Download Sources of magnetic field

Sources of magnetic field
Lots of details so far on how Magnetic Fields exert forces.
But, where did those Magnetic Fields come from?
B-fields are created by moving charges (currents).
This seems quite opposite to what we might expect!
Field created by a moving charge
Magnetic Field
r created by a single moving charged particle at
a position r
v µ 0 qvv × rˆ
B=
4π r 2
Constant µ0 = 1.257 x 10-6 Tesla meter/Amp.
r̂
r
is a vector in the direction of r , but with magnitude = 1
(often referred to as a unit vector). r̂ direction is from charge
to where you are evaluating B !
Field created by a moving charge
v
µ qv × rˆ
B= 0
4π r 2
v
Clicker Question
r
Point P
A new Right Hand Rule…
Thumb in direction of the motion, and fingers
then curl in direction of the B-field.
v µ 0 qvv × rˆ
B=
4π r 2
What is the direction of the Bfield at point P indicated?
A) Up and to the Left
B) Down and to the Left
C)Out of the Page
D)Into the Page
E) None of the Above
Example: Forces between two moving protons
r
1 e2
FE =
yˆ
4πε 0 r 2
v
µ ev × rˆ µ0 ev
=
zˆ
B= 0
4π r 2
4π r 2
v
r
r
r
FB = e( −v ) × B =
µ0 ev ⎞
zˆ ⎟ =
2
r
π
4
⎝
⎠
⎛
e( −vxˆ ) × ⎜
µ0 ev
µ0 e 2 v 2
( xˆ × zˆ ) =
e( − v )
yˆ
4π r 2
4π r 2
Magnetic field of a current element
Now the B-field created by many moving charges (i.e. current
flowing in a wire).
r
v
dB =
I
dl
r
X dB
µ0 dQvd × rˆ
=
r2
4π
r
µ0
v × rˆ
( nqAdl ) d 2 =
4π
r
r
µ0 I dl × rˆ
4π r 2
Biot-Savart first discovered this
experimentally.
Magnetic field of a current element
r
µ I dl × rˆ
dB = 0
4π r 2
v
I
Magnetic field of a current-carrying wire
v
µ I dL × rˆ
dB = 0
4π r 2
v
v
µ I dL × rˆ
Btot = ∫ dB = 0 ∫ 2
4π
r
v
v
This can be a very difficult integral to evaluate.
Magnetic field from an infinite straight wire
v
µ I dL × rˆ
dB = 0
4π r 2
v µ I dy
dB = 0 2 sin θ (into the page)
4π r
v
r = y +x
Now a little
geometry sin θ = x / r = x / x 2 + y 2
2
v
dB =
2
2
µ0 I
xdy
4π (x 2 + y 2 )3 / 2
Magnetic field from an infinite straight wire
(cont.)
v
dB =
v
v
µ0 I
xdy
4π (x 2 + y 2 )3 / 2
B = ∫ dB =
v
B=
+∞
µ0 I
xdy
2
4π ( x + y 2 ) 3 / 2
−∞
∫
µ0 I
(into the page)
2πx
Magnetic field from an infinite straight wire
(cont.)
This is a key result!
v µ0 I
| B |=
2πR
B-field a perpendicular distance
x away from an infinite (or very
long) wire.
Clicker Question
A long wire has a current moving as shown.
What is the direction of the B-field created by the wire just
below the wire?
L
A) Into the Page
B) Out of the Page
C)To the right
D)Down
E) None of the Above
i
r
B=?
v
µ i dL × rˆ
dB = 0
4π r 2
v
Clicker Question
A long wire has a current moving as shown.
What is the direction of the B-field created by the wire just
above the wire?
B=?
A) Into the Page
B) Out of the Page
C)To the right
D)Down
E) None of the Above
r
L
i
v
µ i dL × rˆ
dB = 0
4π r 2
v
Example with numbers
Power line has 500 Amps going through it.
What is the B-field strength 15 meters below
on the ground?
(1.26 × 10-6 Tm / A)(500 A)
µ0 I
B=
=
= 6.7 × 10−6 T
2πR
2π (15m )
Interaction between two current carrying wires
The two wires may exert forces on each other through Magnetic Interactions.
I
I’
• Think of the Red Wire as creating
a B-field.
• Then think of that B-field creating
a force on the moving charges
(current) in the Blue Wire.
