Ben Gurion University of the Negevwww.bgu.ac.il/atomchip
Week 7. Sources of the magnetic field – Biot-Savart law • parallel currents • Ampère’s law • Gauss’s law for B • towards Maxwell’s equationsSource: Halliday, Resnick and Krane, 5th Edition, Chap. 33.
Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shay Inbar
Physics 2B for Materials and Structural EngineeringPhysics 2B for Materials and Structural Engineering
Biot-Savart law
The last lecture was all about how moving charges respond to magnetic fields. This lecture is all about how moving charges create magnetic fields!
In 1820, the Danish scientist Hans Christian Oersted gave a lecture-demonstration about how an electric current in a wire generates heat. For another part of the demonstration, he had a compass nearby. While lecturing, he noticed to his surprise that every time the electric current was switchedon, the compass needle moved. He said nothingbut, after a few months of experiments, published his discovery (in Latin) with no explanation.
Biot-Savart law
Within months, Jean-Baptiste Biot and Félix Savart had made careful quantitative experiments on the force exerted by an electric current on a nearby magnet. They summarized their results in the law named after them.
ds
dBr
Biot-Savart law
Each current element ds contributes to the magnetic field B(r)
a contribution where μ0 = 4π × 10–7 T·m/A .
“the permeability (פרמאביליות, חלחלות) of free space”
, ˆ
4 20
r
dId
rsB
ds
dBr
Biot-Savart law
Each current element ds contributes to the magnetic field B(r)
a contribution where μ0 = 4π × 10–7 T·m/A .
• The vector dB is perpendicular to ds and to r.• The vector dB is proportional to I, to 1/r2 and to |ds|.• When ds and r are parallel, dB vanishes. • When ds and r are not parallel, dB is perpendicular
to both of them, according to the right-hand rule, and proportional to the sine of the ds → r angle.
, ˆ
4 20
r
dId
rsB
dB
ds
dBr
Right-hand rule (for the Biot-Savart law):
ds
r dB
Biot-Savart law
Example 1: A point makes angles θ1 and θ2 with the ends of a
straight wire carrying current I. The distance of the point from the wire is a. What is B?
ds = dx
dB
a
θ1 θ2x
θ
y
r
Answer: Introduce x- and y-axes. Note that r = (0,a) – (x,0) = (–x, a), that a = r sin θ and thus r = a/sin θ. Note also that x = –r cos θ =
–a cot θ, so dx = a dθ /sin2 θ.
Biot-Savart law
Example 1: A point makes angles θ1 and θ2 with the ends of a
straight wire carrying current I. The distance of the point from the wire is a. What is B?
Answer: The magnetic field is the integral
but dx/r2 = dθ/a, so
sin
4 ˆ
ˆ
4)(
20
20
dx
rI
r
dId
z
rsBrB
ds = dx
dB
a
θ1 θ2x
y
θ
r
Biot-Savart law
Example 1: A point makes angles θ1 and θ2 with the ends of a
straight wire carrying current I. The distance of the point from the wire is a. What is B?
Answer: The magnetic field is the integral
. coscos 4
ˆ
sin 4
ˆ)(
210
0 2
1
aI
da
I
z
zrB
ds = dx
dB
a
θ1 θ2x
y
θ
r
Biot-Savart law
Example 2: What is B if the wire is straight and infinitely long?
Answer: Then θ1 = 0 and θ2 = π so
. 2
ˆ
coscos 4
ˆ)(
0
210
aI
aI
z
zrB
ds = dx
dB
a
θ1 θ2x
y
θ
r
Biot-Savart law
In this example, a different version of the right-hand rule is useful: If your thumb shows the direction of the current, your fingers curl in the direction of the magnetic field.
ds = dx
dB
a
θ1 θ2x
y
θ
r
Biot-Savart law
Example 3: What is B a distance x along the symmetry axis of a circular loop of radius R carrying current I?
Answer: The magnitude of dB is dB = μ0 I ds/4πr2 , where
r2 = x2 + R2, but only the x-component of dB survives. We multiply dB by cos θ = R/r and replace ds by 2πR to get
x
.][ 2 2/322
20
Rx
IRB
y
rdB
Rθ
θ
Parallel currents
Now let’s look at magnetic fields and moving charges as a complete system. Our first question is simple but important: Two straight, infinitely long parallel wires, of separation a, carry currents I1 and I2 . What is their mutual force per unit
length?
