Download Sources of Magnetic Fields Chapter 28

Chapter 28
Lecture 13
Sources of Magnetic Fields
Announcements
— Quiz in discussion sections tomorrow
[covering Ch 26 and 27]
— Ch 28 HW will be assigned soon
— Grading of midterm is completed…
Notes on the Midterm
Problem 1:
- Similar to problems from Lecture and Ch 21 and Ch 23 HW;
• Difference between electric field (E, a vector) & electric potential (V, not a vector).
• Cons. of Energy; kinematic equations don't apply when acceleration isn't constant.
[From Lec 5]
Problem 2:
- Similar to problems from Lecture and Ch 26 HW;
• Resisitors in series & parallel
• Application of Kirchhoff’s loop and junction rules?
Midterm Exam
Grades posted to EEE.
Max score = 35.
Partial credit awarded.
From Lecture 12…
Magnetic field of a moving point charge
The constant µ0 is called the permeability of free space : µ0 = 4π x 10-7 T·m/A.
+
From Lecture 12…
Magnetic field of an infinite current-carrying wire
~ =
B
µ0 I
2a
p
k̂
2
2
4⇡ x x + a
If we let the current element become infinitely long (a!∞),
the magnitude of the field at a distance r (x = r) becomes:
µ0 I
B=
2π r
Magnetic field near a straight long wire
As always, the direction of the field is given by the righthand rule: the field lines loop around the wire in concentric
circles.
Magnetic field of a circular current loop
A circular conductor of radius a carries a current I. Find B field at
some point on the axis of the loop, a distance x from the center.
d~l
r
a
x
~
dB
r̂
1. Break the loop down into tiny segments of length dl.
2. For each segment, dB will be perpendicular to both r and dl.
3. The y- and z-components of the B field will cancel out due to symmetry
4. The x-components magnetic field from each current element will add
Magnetic field of a circular current loop
d~l
A circular conductor of radius a carries a current I. Find B field at
some point on the axis of the loop, a distance x from the center.
r 2=
x 2+
a2
r
a
x
~
dB
Magnitude is the same all around loop:
r̂
µ0 I dIl
dB =
4⇡ x2 + a2
Direction changes
1. Break the loop down into tiny segments of length dl.
2. For each segment, dB will be perpendicular to both r and dl.
3. The y- and z-components of the B field will cancel out due to symmetry
4. The x-components magnetic field from each current element will add
Magnetic field of a circular current loop
d~l
a
r
a
x
x
~
dB
µ0 I
dl
dB =
4⇡ x2 + a2
dBx
Magnetic field of a circular current loop
d~l
a
r
a
x
x
~
dB
µ0 I
dl
dB =
4⇡ x2 + a2
µ0 I
dl
a
dBx = dB cos ✓ =
4⇡ x2 + a2 (x2 + a2 )1/2
dBx
Magnetic field of a circular current loop
d~l
a
r
a
x
x
~
dB
µ0 I
dl
dB =
4⇡ x2 + a2
µ0 I
dl
a
dBx = dB cos ✓ =
4⇡ x2 + a2 (x2 + a2 )1/2
dBx
Magnetic field of a circular current loop
d~l
a
r
a
x
x
~
dB
µ0 I
dl
dB =
4⇡ x2 + a2
µ0 I
dl
a
dBx = dB cos ✓ =
4⇡ x2 + a2 (x2 + a2 )1/2
dBx
y-component will cancel out around loop
Magnetic field of a circular current loop
d~l
a
r
a
x
x
Bx =
Z
µ0 I
dl
dB =
4⇡ x2 + a2
~
dB
µ0 I
dl
a
dBx = dB cos ✓ =
4⇡ x2 + a2 (x2 + a2 )1/2
dBx
y-component will cancel out around loop
µ0 I
dBx =
4⇡

a
(x2 + a2 )3/2
Z
2⇡a
dl
Magnetic field of a circular current loop
d~l
a
r
a
x
x
Bx =
Z
µ0 I
dl
dB =
4⇡ x2 + a2
~
dB
µ0 I
dl
a
dBx = dB cos ✓ =
4⇡ x2 + a2 (x2 + a2 )1/2
dBx
y-component will cancel out around loop
µ0 I
dBx =
4⇡