Interaction between two current carrying wires
v
v
v
r
µ0 I
rˆ
2πr
F = I'L× B
B=
F = I ' LB =
µ0 II ' L
2πr
Clicker Question
I
I
What is the direction of the Force
acting on the Blue Wire?
v
v
v
F = IL × B
A) Up
B) Right
C)Left
D)Into the Page
E) Out of the Page
i1
i2
The B-field from the Red Wire at the
location of the Blue Wire is into the
page.
v
µi
B1 =
X
2πR
(into the page)
Then the Force on the Blue Wire is to
the left.
v
R
01
v
v
F2 = i2 L × B1 = i2 L
µ0i1
(left)
2πR
Clicker Question
i
i
What is the direction of the Force
acting on the Red Wire?
v
v
v
F = IL × B
A) Up
B) Right
C)Left
D)Into the Page
E) Out of the Page
i1
i2
The B-field from the Blue Wire at the
location of the Red Wire is out of the
page.
v
µi
B2 =
0 2
2πR
(out of the page)
Then the Force on the Red Wire is to
the right.
.
v
R
v
v
F1 = i1 L × B2 = i1 L
µ 0i2
(right)
2πR
Interaction between current-carrying wires
i
i
Wires with parallel currents attract each other.
i
What happens if we flip the direction of one current?
i
Wires with antiparallel currents repel each other.
Try following the procedure we just outlined to
confirm this for yourselves.
Field through a circular loop
What about the B-field at the center of a circular loop of wire
of radius a and current I?
dB =
dl
µ0 I dl µ0 I
=
2
2
4π r
4π ( x + a 2 )
dBx = dB cos θ =
Bx = ∫ dBx = ∫
µ0 I
adl
2
4π ( x + a 2 )3 / 2
µ0 I
µ0 Ia
adl
=
dl
2
2 3/ 2
4π ( x + a )
4π ( x 2 + a 2 )3 / 2 ∫
Bx =
µ0 Ia 2
2( x 2 + a 2 ) 3 / 2
Field through a circular loop
Clicker Question
Which point A or B has the larger magnitude Magnetic Field?
I
I
B
A
r
µ I dl × rˆ
dB = 0
4π r 2
v
A
B
C : The B-field is the same at A and B.
Answer: Case B has the larger magnetic field. Use the
Biot-Savart Law to get the directions of the B-field due to
the two semi-circular portions of the loop. In A the two
fields oppose each other; in B they add.
Clicker Question
In the limit as x >> a, the expression
for the B-field becomes?
A) B =
B) B =
C)
µ0
Ia 2
B=
2 ( x 2 + a 2 )3 / 2
µ0 Ia 2
B=?
2 x3
µ0 Ia 2
x
2 (x + a )
2
B=
2
µ0 I
a
2 a
I
Magnetic Dipole moment
v
B=
µ0 Ia 2
2 x3
xˆ
µ Iπa 2
µ IA
B = 0 3 xˆ = 0 3 xˆ
2π x
2π x
v
v
µ = IA
v
B=
B=?
* A = area of loop
Magnetic Dipole Moment
v
µ0 µ
2π x 3
a
I
x
Magnetic Dipole moment
Any current loop looks like a Magnetic Dipole far away.
v µ0 µv
B=
2π x 3
The B-field drops as the distance^3
And depends on the Magnetic Dipole
Moment.
v
v
µ = IA
Clicker Question
A square loop of side length a of wire carrying current I is in a
uniform magnetic field B. The loop is perpendicular to B (B
out of the page). What is the magnitude of the net force on
the wire?
v
v v
A: IaB
F = IL × B
B
B: 4IaB
C: 2IaB
D: 0
E: None of these
I
Clicker Question
The same loop is now in a non-uniform field.
r
B = Bz$ , where B = B(y) = A ⋅ y
where A is a constant. The direction of the net force is?
B stronger
y
B
A
C
x
B weaker
D
E: net force is zero
Uniform fields produce zero net
force!!!
In a uniform B-field, regardless of the orientation between the
B and the Magnetic Moment of the loop µ, the net force is
always zero.
However, that does not mean the net
torque is zero!
Magnetic torque
If µ is not parallel to B, then there is
a net torque.
µ
v
v
v
v
v
τ = ∑r ×F
v
v
τ = ∑ r × ( IL × B ) = 2[( L / 2) I LB sin θ ]
v
τ = IL2 B sin θ
µ
v
v
v
τ = IA × B
v
v v
τ = µ×B
The DC motor
v
τ = µ×B
v
v
Torque wants to twist the
loop so that µ and B align.
Clicker Question
Two loops of wire have
current going around in
the same direction.
The forces between the
loops is:
i2
A)Attractive
B)Repulsive
C)Net force is zero.
i1
Magnetic Field Lines and Flux
Gauss’s Law for Magnetic Fields
Magnetic monopoles (so far) do not exist!!!
