Faraday’s law of induction
In the diagram above, a metal bar with length L is pulled to the right at a speed v in a uniform magnetic
field, B. At the top and bottom, the bar makes electric contact with a U-shaped wire which includes a
lightbulb with resistance, R. This situation is often called motional EMF.
If we imagine the bar filled with positive charges, those positive charges are moving in a magnetic
field, so they will experience a magnetic force. With your right hand, if you point your index finger
with the motion of the sliding bar (and the charges it carries), rotate your middle finger to align with
magnetic field into the screen, your thumb will indicate the charges are pushed up the bar.
The charges will then continue to circulate counter-clockwise, through the lightbulb, making it glow.
But now that current is flowing upwards in the bar, there is a second magnetic force. Point your right
hand index finger upwards with the current, rotate your middle finger into the screen, and your thumb
will point left with a force opposite the velocity of the bar.
So to keep the bar sliding at a constant velocity, there must be some rightward input force, Fin.
By conservation of energy, the power input by this input force will match the power output of the
lightbulb.
Fin·v = I·ΔV
This input force is just as strong as the magnetic force, I·L·B, so
(I·L·B)·v = I·ΔV
For motional emf, ΔV = B·L·v
or
E = B·L·v
Question 1: What voltage would be produces if a 10cm metal bar is slid at 2.0m/s in a magnetic field
of 3.0T?
This is a specific example of a much more general law, called Faraday’s law of induction:
E=-
𝑑𝛷𝐵
𝑑𝑡
Where E is an emf, or voltage
ΦB is magnetic flux, similar to electric flux, but ΦB = ∬ 𝑩▪𝒅𝑨
This can be read: induced emf is the negative of the rate of change of magnetic flux
We can see how this is related to the motional emf equation as follows:
For a constant magnetic field, dΦB = B·dA where dA is a change in the enclosed area (surrounded by
the circulating current), or L·dx
𝑑𝑥
So E = B·L· 𝑑𝑡 = B·L·v
There is a principle associated with Faraday’s law known as Lenz’s law and it is stated this way:
The magnetic field created by an induced current will oppose the change in magnetic flux which
created it.
For the motional emf example, this is how that works:
As the bar slid right, the enclosed area included a greater and greater magnetic flux into the page.
Therefore, by Lenz’s law, the induced current must generate a magnetic field opposing this, out of the
page.
We determined the sliding bar would generate a current circulating counter-clockwise. If you curl
your right hand counter-clockwise, your thumb will point out of the screen in the same direction as the
induced magnetic field in the enclosed area, as required.
A simpler way of saying this is: the bar sliding right causes a gain of x’s, therefore the induced current
must produce its own dots. Point your thumb out of the page with the dots that must be created and
your right hand curls counter-clockwise with the induced current.
If you have a wire loop in a magnetic field, there are a few ways to change the enclosed flux as time
passes.
1. Make the magnetic field stronger or weaker
2. Make the ring smaller or larger
3. Rotate the ring around the y or x axis, making the field and area vector less parallel
4. Pull the ring out of the field of then push it back into the field
Suppose the ring shrinks, causing a loss of x’s. By Lenz’s law, we must put in our own x’s to oppose
this. If you point your thumb into the screen with the x’s that need to be created, your right hand curls
clockwise with the current that must be induced.
Suppose the field gets stronger, implying a gain of x’s. By Lenz’s law, we must put in our own dot’s
(an induced current can’t create a loss of x’s) to oppose the gain of x’s. If you point your thumb out of
the screen with the dots that need to be created, your right hand curls counter-clockwise with the
current that must be induced.
Question 2: In the diagram above, if the ring expands in size, in what direction would be the induced
current?
Even if there wasn’t a physical ring there, something should be happening in that space that could
cause a current. If we use the equation |ΔV| = ∫ 𝑬 · 𝒅𝒔 and replace |ΔV| with the induced emf, we get
-
𝑑𝛷𝐵
𝑑𝑡
= ∫ 𝐸 · 𝑑𝑠
or
𝑑𝛷𝐵
𝑑𝑡
= - ∫ 𝐸 · 𝑑𝑠
Let’s make sure the signs are correct. If the field is into the page and increasing, then
𝑑𝛷𝐵
𝑑𝑡
negative.
This implies ∫ 𝑬 · 𝒅𝒔 is positive. The current the electric field could drive would be counterclockwise, so the electric field must be counter-clockwise also. When we take steps of ds counterclockwise, all the dot products have cos 0º, which makes ∫ 𝑬 · 𝒅𝒔 positive.
Suppose the loop has an area, A, the magnetic field is B, and the loop rotates around the y-axis at an
angular velocity, ω. If the area vector begins pointed into the page, then
ΦB = B·A·cosθ
=
B·A·cos(ωt)
By Faraday’s law,
E=-
𝑑𝛷𝐵
𝑑𝑡
E = - [-ω·B·A·sin(ωt)]
E = B·A·ω·sin(ωt) and if the coil has N number of loops, the voltages are summative, so
E = N·B·A·ω·sin(ωt)
Question 3: If you double the rate at which a generator spins, what happens to the voltage it produces?
𝑑𝐼
Lastly, consider a wire carrying a current that is increasing with a rate 𝑑𝑡. Next to this wire is a loop
containing a resistor with resistance, R. What is the current in this loop?
First, we know from a previous derivation that the magnetic field from a long, straight wire is:
B=
𝜇0 ·𝐼
2𝜋·𝑅
Now we look at a sliver of the wire loop and find the flux in that sliver, taking the width of the sliver to
be dx.
𝜇0 ·𝐼
dΦB = B·dA =
2𝜋·𝑥
·c·dx
We integrate these slivers from x = a to x = a+b
𝑎+𝑏 𝜇0 ·𝐼
· c · dx =
2𝜋·𝑥
ΦB = ∫𝑎
So
𝑑𝛷𝐵
𝑑𝐼
=
𝜇0 ·𝑐
2𝜋
𝑎+𝑏
·ln(
𝑎
𝜇0 ·𝐼·𝑐
2𝜋
|E| =
2𝜋
𝑎+𝑏
·ln(
𝑎
)·
𝑎
)
)
Then, by the chain rule, E = 𝜇0 ·𝑐
𝑎+𝑏
·ln(
𝑑𝐼
𝑑𝛷𝐵
𝑑𝑡
=-
𝑑𝛷𝐵 𝑑𝐼
𝑑𝐼
·
𝑑𝑡
𝑑𝑡
So, in the loop,
I=
𝜇0 ·𝑐
2𝜋·𝑅
𝑎+𝑏
·ln(
𝑎
)·
𝑑𝐼
𝑑𝑡
Question 4: If the current in the left-hand wire in increasing, in what direction is the induced current in
the loop?