Videos
SP212
Ch. 29 - Magnetic Fields due to Currents
Simpsons http://www.lghs.net/ourpages/users/
dburns/ScienceOnSimpsons/Clips.html
10 Minute Compilation http://www.youtube.com/
watch?v=V-M07N4a6-Y&feature=share&list=
PLzQYvo_Tb2BmD-jIOvpTaJag3Lmth0if0
Maj Jeremy Best USMC
Physics Department, U.S. Naval Academy
February 23, 2016
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
1 / 25
The Biot-Savart Law
T m
Tesla meter
=
A
Amp
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
2 / 25
The Biot-Savart Law
In the last lab, we learned that electric currents produce
magnetic fields. Let’s make that statement precise.
First, we formally introduce µ0 , the permeability of free
space.
µ0 = 4π × 10−7
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
In chapter 22, we found a lot of electric
fields by setting up a little bit of electric
field due to a little bit of charge, and
then integrating over all the charge.
Now we’re going to do the same, but
with little bits of current.
~ =
dB
3 / 25
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
µ0 i d~s × r̂r
4π r 2
February 23, 2016
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The Biot-Savart Law
Using Biot-Savart
In the simplest case, an infinitely long
straight wire carrying a uniform current
i, the magnetic field is oriented in a
circle around the wire , pointing in a
direction given by the right hand rule.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
5 / 25
Using Biot-Savart
We play the same game we played in chapter 22:
Write down the general form of the law
Replace the things you don’t know (usually r ,
maybe d~s) with the actual geometry of the problem
Integrate
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
6 / 25
Using Biot-Savart
Z
B=
=
and we see that
p
r = s2 + R2
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
µ0
4π
dB
Z
B=
i ds sin θ
r2
sin θ = √
=
R
s2 + R2
February 23, 2016
7 / 25
=
=
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
Z
µo i ∞
R ds
4π −∞ (s 2 + R 2 )3/2
∞
µo i
s
√
4πR
s 2 + R 2 −∞
µo i
[1 − (−1)]
4πR
µ0 i
2πR
February 23, 2016
8 / 25
An Arc of Current
Force Between Two Parallel Currents
For a semi-circular arc of current
i and radius R, subtending an angle φ, we find that the
field at the center of the arc has magnitude
B=
So if the current in a wire creates a magnetic field , and
the current in another wire is a bunch of moving charges,
two parallel wires should exert forces on each other.
Consider two parallel wires (a and b), carrying currents ia
and ib , separated by a distance d. First, we find the field
produced by a at the location of b.
Ba =
µ0 iφ
4πR
µ 0 ia
2πd
Note that φ must be in radians!
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
9 / 25
Force Between Two Parallel Currents
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
10 / 25
Force Between Two Parallel Currents
That field produces a force on wire b:
We use the right hand rule to find the direction of ~Fba :
If the currents are parallel, the force is attractive
If the currents are anti-parallel, the force is repulsive
~a
F~ba = ib ~L × B
:1
◦
90
Fba = ib LBa
sin
ib Lµ0 ia
=
2πd
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
11 / 25
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
12 / 25
Ampere’s Law
Ampere’s Law
Just like we learned to avoid the brute force integration
of chapter 22 using Gauss’s Law, there is an elegant way
of finding the magnetic field for situations with a great
deal of symmetry: Ampere’s Law.
I
~ · d~s = µ0 ienc
B
Looks like our old friend:
I
~ = (1/0 )qenc
E~ · d A
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
13 / 25
Ampere’s Law
Just like Gauss’s Law, Ampere’s Law is
always true , but it’s only useful when
we can exploit symmetry to evaluate
that integral. Let’s revisit the infinite
straight wire:
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
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February 23, 2016
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Find the Physics
I
µ0 ienc =
~ · d~s
B
I
=
B cos θ ds
I
= B ds
µ0 ienc = B(2πr )
µ0 i
B=
2πr
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
15 / 25
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
The Solenoid
The Solenoid
One of the more commonly encountered magnetic
systems is the solenoid - a long, tightly wound helical
coil of wire.
The solenoid is useful because it produces a very
uniform field inside . The magnetic field of a solenoid
can be found using Ampere’s law to be:
B = µ0 ni
Where n is the number of turns per unit length of the
solenoid
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
17 / 25
Solenoid
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
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The Magnetic Field of a Toroid
A solenoid can be used in your car, to engage the starter
motor, sliding the gear forward to mesh with the teeth of
the flywheel. It can also be used to make a fun (but
dangerous) coil gun.
If we bend a solenoid around until it forms a closed ring
, we have a toroid . The magnetic field inside a toroid
is:
B=
µ0 iN
2πr
Note that the field of a toroid is not uniform through
its cross section, and N is now the total number of
turns, not the turns per unit length.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
19 / 25
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
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Toroidal Magnetic Fields
The Field of a Current-Carrying Loop
A Toroidal magnetic field can be used to contain a
superheated plasma used in a fusion reactor. (for
instance)
Earlier, we showed that a current carrying loop had a
dipole moment and acted like a bar magnet, aligning
with an external magnetic field. There is a torque on it
given by:
~
~ =µ
τ
~ ×B
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
21 / 25
The Field of a Current-Carrying Loop
µ0 µ
~
2πz 3
Where z is the distance from the center of the loop,
measured along the central axis. Don’t confuse
µ
~ = NiA (the dipole moment), with µ0 , the permeability
of free space!
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
February 23, 2016
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Current-Carrying Loop
Also like a bar magnet, we now see that the current
must create its own magnetic field. That field is:
B(z) =
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
23 / 25
There are two ways to look at a current carrying coil as a
magnetic dipole:
1
It experiences a torque when we place it in an
external magnetic field .
2
it generates its own intrinsic magnetic field which
is given by the previous equation for points along its
z axis.
A current carrying loop will rotate in an external
magnetic field just like a bar magnet.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
24 / 25
Wiley Plus Homework
Chapter 29: Questions 4, 7. Problems: 8, 21, 43, 45, 50,
56, 57.
Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
SP212
February 23, 2016
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