There are no sources of magnetic flux
=>Flux through a
closed surface:
r
r
∫ B ⋅ dA = 0
Ampere’s law
Is there something like Gauss’ Law for Electric Fields that
helps us solve for Magnetic Fields in a simpler way for cases
with nice symmetries?
∫
loop
v v
B ⋅ dl = µ0 I thru
For any closed imaginary loop where the current is
constant, the above relation is true.
Clicker Question
∫
loop
v v
B ⋅ dl = µ0 I thru
We need a sign convention for I(thru).
Place the fingers of your right hand around the imaginary loop
and your thumb points in the direction of positive I(thru).
What is I(thru) in the below case where all three wires have 5 A?
A) I(thru) = +15 Amps
X
B) I(thru) = +5 Amps
X
C) I(thru) = -5 Amps
D) None of the above
Ampere’s law for a straight wire
i
Place an imaginary circular loop of radius R
around the wire.
v v
∫ B ⋅ dl
loop
Vector dot product!
Clicker Question
Look at the previous wire from the view where the wire current
is coming out of the page.
The Ampere circular loop is drawn in blue.
What is the relation of the vector dl that
one integrates around the loop as shown
and the Magnetic Field vector?
A)Parallel
B)Anti – Parallel
C)Perpendicular
D)Depends where in the loop for dl
Thus, this is a special symmetry case where
all around the loop the two vectors are
parallel.
v v
B || dl
In fact, also due to symmetry, we know that
the magnitude of the B-field is also constant.
v
| B | constant around the loop
i
dl
i
Ampere’s law for a straight wire
v v
v v v
∫ B ⋅ dl =| B | ∫ dl =| B | 2πa
loop
i
loop
Where a is the radius of the loop.
∫
v
loop
v
B ⋅ dl = µ0 I thru
v
| B | 2πa = µ0 I
v µ0 I
| B |=
2πa
Clicker Question
A long straight copper wire has radius b and carries a
constant current of magnitude i. The current density of
magnitude J=i/(πb2) is uniform throughout the wire. What is
the current contained in the circular loop £, with radius r < b,
centered on the wire's center as shown?
r
A) i
b
3
C) i r 3
b
r2
B) i 2
b
D) None of these.
b
r
J = constant
v v
∫loop B ⋅ dl
Clicker Question
= µ0ithru
A long straight copper wire has radius b and carries a
constant current of magnitude i. The current density of
magnitude J=i/(πb2) is uniform throughout the wire.
How does the magnitude of the B-field a distance r < b from
the center of the wire depend on r?
b
A) B ∝ r
C) B ∝ 1/r
E) None of these
B) B = constant
D) B ∝ 1/r2
J = constant
For r < b:
b
r
∫
loop
r
v
v
B ⋅ dl = µ 0ithru
v
| B | 2πr =
J = constant
v
r
2πb 2
v
µi
| B |= 0
2πr
| B |= µ 0i
For r >= b:
|B|
b
r
2
⎛ r
µ0 ⎜⎜ i 2
⎝ b
⎞
⎟
⎟
⎠
∫
Clicker Question
loop
v
v
B ⋅ dl = µ0ithru
We have determined that for the closed loop as drawn below the
integral on the left is -10 Tesla meters. What is the direction of
the current in the red wire shown below?
A)
B)
C)
D)
E)
Into the Page
Out of the Page
Left
Up
Down
Field of a solenoid
A single wire tightly coiled up into loops.
Since it is a single wire, the current magnitude is the same in
all parts of the coil.
Field of a solenoid
What is the B-field from a solenoid of N turns and length L?
Length = L
Current
Direction
n = N/L
Number of Turns = N
Field of a solenoid
Side View
v
v
B ⋅ dl = µ0 I thru
Zero B Outside
∫
Uniform B Inside
| Binside | l = µ0 ( nlI )
loop
v
# loops in length l
X X X X X X X X
l
v
| Binside |= µ0nI = µ0
N
I
L
B-Field Inside a Solenoid
Clicker Question
Three long straight solenoids all of
length L and all with the same
(large) number of closely-packed
turns N, all with the same current i,
have different cross-sections as
shown.
A
B
C
Which solenoid has the largest field |B| at its center?
A) A
B) B
C) C
D) All three have the same magnitude magnetic field
Answers: The field is the same magnitude and uniform for all
three solenoids. The field within a solenoid is B = µni. This
depends only on the current i and the turns per length n. This
formula does not depend on either the cross-sectional shape
of the solenoid or the position within the solenoid.
