Transcript Slide 1

Today’s agenda:
Magnetic Field Due To A Current Loop.
You must be able to apply the Biot-Savart Law to calculate the magnetic field of a current
loop.
Ampere’s Law.
You must be able to use Ampere’s Law to calculate the magnetic field for high-symmetry
current configurations.
Solenoids.
You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and
toroids. You must be able to use the magnetic field equations derived with Ampere’s Law
to make numerical magnetic field calculations for solenoids and toroids.
Magnetic Field of a Current Loop
A circular ring of radius a carries a current I as shown.
Calculate the magnetic field at a point P along the axis of the
ring at a distance x from its center.
Draw a figure. Write
down the starting
equation. It tells you
what to do next.
μ 0 I d  rˆ
dB =
4π r 2
y
I
a
x
x
z
P
Magnetic Field of a Current Loop
A circular ring of radius a carries a current I as shown.
Calculate the magnetic field at a point P along the axis of the
ring at a distance x from its center.
Complicated diagram!
You are supposed to
visualize the ring
lying in the yz plane.
dl is in the yz plane. rˆr
is in the xy plane and
is perpendicular to dl.*
Thus d  rˆ = d .
μ 0 I d  rˆ
dB =
4π r 2
y
I
dl

rˆ
dB
dBy
r
a
90-
x
z

P dBx
Also, dB must lie in the xy plane* (perpendicular to dl) and is
perpendicular to r.
x
*Only when dl is centered on the y-axis!
μ 0 I d  rˆ
dB =
4π r 2
y
I
μ0 I d
dB =
4π r 2
μ0 I d
dB =
4π  x 2  a2 
dl ˆ
r
dB

dBy
r
a
90-
x
z

P dBx
x
μ0 I d
μ0 I d
a
dB x =
cos =
2
2
4π  x  a 
4π  x 2  a2   x 2  a2 1/2
μ0 I d
μ0 I d
x
dB y =
sin =
2
2
4π  x  a 
4π  x 2  a2   x 2  a2 1/2
By symmetry, By will be 0. Do you see why?
Use symmetry to find By. Don’t
try to integrate dBy to get By.
See here for the reason.
y
I dl
rˆ
r
dBy
dBz
x
z
P dBx
When dl is not centered at z=0, there will be a z-component
to the magnetic field, but by symmetry Bz will be zero.
x
y
μ0
Iad
dB x =
4π  x 2  a2 3/2
Bx =
 dB
I
dl ˆ
r
dB

dBy
r
a
x
90-
ring
x
z

P dBx
I, x, and a are constant as you integrate around the ring!
μ0
Ia
Bx =
4π  x 2  a2 3/2
Bx =
μ 0 I a2
2x  a
2

2 3/2
μ0
Ia
ring d = 4π x 2  a2 3/2 2a


This is not on your starting equation sheet. I will add
it if we have homework or a test problem where I
judge that you need it.
x
y
At the center of the
ring, x=0.
I
B x,center =
2
μ0 I a
2 a
dl ˆ
r
dB

r
a

dBy
2 3/2
90-
x
z

P dBx
μ 0 I a2
μ0 I
B x,center =
=
2a3
2a
For N tightly packed concentric rings (a tight coil)…
μ NI
B x,center = 0
2a
This is not on your starting equation sheet, but I will
add it if we have homework or a test problem where
I judge that you need it. For homework, if a
problem requires this equation, you need to
derive it!
x