Discussion Question 9A
P212, Week 9
Ampere’s Law in a Nutshell
Ampere’s Law is the magnetic analogue of Gauss’ Law.
Gauss’ Law relates the integral of the electric field through a
closed “Gaussian surface” to the total charge enclosed by that
surface.
E dA
Qenc
0
Ampere’s Law relates the integral of the magnetic field around a closed “Amperian loop”
to the total current enclosed by that loop.
Just like Gauss’ Law, Ampere’s Law provides the simplest method for determining the magnetic
field of a known current distribution … but it can only be used in a practical way if the problem
has enough symmetry. It all boils down to choosing a suitable Amperian loop. The
mathematical curve we choose has to have these properties:
The magnetic field must have a constant magnitude on our curve.
The magnetic field must make a constant angle with our curve (or portions thereof).
Otherwise, we’ll never be able to extract the B field we want from that line integral!
The other part of the procedure is finding the current enclosed by our Amperian loop:
In your mind, imagine that the Amperian curve is a loop of coat-hanger wire, and
visualize stretching a rubber sheet across it. The enclosed current is simply the total
amount of current punching through that rubber sheet.
Procedure
1. Visualize and sketch the magnetic field, using all the symmetry that the problem offers.
Without knowing the direction and spatial behavior of the field beforehand, it is
impossible to solve for the field using Ampere’s Law!
2. Based on the field geometry, choose a suitable Amperian loop that passes through the
field point of interest.
3. Determine the total current enclosed by your Amperian loop.
4. Finally, evaluate the line integral over the loop to determine B.
Problem Classes
There are four classes of current distribution that can be analyzed easily with Ampere’s Law:
1. Straight wires and cylinders of infinite length
3. Sheets or slabs of infinite area
2. Solenoids of infinite length
4. Toroids
That’s it! All the Ampere’s Law problems you will encounter will be constructed from these
basic systems. So let’s go through them! Follow the steps outlined above to find the magnetic
field B (both magnitude and direction!) due to the four current-carrying objects shown below.
Basic Object #1: A straight wire of radius R and infinite length, carrying a uniformlydistributed current I coming out of the page
find B at a distance d > R from the axis of the
wire
x
(i.e. outside the wire … we’ll do the interior in the next discussion question).
I
x
R
d
R
I
r>R
x
r<R
Make the Amperian loop be a circle of radius d .
I
I
Bd
B2 d
B
2 d
Now find B at a distance d < R from the axis of the wire (i.e. Inside the wire).
I
which means the current inside of an Amperian
R2
I d2
I d2
Hence
Bd
2 dB
R2
R2
The current density per unit area J
loop of radius r is Iinside
and B
1
2 d
I d2
R2
r2J
Id
2 R2
Basic Object #2: An infinitely-long solenoid with n turns of wire per meter. The turns of the
solenoid are circular with radius R, and the wrapped wire carries a current I
find B at a
distance d from the axis of the solenoid, considering points both inside and outside the coils.
Consider this Amperian loop. Since there is no field outside the loop.
Bd
BL
I inside where I inside
I nL
B
I n By righthand
L
rule this points along +zˆ direction.
I
I
Basic Object #3: A sheet of infinite area carrying a total current I into the page. The current is
distributed uniformly across the very-large width w of the sheet.
find B at a distance d both
above and below the sheet.
y
r
I
L
x
r
w
I
Consider this Amperian loop.
B
Bd
B2L
I inside
I
L
w
I
with directions as shown in the figure.
2w
Basic Object #4: A toroid with N square windings of side a. The radius of the toroid (from the
axis of the toroid to the center of the coils) is R, and the wrapped wire carries a current I in the
direction shown on the figure.
find B at a distance d from the axis of the toroid, considering
points both inside and outside the coils.
R
Consider an Amperian loop centered on the toroid axis of
NI
radius d . B d
B2 d
I inside
NI
B
2 d
Inside of the coil pack there is no current. Outside the
inner current (down) cancels the outer current (up)
a
I
Note: The shape of the coils in the solenoid and the toroid actually makes no difference, as long
as it’s constant. Can you explain why?
Only the cross-section of the wires counts, so as long as they’re consistent, the B-field will be
equivalent
Discussion Question 9B
P212, Week 9
Magnetic Field due to an Asymmetric Hollow Cylinder
A conducting cylinder is oriented parallel to the z axis and carries a uniform current I in the
negative z direction (into the page). This cylinder is hollow, however, with a cylindrical bore
centered on the point Q shown in the figure. The radius of this bore is R1, while the outer radius
of the cylinder is R2.
y
P
I = 2.5 A into page
R1 = 5 cm
R2 = 12 cm
point P: y = 15 cm
point Q: y = -3 cm
R2
x
R1 Q
(a) Calculate the magnetic field at the point P on the y axis.
(i) This appears to be a horrendous problem, with no symmetry! But really, the hollow
cylinder is a superposition of two solid cylinders ... and a solid cylinder of current is
something we can deal with. How could you treat this system so that it consists of two solid
cylinders?
Two separate cylinders, one w/ I into the page that has radius R2, and one centered at Q, w/
radius R1 and current I out of the page
(ii) Calculate the field at point P due to each of the two cylinders, then add the two
contributions together. Remember to work by components since you are adding field
vectors.
I
area
The current density is J
cylinder with current Iouter
with I inner
I - I outer
I
2
2
2
1
R
R22 J
R )
66.87 A/m 2 . We think of this as outer
3.025 amps into the plane and an inner cylinder
2.5 3.025
0.5252 or +0.5252 out of page .
