Post on 25-Sep-2020
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:
Keep Reading
Mastering Physics for Wednesday
Homework for Friday