The field due to one current (say, I1) is μ0 I1/2πa. It exerts a
magnetic force (μ0 I1/2πa) I2L. on any section of length L in the
other wire. Therefore the magnetic force per unit length FB /L
is FB /L= μ0 I1I2/2πa. This force per unit length is invariant if
we interchange I1 and I2. If we check the direction of each
force, we find that the mutual forces are equal and opposite, as Newton’s third law requires.
Parallel currents
Do parallel currents attract or repel each other? What about anti-parallel currents?
I1
I2
L
a
a
B2
F2
Parallel currents
A large current begins to flow through a spring. Does the spring expand or contract?
Parallel currents
We calculated the magnetic force per unit length to be
Now let’s try some numbers. Assume I1 = I2 = I, a = 1 m, and
FB/L = 2 × 10–7 N/m. Since μ0 = 4π × 10–7 T·m/A, we have
Since 1 T = N /A·m (recall FB= qv × B), we obtain I2 = 1 A2
and an operational definition of the ampere:
When the force per unit length between two long, parallelwires separated by 1 m and carrying identical currents I is
FB = 2 × 10–7 N/m, the current in each wire is 1 A.
. 2
210
a
II
L
FB
. (T/A)102m/N 102 277 I
Parallel currents
We likewise obtain an operational definition of the coulomb:
When a wire carries a steady current of 1 A, the amount of charge that flows through the wire in 1 s is 1 C.
I = 0
I
Bdr
Ampère’s law
Oersted’s discovery that an electric current deflects a compass needle suggests the following experiment:
B(r) depends only on r. From the Biot-Savart law, B = μ0 I/2πr
and so in this case, at least, .)( 0Id rrB
I = 0
I
Bdr
Ampère’s law
Oersted’s discovery that an electric current deflects a compass needle suggests the following experiment:
Ampère discovered that this law holds in general, and the
equation is called Ampère’s law.Id 0)( rrB
1 A 5 A
2 Aa
b
c
d
Ampère’s law
Example 1: Rank for the four closed curves a, b, c, d
using Ampère’s law.
rrB d )(
Ampère’s law
Example 2: What is B inside and outside an infinitely long conducting wire of radius R, carrying uniform current I?
Answer: By symmetry, B must circulate around the axis of the wire. Outside the wire, we choose a concentric circle with radius r > R. Then Ampère’s law tells us that 2πrB = μ0 I so
B = μ0 I/2πr.
B I
Ampère’s law
Example 2: What is B inside and outside an infinitely long conducting wire of radius R, carrying uniform current I?
Answer: By symmetry, B must circulate around the axis of the wire. Inside the wire, we choose a concentric circle with radius r < R. Then Ampère’s law tells us that 2πrB = μ0 I(πr2/πR2) so
B = μ0 Ir/2πR2.
IB
Js
Ampère’s law
Example 3: What is B for an infinite current sheet carrying surface current density Js?
BB
Answer: We use the rectangular
loop shown to evaluate
The top and bottom sides don’t contribute, while the right and left sides each contribute BL, where L is the height of the rectangle. We have 2BL =μ0 JsL, so B =μ0 Js/2.
. )( rrB d
Ampère’s law
Example 4: Ampère’s law, like Gauss’s law, is especially useful in cases of high symmetry. Let’s apply Ampère’s law to find the field of a long solenoid, which would be awkward using the Biot-Savart law.
Experiment shows that Boutside such a solenoid isvery weak.
What is the field B inside the solenoid?
B
W
L1 3
4
2
Ampère’s law
Example 4: Ampère’s law, like Gauss’s law, is especially useful in cases of high symmetry. Let’s apply Ampère’s lawto find the field of a long solenoid, which would be awkward using the Biot-Savart law.
Consider the integral of B·draround a rectangular path asshown, in the direction shown. On Side 3 , B is negligible. On Sides 2 and 4 the integrals of B·ds sum to 0. So B·dr on Side 1 is what must equal μ0 times
the current through the rectangle.
B
W
L1 3
4
2
Ampère’s law
Example 4: Ampère’s law, like Gauss’s law, is especially useful in cases of high symmetry. Let’s apply Ampère’s lawto find the field of a long solenoid, which would be awkward using the Biot-Savart law.
We have BL = μ0 NI, where N
windings, each carrying current I, pass through the rectangle. Letting n = N/L denote the number of windings per unit length, we conclude that B = μ0 nI. This is
the magnetic field anywhere inside the solenoid; it is uniform.
B
W
L1 3
4
2
Gauss’s law for B
We have seen that the field lines of E and B are similar. One striking difference is that there are no magnetic monopoles.
This similarity of the field lines is not just a coincidence. We will see that the fundamental equations for E and B are very similar.