We now use expression for infinite wire:
B=
2e - 7 I
r
ˆx
2e-7 3.025
0.15
ˆx
2e-7 0.5252
0.15+0.03
ˆx
3.450 ˆx T
(b) Calculate the magnetic field at the point Q at the center of the hollow bore.
y
P
R2
x
R1 Q
I = 2.5 A into page
R1 = 5 cm
R2 = 12 cm
point P: y = 15 cm
point Q: y = -3 cm
Again, compute the field due to each of the two cylinders separately, then add the fields
together. This time, one of the contributions will be very easy .... but one of them requires a
little more thought. What is the enclosed current that goes into Ampere's Law?
The inner cylinder contributes no B. The current from the outer
cylinder is I outer
B
2e - 7 I
r
ˆx
J
0.032
0.1891 A. We next use
2e - 7 0.1891
0.03
ˆx
1.26
ˆx
Faraday's Law: Moving Loop
A one-turn current loop in the shape of an equilateral triangle with sides a lies in the xy plane
and moves with velocity v in the +x direction. (A magic external force keeps the velocity
constant ) The loop has a net resistance R . It passes through a region of constant, spatially
uniform magnetic field B that points in the -z direction (into the page) and extends from x = 0 to
x=d.
y
a
a
a
v
x
0
x
d
We will specify the loop’s position using the x-coordinate of the triangle’s tip (as shown in the
figure). Now, before we do any calculations, let’s think physically …
x
Remember, induced EMF only occurs when the magnetic flux through a loop is changing.
entering: x = 0 to 3*a/2
exiting: x = d to d + 3*a/2
I
x
Let positive I indicate current flow in the clockwise direction.
I
t
In computing the flux, we take the normal to be in the
direction of the positive current direction according to
the righthand rule. This direction would be into the page
BA
of the paper and hence the flux is positive
d
where A is the area of the loop in the field.
.
dt
Hence as the triangle enters the loop is increasing and
0
0. As the triangle exits the loop is decreasing and hence
0 I 0. The rate of change of is proportional to the
velocity times the base of the triangle and hence you get the
ramped shape.
I
y
a
a
a
v
x
0
x
d
I
x
This expression should be valid from the moment the tip
of the triangle enters the magnetic field until the base of the triangle enters the field.
(i) To obtain the induced current, you first need an expression for the magnetic flux
through the loop as a function of the loop's position x. The B-field part of this expression
is easy ... the area part requires some thought. (Hint: what is the height of an equilateral
triangle of side a?)
(ii) Now take the time derivative of your expression, to get an EMF.
(iii) The final step is to turn your induced EMF into a current .
The key insight is to realize that the part of the loop covered
by the B field is still an equilateral . As the figure shows
the height of this triangle is x and the base is 2 x tan 300.
1
Hence the area is A= base height
2
0
30
x
The flux is then
d
dt
I
R
B dx 2
3 dt
2 Bxv
3R
x2
3
B dx 2 dx
3 dx dt
2
x tan 30
2B
xv
3
0
x2
3
Faraday's Law: Rotating Loop
rad/s
y
R=4
Consider a rectangular loop of
wire with width w, height h, and
total resistance R. The region of
space occupied by the wire loop
contains a uniform magnetic field
B pointing out of the page.
positive I
=
B = 1.2 T
h = 3 cm
clockwise
x
A motor attached to the loop
w = 5 cm
keeps it rotating about the vertical
axis with constant angular
velocity . At time t = 0, the loop is parallel to the paper (as shown in the figure).
(a) Plot the magnetic flux B through the loop as a function of time. Find the maximum value
(amplitude) of the flux, and write it on your plot.
1.8 mT m 2
max
z
B
y
x
y
x
t
3
t t
ˆ
The is easiest to see in the x-z plane. I show the view of an observer that is looking along the
axis of rotation (the y axis). The normal ˆ is
to the x-y plane in a direction given by
the right hand rule for our current loop. At t 0 this lies along the -z axis. The sense of the rotation
has ˆ rotating in the -x direction as shown. By trig ˆ
sin t ˆx cos t ˆz
B (area ˆ ) =B(area) ˆz ˆ
-B area cos t
( 1.2T )( 0.03 0.05 m2 )cos t max 14.4 mV
1.8 10 3 cos t T m 2
3
t
Faraday's Law: Rotating Loop
(b) Plot the induced EMF E through the loop as a function of time. Find the maximum value
(amplitude) of the EMF, and write it on your plot.
d
d
1.8 10 3 cos t
1.8 10 3 sin t
dt
dt
1.8 10 3 8 sin t
14.4 10 3 sin t V
(c) Plot the induced current I through the loop as a function of time. Find the maximum value
(amplitude) of the current, and write it on your plot.
max=
I
R
14.4 10 3
sin t
4
3.6 mA
3.6 sin t mA
3
(d) Plot the torque provided by the motor to keep the loop rotating at a constant angular
velocity. Find the maximum value (amplitude) of the torque, and write it on your plot.
points along the plane normal.
z
I (area ) sin 180
mag
I (area)
B
x
y
t
mag
max=6.48
sin t yˆ
3
mag
t
3.6 10 sin t
0.05 0.03
N•m
1.2 sin t yˆ
6.48 sin 2 t yˆ N m
The motor must supply
motor
6.48 sin 2 t yˆ N m
to cancel the magnetic torque and keep
constant
motor
is in the direction of motion
and hence the motor is supplying work.
2
3
2
2
t