnetic Field of a Moving ChargeMagnetic Field of a CurrentMagnetic Force on Parallel ConductorsSummary

B-Field of an Infinitesimal Segment of Current

d ⇥B =µo

4�

Id⇥l ⇥ r̂

r2

Can integrate over all current-carrying segments:

⇥B =µo

4�

ZId⇥l ⇥ r̂

r2

“Biot-Savart Law”

B-Field of an Infinitesimal Segment of Current

B-Field due to Current in a Conductor

a

-a

r =p

x

2 + y

2

x

y

d�lr̂

d �Bx

B-Field is “into the page”

B =µo

I

2�r

Best to do the math on the

board...

For Infinite (Really Long) Wire:

Today’s Class

Magnetic Field of a Moving ChargeMagnetic Field of a CurrentMagnetic Force on Parallel ConductorsSummary

Force Between Parallel Conductors

B =µo

I

2�r

Given the B-field due to a current-carrying wire:

Force on wire with Current I’: �F = I 0�L⇥ �B

F = I 0LB =µo

II 0L

2�r

F

L=

µo

II 0

2�r

II’ F

Force Between Parallel Conductors

II’ F

II’ F

On Board, let’s Consider the R-H-R

.B due to I

xB due to -I

Definition of the Ampere

Two Infinitely-Long Parallel Conductors,Separated by 1 Meter,

Each carrying current 1 Ampere,Each experience exactly 2x10-7 N

of Magnetic Force.(Plug this into Force Equation!)

II’ F

.B due to I

Summary / Next Class:

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