8.022 (E&M) – Lecture 13
Topics:
„
B’s role in Maxwell’s equations
„
Vector potential
„
Biot-Savart law and its applications
What we learned about
magnetism so far…
„
Magnetic Field B
„
„
„
Experiments: currents in wires generate forces on charges in motion
Force exerted on charge q with velocity v:
„
Explanation: there must exist a magnetic field B
v
×B
c
Special Relativity: B is just E seen from another reference frame…
Ampere’s Law:
∫ B i ds =
C
„
F = q
„
4π
I encl
c
Application: B generated by current in a wire:
B=
2I
ϕ̂
cr
1
Divergence of B
„
„
2I
Consider the B produced by a wire of current: B = ϕˆ
cr
Calculate its divergence in Cartesian coordinates:
Given r = x 2 + y 2 and ϕˆ = yˆ cos ϕ - xˆ sin ϕ =
B=
„
„
2 I ⎛ xyˆ
yxˆ ⎞
⎜
⎟ ⇒
cr ⎝ x 2 + y 2 x 2 + y 2 ⎠
∇i B =
xyˆ
x2 + y2
-
yxˆ
x2 + y2
⇒
2 I ⎛ 2 yx
2 xy ⎞
⎜
⎟ =0
cr ⎝ ( x 2 + y 2 ) 2 ( x 2 + y 2 ) 2 ⎠
This is a general property of the magnetic field: ∇i B = 0
Similar equation for E: ∇i E = 4πρ
„
„
The divergence of E is related to the density of electric charges
The divergence of B must be related to the density of magnetic charges
Æ Magnetic monopole don’t exist
(There may be magnetic monopoles leftover from the Early Universe, but never observed
experimentally so far)
G. Sciolla – MIT
8.022 – Lecture 13
3
Ampere’s law in differential form
„
Apply Stoke’s theorem to Ampere’s law:
∫ Bids =
C
4π
I encl
c
∫ Bids = ∫ ∇ × BidS =
S
C
⎛
∫ ⎜⎝ ∇× B −
S
4π
c
∫
S
J idS
4π ⎞
J ⎟idS = 0 for any surface
c ⎠
Æ Ampere’s law in differential form: ∇ × B =
G. Sciolla – MIT
8.022 – Lecture 13
4π
J
c
4
2
Toward Maxwell’s equations
„
Let’s collect all the equations in differential form that we
found so far:
⎧ ∇ i E = 4πρ
⎪
⎪⎪ ∇ i B = 0
⎨∇ × E = 0
⎪
⎪ ∇ × B = 4π J
⎪⎩
c
„
Å Relates E and charge density (ρ) - Gauss
Å No magnetic monopoles! Å E is a conservative field Å Relates B and its sources (J) - Ampere
Not complete Maxwell’s equations yet, but we are getting
closer…
G. Sciolla – MIT
8.022 – Lecture 13
5
Vector potential A
„
Definition of potential for electric field:
„
„
„
„
φ(P) = work needed to move a unit charge from reference to P
Relationship between φ and E: E = − ∇ φ
Hidden advantage:
If E=-∇φ ⇒ ∇ × E ≡ 0 because ∇ × (∇φ ) = 0 ∀φ
Can we introduce something similar for B?
„
„
Goal: enforce div B=0
Since ∇i∇ × f = 0 for any f, we define
B ≡ ∇×A
„
„
A is called “vector potential” in analogy with φ
A is not connected to work or energy (but to angular momentum)
G. Sciolla – MIT
8.022 – Lecture 13
6
3
Non Uniqueness
„
„
Electrostatics: given a charge distribution and boundary conditions Æ
potential φ is uniquely identified
Magnetism: does it work the same for A? No, there are infinite
number of A corresponding to a single B
„
Example: B = B 0 zˆ. Find A that creates this B field.
Q: what current creates this B?
Requirements:
⎧
∂A z ∂A y
−
=0
⎪B x =
∂y
∂z
⎪
⎪
∂A x ∂A z
−
= 0 ⇒ Possible solutions:
⎨B y =
∂z
∂x
⎪
⎪
∂A x ∂A y
−
= B0
⎪B z =
∂y
∂x
⎩
„
⎧
A = −yB 0 xˆ
⎪
A = xB 0 yˆ
⎪
⎪
⎨
B
⎪ A = 0 (−yxˆ + xyˆ)
2
⎪
⎪⎩ A = ...infinite others!
We are given one “coupon” to simplify equations when needed
G. Sciolla – MIT
8.022 – Lecture 13
7
Poisson’s equation for A
„
Electrostatics:
⎧⎪E = −∇φ
⇒ ∇2φ = −4πρ Poisson's equation
⎨
⎪⎩∇iE = 4πρ
„
Magnetism:
⎧B = ∇ × A
⎪
⎨
4π
J
⎪∇ × B =
c
⎩
⇒
∇×∇×A =
4π
c
J ⇒ ∇ (∇ i A ) − ∇ 2 A =
We used the identity: ∇ × ∇ × A = ∇ (∇ i A ) − ∇ 2 A
„
Use your coupon now!
Choosing ∇i A=0 ⇒
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∇2 A = −
8.022 – Lecture 13
4π
c
(Pset#7)
4π
J
c
/
J
8
4
Solving Poisson’s equation for A
H o w d o y o u s o lv e ∇ 2 A = −
4π
c
J ?