Toroidal solenoid
Using Ampere’s Law:
If the “thickness” of the
core is very small:
2πrB = µ0 NI → B = µ0
N
I
2πr
N
≈ n → B = µ0nI
2πr
How does a rail gun work?
v
v
v
F = IL × B
Projectile
Conducting
Gas
X B
I
Rail Guns are by far the most spectacular type of
electromagnetic accelerators ever developed. They hold the
record for fastest object accelerated of a significant mass,
for the 16000m/s firing of a .1 gram object by Sandia
National Research Laboratories' 6mm Hypervelocity
Launcher, and they can also propel objects of very sizeable
masses to equally impressive velocities, such as in the
picture, where Maxwell Laboratories' 32 MegaJoule gun
fires a 1.6 kilogram projectile at 3300m/s (that's 9
MegaJoules of kinetic energy!) at Green Farm research
facility.
Natural magnetism: Where is the
B-field coming from?
v
µ qv × rˆ
B= 0
4π r 2
v
r
µ I dl × rˆ
dB = 0
4π r 2
v
Moving charges (currents) create B-fields.
Where are the moving charges?
Natural magnetism: Where is the
B-field coming from?
Atoms have moving electric charges.
Natural magnetism: Where is the
B-field coming from?
Consider a single electron moving in a circular orbit around
the nucleus.
ev
(1.6 × 10−19 C )( 3 × 106 m / s )
I=
=
= 7.6 × 10−4 A
−10
2πR
2π (10 m)
What is the B-field at the center (at the nucleus)?
v
B=
µ0 I
2R
≈ 5Tesla
Electron in orbit,
but current in
opposite direction.
Natural magnetism
In most materials all the electrons in orbit have random
orientations.
Superposition of B-field vectors over many atoms gives B=0
Ferromagnetism
In Ferromagnetic materials (Fe, Ni, Cr, some alloys
containing these metals too), the atomic currents can all
orient the same way, making a net B-field.
B
B
B
B
Ferromagnetism
Sometimes the material is fragmented into many domains
(top) and is thus unmagnetized. Domains that are
magnetized in the direction of the field grow larger.
Permanent magnets
In all permanent magnets (just like a current loop)
the B-field points away from one end (North)
and towards the other end (South).
Very important convention.
Thus, these magnets act just like a current loop!
i
Clicker Question
What is the current direction
in this loop? And which side
of the loop is the north pole?
i
A. Current clockwise; north pole on top
B. Current counterclockwise, north pole on top
C. Current clockwise; north pole on bottom
D. Current counterclockwise, north pole on bottom
Electron “spin”
Atoms have a Magnetic Dipole Moments from the orbit of the
electrons.
v
v
µ (orbit) = IA
But, amazingly electrons themselves also have a Magnetic
Dipole Moment.
v
µ (electron or spin)
Is the electron really a fundamental point-like particle?
Or does it have little charges moving around inside it?
Or could the electron be a rotating sphere of charge?
Electron “spin”
We know from the world’s best microscopes (particle physics
accelerators) that the electron is less than 10-18 meters
across.
Thus, knowing the Magnetic Dipole Moment, we can
calculate how fast the outside of the sphere must be rotating.
v = 2.5 ×1014 meters/second
Electron “spin”
This velocity is faster than the speed of light, and thus not
allowed in Einstein’s Theory of Relativity.
v = 2.5 × 1014 m/s >> 3 × 108 m / s (speed of light)
Thus, it may be that electrons just have an intrinsic property
of Magnetic Dipole Moment.
This is not understood at any more fundamental level today.
It turns out in permanent magnets, the
atoms have many electrons in orbits with
different orientations.
Thus, it is often the intrinsic magnetic
dipole moment of the electrons that is
important and tends to be aligned into
domains.
i
Clicker Question
A permanent bar magnet is broken in half. Do the pieces
attract or repel?
1
i
2
A: Attract
B: Repel
C: Neither! There is no net force.
i
Answer: They attract. The bar magnet can be thought of as a
coil of current. Parallel currents attract. The currents on the
ends are parallel so there is an attraction.
Clicker Question
A permanent bar magnet is broken in half. The two pieces
are interchanged, keeping their orientations fixed, as shown
below. Do the pieces attract or repel?
2
1
2
A: Attract
B: Repel
C: Neither! There is no net force.
1
Answer: They attract. The bar magnet can be thought of as a
coil of current. Parallel currents attract. When the magnet is
broken and rearranged, the currents on the ends are parallel
so there is an attraction.
Opposite Poles attract.
Similar Poles repel.
This is not new physics. It
is identical to the physics of
our current loops.
Now one can see why we
can never break a magnet
and get just one pole!