Ampere’s law
To determine the magnetic field created by a current, an equation much easier to use than Biot-Savart
is known as Ampere’s law.
μ0·Ienclosed = ∮ 𝑩▪𝒅𝒔
𝑇·𝑚
As before, μ0 is the permeability of free space, 4π x 10-7 𝐴
I is electric current
B is the magnetic field vector
ds is a small step in space
The integral symbol, summa, with a loop indicates a path integral, meaning the sum of all of the little
steps must be a closed loop
The right-hand side is a dot product, but all of the problems we do will have angles of either 0º or 180º
between B and ds, so we only need to consider if the two are parallel or anti-parallel if we follow the
convention that the ds steps are taken counter-clockwise.
Suppose we have a wire which carries a current into the page, along the negative z-axis.
We can use the cross product of Biot-Savart to determine the direction of the field at each of these four
surrounding points. With the index finger into the page with the current flow, rotate the middle finger
to point a and you will find your thumb points down. Rotate the middle finger to d and the thumb
points left, to c and the thumb points up, to b and the thumb points right.
In general, the current creates a field that circulates around the wire clockwise. A simple shortcut for
this is to point your right hand thumb with the current flow and see that your fingers naturally curl
clockwise (the negative z-axis is tied to negative rotation in the xy-plane).
This circulating magnetic field is true at any distance from the wire, though the strength is weaker with
distance, so a fuller diagram would look like this:
Question 1: What would be the direction of the magnetic field if the current was flowing out of the
page?
Now let’s apply Ampere’s law by drawing a loop around the wire.
μ0·Ienclosed = ∮ 𝑩▪𝒅𝒔
The Ienclosed is just the lone current, I, but along the negative z-axis, so –I.
If we look at the little step of ds on the right side, it is antiparallel to the magnetic field, so we have
cos180 from the dot-product, or -1.
So now we have
μ0·(-I) = ∮ |𝐵| · |𝑑𝑠| · (−1)
By radial symmetry, the magnetic field strength should be the same everywhere around the loop, so we
can pull |B| out of the integral. We can pull out the -1 as well, leaving only ∮ |𝑑𝑠|. Conceptually, this
is just the circumference of the loop, 2πR.
This gives us
μ0·(-I) = |B|·(-1)·(2πR)
So the magnitude of the magnetic field from a long, straight wire is:
B=
𝜇0 ·𝐼
2𝜋·𝑅
where R is the distance from the wire
Question 2: If you double the distance from a given current-carrying wire, what happens to the strength
of the magnetic field?
What if we had, instead of a thin wire, a current, I, of uniform density traveling through a cylinder of
radius, a?
The magnetic field for R > a is going to be what it was for the last example. The process of Ampere’s
law is exactly the same.
What is the magnetic field for R < a? We’ll simply draw an Amperian loop of radius R inside the
cylinder.
μ0·Ienclosed = ∮ 𝑩▪𝒅𝒔
Now the enclosed current isn’t the total current, I. It is only some fraction of that current, the fraction
being the area within the green Amperian loop over the area of the entire cylinder.
𝜋·𝑅 2
So Ienclosed = I·
𝜋·𝑎2
= I·
𝑅2
𝑎2
but, again, negative along the negative z-axis.
The right-hand side of Ampere’s law follows the same process as the previous example and will be
|B|·(2πR)·(-1)
𝑅2
So μ0·(- I·𝑎2 ) = |B|·(2πR)·(-1)
The magnitude of the magnetic field inside the cylinder is then
B=
𝜇0 ·𝐼 𝑅
·
2𝜋 𝑎2
This equation makes sense when R drops to zero and agrees with the previous equation when R = a.
Question 3: If you graphed magnetic field versus radius from the center of the wire, what would that
graph look like from R = 0 to R = a?
There are two common structures of coiled wire that can be used to create very strong magnetic fields,
a solenoid and a torus.
A solenoid is just a cylindrical coil of wire, something like an insulated Slinky.
Let’s say that looking at the solenoid from point P, you see a current I traveling clockwise in the
solenoid. In this diagram, that means current is flowing out of the page on the top side of the solenoid
and then flowing back into the page on the bottom side of the solenoid.
Knowing that, let’s draw a cut-away view.
Now let’s take a look at one ring of the solenoid and what direction of magnetic field it would produce
at point H.
For the bottom of the ring, point your index finger into the screen with the current and rotate your
middle finger towards H. You will see your thumb points down and right with the magnetic field.
For the top of the ring, point you index finger out of the screen with the current and rotate your middle
finger towards H. You will see your thumb points up and right with the magnetic field.
The vertical components add to zero, leaving a field only to the right. All the other rings will have a
similar effect at H, creating a very strong rightward field there. Likewise for any other point inside the
solenoid.
If you put H outside the solenoid, you can see the two magnetic fields almost add to zero. They get
closer and closer to adding to zero as the coil radius R becomes smaller and smaller relative to the
solenoid length, L. For an idealized solenoid, this external magnetic field is taken to be zero.
So now let’s apply Ampere’s law to the red loop shown below.
We’ve just stated that the magnetic field is essentially zero outside the solenoid, so B▪ds is going to be
zero from a to b and also the portions of b to c and d to a outside the solenoid. But even for the
portions of b to c and d to a which are inside the solenoid, the little steps of ds are perpendicular to the
B field (which is we’ve shown is pointing right), so B▪ds is going to be zero.
So the only relevant path is from c to d.
μ0·Ienclosed = ∮ 𝑩▪𝒅𝒔
The enclosed current is the current in the solenoid, I, times the number of loops we’ve included in the
red box, which we will designate as N.
∫ 𝐁▪𝐝𝐬 from c to d is just |B|·|x|·cos0º = B·x
So μ0·I·N = B·x
𝑁
B = μ0·I· 𝑥
where
𝑁
𝑥
For an ideal solenoid, B
is coils per length, often written as n
= μ0·I·n
where n is coils per length
Question 4: What are two ways of increasing the magnetic field inside a solenoid?
The second common geometry of a coiled wire is a torus or toroid. This is basically a solenoid which
has had the two ends connected to form the shape of a bagel.
If we imagine that the current flows into the screen along the outer radius of the toroid and then out of
the screen on the inner radius of the toroid, the cut-away view would look like this:
We’ll draw an Amperian loop inside the toroid with a radius R and assume the toroid has N number of
turns and carries a current, I.
μ0·Ienclosed = ∮ 𝑩▪𝒅𝒔
Ienclosed is simply I·N
The loop in enclosing a current out of the page and if you point your right hand thumb with that
current, your fingers curl counter-clockwise with the magnetic field. So the dot product will involve
cos0º. And the sum of the ds steps is just the circumference of the Amperian loop.
∮ 𝑩▪𝒅𝒔 = |B|·(2π·R)·(cos0º) = B·2πR
μ0·I·N = B·2πR
So for a toroid, B
=
𝜇0 𝐼·𝑁
2𝜋·𝑅
where N is the number of turns in the toroid