4π
⎧ 2
⎪∇ Ax = − c J x
⎪
4π
⎪ 2
J
⎨ ∇ AY = −
c Y
⎪
⎪ 2
4π
⎪ ∇ AZ = − c J Z
⎩
T h i n k o f it i n c a rte s i a n c o o rd i n a te s :
R e m e m b e r P o i s s o n 's e q u a t i o n ∇ 2 φ = − 4 πρ
a n d its s o l u t i o n φ =
ρ
∫r
dV
V
S a m e a s o u r n e w e q u a t i o n i f re p la c e φ → A a n d ρ →
A =
F o r c u rre n t f l o w i n g i n a w i re :
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I
c
∫
w ire
J
1
⇒ A =
c
c
J
∫
V r
dV
dl
r
9
8.022 – Lecture 13
Biot-Savart Law
Find B produced from current know ing that A =
B= ∇ × A= ∇ ×
I
c
∫
wire
dl
I
=
r
c
∫ ∇×
wire
I
c
∫
wire
dl
.
r
dl
r
Using the fact that ∇ × (ab)=a( ∇ × b)+( ∇ a) × b:
=
⎤ I
1
I ⎡ 1
( ∇ × dl ) + ∇ × dl ⎥ =
⎢ ∫
c ⎣wire r
r
⎦ c
Since ∇×d l=0 and ∇
1
r
=−
⇒
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1
∫
wire r
(∇ × dl ) + ∇
1
r
× dl
rˆ
:
r2
B =
I
c
∫ dl
wire
8.022 – Lecture 13
×
rˆ
r2
10
5
Biot-Savart Law: illustration
„
Biot-Savart:
dB =
I
rˆ
dl × 2
c
r
dl
rˆ
„
„
dB
dB is perpendicular to current and to radial direction
E.g.: if you have dl // x, r // y Æ B//z
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11
8.022 – Lecture 13
z
Application of Biot-Savart:
B from loop of current
„
Calculate B created by a loop of current
„
„
„
x
y
90-θ
Radius: R
Distance from center of the loop: z
Solution on axis
„
„
„
B=
Apply Biot-Savart
Determine direction of dB
Symmetry Æ only component // z survives
∫ ( dB )
w ire
z
=
I
∫
w i re cr
2
dl × rˆ sin θ
dl × rˆ = dl = R d ϕ ; sin θ = R / r ; r =
B =
2π
2 π IR 2
I
ˆ
ˆz
sin
z
=
R
d
θ
ϕ
2
2
2 3/2
∫0
) 13
+ –zLecture
crSciolla
c (R8.022
G.
– MIT
R2 + h2
⇒
B loop
c ente r
=
2π I
zˆ
c R 12
6
Application of Biot-Savart:
B from solenoid
„
„
What if we stack a N rings over a length L?
Use result of single loop + superposition:
2πR 2
Single ring: dB =
dI
c (R2+z2 )3/2
Integrate on all rings (in the middle of the solenoid)
2πR 2
2π nI
nIdz =
-L /2 c (R 2 + z 2 )3/2
c
B =∫
=
„
L /2
2π nI
2L
c
L + 4R 2
R 2dz
∫-L /2 (R 2 + z 2 )3/2
L /2
With n=N/L
2
For L>>R: B = 4πnI
c
G. Sciolla – MIT
8.022 – Lecture 13
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Solenoid and Ampere’s law
„
„
One can prove that B outside the solenoid is =0
Ampere can be used to simply prove that B does not
depend on r:
∫
rectangle
B idl =
4π
c
I encl
Since B is //z and present only inside the solenoid:
4π
4π N
4π
I = nI no dependence on R
B(r)L= NI ⇒ B (r ) =
c
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c L
c
8.022 – Lecture 13
14
7
Solenoid’s magnetic field: demos
„
Expected:
B = 4π nI
c
I
I
„
Can we test this experimentally?
„
„
„
G12: B from a single wire using iron filings
G13: B from 2 wires
G16: B inside solenoid
G. Sciolla – MIT
8.022 – Lecture 13
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More demos on magnetic fields
„
More demos:
„
„
„
G14: map B around a wire using a compass
G9a: collapsing solenoid
„ Can you explain what’s happening?
G18: Long solenoid
„ Long solenoid with N
turn=2760, I=4.5 mA, length = 46 cm
„
„
B =
„
(R=10 Ω, L=128 mH)
What is B?
4π
c
nI =
4π 2760
4.5 = 230i10 −8 Gauss ???
3i1010 50
Verify with Hall probe
G. Sciolla – MIT
8.022 – Lecture 13
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8
Thompson’s experiment: variation
„
„
Variation on a theme: instead of canceling effects of E and
B, one could tune the fields and measure the radius of
curvature of the electron beam.
Parameters of the problem:
„
„
„
„
V= 300 V
I= 1.4 A
R= 5 cm
Solution:
„
e/m= 2.02 x 1011 C/Kg (cfr: 1.76 x 1011 C/Kg)
G. Sciolla – MIT
8.022 – Lecture 13
17
Summary and outlook
„
„
Today:
∇iB = 0
„
Toward Maxwell’s equations:
„
Vector Potential:
B ≡ ∇×A
„
Biot-Savart Law:
dB =
and
∇× B =
4π
J
c
I
rˆ
dl × 2
c
r
Next time:
„
What happens when B varies in time?
„ Faraday’s and Lenz’s laws and their applications
G. Sciolla – MIT
8.022 – Lecture 13
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