a
(x2 + a2 )3/2
µ0 Ia2
Bx =
2(x2 + a2 )3/2
Z
2⇡a
dl
Magnetic field of a circular current loop
I
r 2 = x2 + a2
x
µ0 Ia2
Bx =
2(x2 + a2 )3/2
Magnetic field on the axis of a solenoid
µ0 Ia2
Bx =
2(x2 + a2 )3/2
Field along axis for 1 loop.
If we have a coil of N closely-spaced circular loops, each loop adds together:
N µ0 Ia2
Bx =
2(x2 + a2 )3/2
N coils
Off-axis field of a coil
The result we derived is only valid on the axis of a coil.
Off-axis, the field is more complicated, and resembles the field of a bar magnet
Gauss’s Law: Electricity –vs– Magnetism
Recall: Gauss’s law relates the surface integral of the electric flux to the charge
contained inside the surface.
Since there are no magnetic monopoles Gauss’s Law for magnetism is much
simpler. But it doesn’t help us calculate the magnetic field either:
Magnetic flux through close surface
Line integrals: Electricity –vs– Magnetism
We have also seen that the line integral of the electrostatic field E around any closed path is
zero:
e.g. equipotential surface and conductors
This is equivalent to the statement that the electric field is conservative. The line integral
of E is related to the work done by the electric field on a test charge following the
integration path.
•This is because the force on a test charge is proportional to, and in the same direction, as E.
•The total work done by E on a test charge must be zero for a closed path.
This reasoning does not apply to the integral of the magnetic field B around a closed path.
•Why —> The magnetic force is not parallel to B, and B is not conservative.
•Magnetism is NOT a conservative force.
Line integral of the Magnetic Field around a wire
Consider the line integral of the magnetic field B
in a closed circular loop around an infinite
current-carrying wire:
If the current I in the wire is coming out of the
screen, we showed the magnetic field magnitude
at a distance r from the wire will be:
B=
µ0 I
2π r
So the line integral of B if we integrate counterclockwise is:
counter-clockwise
B and dl are parallel
Ampere’s Law
The line integral of the magnetic field around any closed path is equal to the enclosed current:
Ienclosed is the total current passing through the
surface bounded by the integration path. Will
add either +µ0Iencl or -µ0Iencl depending on the
direction of I with respect to integration path.
We call current positive if the right-hand rule
for the integration path points in the same
direction, or negative otherwise.
• Currents in RHR direction are +
• Currents opposite to RHR direction are –
Ampere’s Law
What does a zero line integral mean?
!" "
If the line integral #∫ B • dl = 0 , it does not mean that the magnetic field is zero over
the integration path.
It means the total current passing through area bounded by the integration
path is zero.
The figure shows two equal currents going into and out of the page.
•The field is not zero anywhere, but the integral over the rectangular path is zero because the
total current through the rectangle is zero.
Example: Using Ampere’s Law
!" "
#∫ B • dl = µ0 I encl
Long cylindrical conductor of radius R, carrying current I.
Symmetry: The magnetic field will have constant
magnitude at a given distance from the center.
Pick: loop to be a circle of radius r
For loops with r > R:
- total current enclosed will just be the total current in the wire: I enclosed = I
For loops with r < R:
- total current enclosed will be the fraction of the cross-sectional area:
I enclosed
πr 2
=I
π R2
Example: Magnetic field of a long cylindrical conductor
Applying Ampere’s Law:
For r < R:
!" "
#∫ B • dl = µ0 I encl
r2
B(r ) (2π r ) = µ 0 I 2
R
µI r
B(r ) = 0 2
2π R
For r > R:
B(r ) (2πr ) = µ0 I
µ0 I
B(r ) =
2πr
Example: Magnetic Field of a Solenoid
A solenoid is a cylindrical wrapping of a single current-carrying wire.
It is effectively N circular coils, with each coil carrying the same current, I.
If the wrapping is tightly spaced and if the cross-sectional area of a solenoid is large
compared to its length, the field inside the solenoid near the midpoint is very nearly
uniform, and the field outside the solenoid near the midpoint is very nearly zero.
With this approximation, we can use Ampere’s Law to find the value of the field
inside.
n = # of coils per unit length = N/L
N coils per length l
Example: Magnetic Field of a Solenoid (cont.)
The figure shows a cut-away view of a plane through central axis of the solenoid.
We choose an integration loop of length L, which encloses N wrappings of the
conductor. The only non-zero contribution to the line integral comes from the path
between point a and point b
n = # of coils per unit length
Recap
Electric
E of point charge:
!"
1 Q
E=
rˆ
2
4πε 0 r
Force law:
!" !"
F = qE
Gauss’s law:
Magnetic
B of point charge:
"
!" µ qv × rˆ
B= 0
4π r 2
Force law:
!" !" Q
encl
ΦE = #
E
•
dA
=
∫
ε0
Conservation of energy:
Coulomb’s law "➔ Gauss’s Law
Ampere’s law:
!" "
#∫ B • dl = µ0 I encl
Biot-Savart "➔ Ampere’s Law