Download PowerPoint

We’ve been working with the effects of magnetic fields
without considering where they come from. Today we
learn about sources of magnetic fields.
Today’s agenda:
Magnetic Fields Due To A Moving Charged Particle.
You must be able to calculate the magnetic field due to a moving charged particle.
Biot-Savart Law: Magnetic Field due to a Current Element.
You must be able to use the Biot-Savart Law to calculate the magnetic field of a currentcarrying conductor (for example: a long straight wire).
Force Between Current-Carrying Conductors.
You must be able to begin with starting equations and calculate forces between currentcarrying conductors.
Magnetic Field of a Moving Charged Particle
Let’s start with the magnetic field of a moving charged particle.
r
It is experimentally
observed that a moving
point charge q gives rise
to a magnetic field.
B
r̂
+
v
μ0 qv  rˆ
B=
.
2
4π r
0 is a constant, and its value is
0=4x10-7 T·m/A

Remember: the direction of r is always from the source point (the thing that
causes the field) to the field point (the location where the field is being measured.
cross products of unit vectors
We are going to be doing lots of cross products of unit vectors.
Here are some handy ways to do the cross product.
The “always works if you can do math” method
C = A B
Example:

 ˆi


C = det  A x

B

 x
ˆj
Ay
By

 ˆi ˆj
ˆk  ˆj = det  0 0


 0 -1


 
kˆ  

Av z  
B z  

kˆ  
 ˆ
1   = i  0 -  -1   = ˆi
0  

cross products of unit vectors
We are going to be doing lots of cross products of unit vectors.
Here are some handy ways to do the cross product.
The right hand rule method
link to image
http://en.wikipedia.org/wiki/Cross_product
cross products of unit vectors
We are going to be doing lots of cross products of unit vectors.
Here are some handy ways to do the cross product.
The “I learned this decades ago and forgot the name for it”
method
i j k i j k
ˆi  ˆj =kˆ
cross products of unit vectors
i j k i j k
ˆj  ˆi = -kˆ
i j k i j k
ˆi  kˆ = -jˆ
i j k i j k
 
ˆj  kˆ = - -iˆ = ˆi
Example: proton 1 has a speed v0 (v0<<c) and is moving along
the x-axis in the +x direction. Proton 2 has the same speed and
is moving parallel to the x-axis in the –x direction, at a distance
r directly above the x-axis. Determine the electric and magnetic
forces on proton 2 at the instant the protons pass closest to
each other.
y
This is example 28.1 in your text.
FE
The electric force is
v0
1 q1q2 ˆ
FE =
r
2
4  r
E
r
1
2
1 e ˆ
FE =
j
2
4  r
Homework Hint: this and the next 3 slides!
2
z
r̂
v0
x
Alternative approach to calculating electric force. This is
“better” because we use the concept of field to calculate both
of electric and (later) magnetic forces.
At the position of proton 2 there is an electric field due to
proton 1.
y
1 q1 ˆ
1 eˆ
E1 =
r=
j
2
2
4   r
4   r
FE E
v0
This electric field exerts a force
on proton 2.
1 eˆ
1 e2 ˆ
FE = qE1 = e
j=
j
2
2
4  r
4  r
2
r
1
z
r̂
v0
x
To calculate the magnetic force: at the position of proton 2
there is a magnetic field due to proton 1.
 q1 v1  rˆ
B1 =
4 r 2
 ev 0 ˆi  ˆj
B1 =
4 r2
y
FE
v0
  ev 0 ˆ
B1 =
k
2
4 r
2
B1
r
1
z
r̂
v0
x
Proton 2 “feels” a magnetic force due to the magnetic field of
proton 1.
FB = q2 v2  B1
FB = ev 0
 
   ev 0 ˆ 
ˆ
i  
k
2
 4 r

y
FE
v0
 e2 v 02 ˆ
FB =
j
2
4 r
2
FB
B1
What would proton 1 “feel?”
r
1
Caution! Relativity overrules Newtonian mechanics!
However, in this case, the force is “equal & opposite.”
z
r̂
v0
x
Both forces are in the +y direction. The ratio of their
magnitudes is
   e 2 v 02 

2 
4

r
FB 

=
FE  1 e 2 

2 
4

r



y
FE
FB
=     v 20
FE
v0
2
FB
B1
Later we will find that
r
1
1
 = 2
c
z
r̂
v0
x
FB v 20
Thus
= 2
FE c
If v0=106 m/s, then
10 

FB
-5
=

1.11

10
FE  3 108 2
6 2
y
FE
Don’t you feel sorry for the poor,
weak magnetic force?
v0
What if you are a nanohuman, lounging on proton 1. You
rightfully claim you are at rest. There is no magnetic field
from your proton, and no magnetic force on 2.
B1
If you don’t like being confused,
Another nanohuman riding on proton 2 would say “I am
close your eyes
at rest, so there is no magnetic force on my proton,
and iscover
your
even though there
a magnetic
fieldears.
from proton 1.”
This calculation says there is a magnetic field and force.
Who
is here,
right?
Take
Physics
2305/107
totolearn
Or see
here,
and here
for a hint
about how
resolvethe
the answer.
paradox.
2
FB
r
1
z
r̂
v0
x