First, there is a version of Gauss’s law for the magnetic field. It states that total magnetic flux out of a closed surface equals 0:
This law can be derived from the law of Biot and Savart.
. 0 AB dB
Towards Maxwell’s equations
The fundamental equations for E and B are very similar.
First, there is a version of Gauss’s law for the magnetic field. It states that total magnetic flux out of a closed surface equals 0:
Second, we have already seen a statement about E that is very similar to Ampère’s law:
. 0 AB dB
. 0)( rrE d
Remember Faraday?
Now we can add a new basic rule about field lines:
•They never start or stop in empty space – they stop or start on a charge or extend to infinity. •They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. • The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction.• There are no closed loops!
E
E
E
E
Towards Maxwell’s equations
The set of four fundamental equations for E and B,
together with the Lorentz force law FEM = q (E + v × B), sum
up everything we have learned so far about electromagnetism!
, 0
0
0
0
AB
rB
rE
AE
d
Id
d
qd
(Ampère’s law)
(Gauss’s law)
Towards Maxwell’s equations
The set of four fundamental equations for E and B,
are correct for all static charges and steady-state currents, i.e. for electrostatics and magnetostatics. But if E and B change in time, two of the equations need correction. Guess which two?
, 0
0
0
0
AB
rB
rE
AE
d
Id
d
qd
(Ampère’s law)
(Gauss’s law)
Halliday, Resnick and Krane, 5th Edition, Chap. 33, Prob. 9:
A long solenoid, centered along the z-axis, has 100 turns per centimeter. An electron moves within the solenoid in a circle of radius 2.30 cm in the xy-plane. The speed of the electron is 0.0460c, where c = 3.00 × 108 m/s is the speed of light. What is the current I in the wire of the solenoid?
Halliday, Resnick and Krane, 5th Edition, Chap. 33, Prob. 9:
A long solenoid, centered along the z-axis, has 100 turns per centimeter. An electron moves within the solenoid in a circle of radius 2.30 cm in the xy-plane. The speed of the electron is 0.0460c, where c = 3.00 × 108 m/s is the speed of light. What is the current I in the wire of the solenoid?
Answer: The cyclotron frequency ω of the electron is
ω = v/r = 0.0460c/0.0230 m = 6.00 × 108 Hz
and we derived ω = eB/m where e is the electron charge and m is its mass. Also B = μ0 nI for the field in a solenoid, hence I =
mω/neμ0 . Putting in the numbers (m = 9.11× 10–31 kg and e =
1.60 × 10–19 C) we obtain I = 0.272 A.
Surface S1
Surface S2
I
–q
q
r
B
∂S
The problem with Ampère’s law
How do we apply Ampère’s law, to this
electrical circuit?
Current flows through Surface S1.
No current flows through Surface S2.
Hence we have aμ0I/2πr = B = 0:
CONTRADICTION!
,0 AJrB dd SS
The problem with Ampère’s law
You may say, “This is silly. Ampère’s law is not like Faraday’s law, no matter what Ampère thought. Forget about S1 and S2! Just consider how much current flows through the
closed loop ∂S .”
Yes, but suppose the closed loop straddles the capacitor:
How much current is flowing through the loop ∂S now?
I
∂S
Maxwell’s fix
This set of four fundamental equations for E and B,
is called Maxwell’s equations!
,
0
000
0
E
B
dt
dId
dt
dd
d
qd
rB
rE
AB
AE
(Ampère’s law as modified by Maxwell)
(Gauss’s law)
(Faraday’s law)
Surface S1
Surface S2
I
–q
q
r
B
∂S
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,
fixes the problem, for the case of a parallel-plate capacitor.
If we considerSurface S1, we have
(2πr)B = μ0I .
, 000 Edt
dId rB
Surface S1
Surface S2
I
–q
q
r
B
∂S
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,
fixes the problem, for the case of a parallel-plate capacitor.
If we considerSurface S2, we have
(2πr)B = μ0ε0 dΦE/dt.
For a parallel-plate capacitor, it stillequals μ0I!
, 000 Edt
dId rB
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,
fixes the problem, for the case of a parallel-plate capacitor.
If we consider Surface S1, we have (2πr)B = μ0I .
If we consider Surface S2, we have (2πr)B = μ0ε0 dΦE/dt.
For a parallel-plate capacitor of area A and plate separation d, ΦE = AE = AV/d = Aq/Cd, and C = ε0 A/d, so ΦE = qA/Cd = q/ε0.
We get dΦE/dt = I/ε0 so (2πr)B = μ0ε0 dΦE/dt = μ0I!
Note I = ε0 dΦE/dt is often called the displacement current.
, 000 Edt
dId rB
