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Wize High School Grade 12 Physics Textbook > Magnetism

Faraday's Law of Induction

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Faraday's Law


In this section, we finally reveal the connection between electricity and magnetism!
  • Faraday’s Law states that, for a given region, if the magnetic flux changes, there will be an induced electromotive force (emf).
  • The magnitude of the induced emf is equal to the rate of change of magnetic flux.
ε=∣ΔΦmΔt∣=∣Δ(BAcos⁡θ)Δt∣\boxed{\varepsilon =\bigg|\cfrac{\Delta \Phi_m}{\Delta t}\bigg|=\bigg|\cfrac{\Delta(BA \cos \theta)}{\Delta t}\bigg|}
  • There are three ways to change the magnetic flux through a region:
  • Change the magnetic field strength, BB
  • Change the surface area, AA
  • Change the angle between the magnetic field and the area, θ\theta
Wize Tip
When it comes to induction, you can usually think of emf as another word for voltage. The difference between emf and voltage is more important when we are talking about batteries with some internal resistance - which is not the case here.

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Example: Faraday's Law


Consider a magnetic field of strength 4.0 T passing through a circular wire loop of radius 2.0 m. The magnetic field passes perpendicularly through the plane of the wire loop.

a) If the magnetic field strength cuts in half over a 1.0 second time interval, what is the induced emf in the wire loop?
b) If you wanted this same magnitude of induced emf from the original setup by changing the radius instead of the magnetic field strength, what would be the new radius of the circle after 1.0 seconds? (Hint: there are two possible answers!)
c) If you wanted this same amount of induced emf from the original setup, but you wanted to keep the magnetic field strength at 4.0 T and keep the wire at 2.0 m, how would you do it (assume a change over 1.0 seconds)?

Part a)

The induced emf is found as follows (only the magnetic field strength is changing):
ε=∣Δ(BAcos⁡θ)Δt∣ε=∣ΔBΔtAcos⁡θ∣ε=∣(−2.0T)(1.0s)(π(2.0m)2)cos⁡(0o)∣ε=8π\begin{aligned} \varepsilon &=\bigg|\cfrac{\Delta(BA \cos \theta)}{\Delta t}\bigg| \\ \varepsilon &=\bigg|\cfrac{\Delta B}{\Delta t}A \cos \theta\bigg| \\ \varepsilon &=\bigg|\cfrac{(-2.0T)}{(1.0s)}(\pi(2.0m)^2)\cos (0^o)\bigg| \\ \varepsilon &=8\pi \\ \end{aligned}
That is, the induced emf magnitude is 8π8\pi volts, or 25.1 V.

Part b)

We need to find the change in radius that will give us the same induced emf. Note that we can either increase or decrease the size of the circle since we are just asking about the magnitude of induced emf.
ε=∣Δ(BAcos⁡θ)Δt∣ε=∣BΔAΔtcos⁡θ∣(8π V)=∣(4.0T)ΔA(1.0s)cos⁡(0o)∣ΔA=∣2π m2∣\begin{aligned} \varepsilon &=\bigg|\cfrac{\Delta(BA \cos \theta)}{\Delta t}\bigg| \\ \varepsilon &=\bigg|B\cfrac{\Delta A}{\Delta t} \cos \theta\bigg| \\ (8\pi~V) &=\bigg|(4.0T)\cfrac{\Delta A}{(1.0s)}\cos (0^o)\bigg| \\ \Delta A &=|2\pi~m^2| \\ \end{aligned}
The original area of the circle is Aoriginal=π(2.0m)2=4π m2A_{original}=\pi (2.0m)^2=4\pi~m^2. The new area of the circle is therefore either 2π m22\pi~m^2 or 6π m26\pi~m^2. We can find the radius of these circles as follows:

Smaller circle:
A=πr22π=πr2r=2\begin{aligned} A=\pi r^2 \\ 2\pi =\pi r^2 \\ r=\sqrt{2} \end{aligned}
Larger circle:
A=πr26π=πr2r=6\begin{aligned} A=\pi r^2 \\ 6\pi =\pi r^2 \\ r=\sqrt{6} \end{aligned}
Part c)

We have already changed the magnetic field strength and the area. Time to change the angle!

ε=∣Δ(BAcos⁡θ)Δt∣ε=∣BAΔcos⁡θΔt∣(8π V)=∣(4.0T)(π(2.0m)2)Δcos⁡θ(1.0s)∣∣Δcos⁡θ∣=0.5\begin{aligned} \varepsilon &=\bigg|\cfrac{\Delta(BA \cos \theta)}{\Delta t}\bigg| \\ \varepsilon &=\bigg|BA\cfrac{\Delta \cos \theta}{\Delta t} \bigg| \\ (8\pi~V) &=\bigg|(4.0T)(\pi(2.0m)^2)\cfrac{\Delta \cos \theta}{(1.0s)}\bigg| \\ |\Delta \cos\theta| &=0.5 \\ \end{aligned}
Note, there's no way for us to make the angle any lower (it is already at zero degrees, with cos⁡θ=0\cos\theta=0). Let's find the angle that corresponds to an increase in the cosine by 0.5:
θ=arccos⁡(0.5)=60o\theta=\arccos(0.5)=60^o
That is, the ring must be rotated so that the area and magnetic field vectors are at an angle of 60 degrees to each other (this is equivalent to saying that the magnetic field forms a 30 degree angle with the plane of the ring).
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Motional EMF


The induced emf becomes useful when it is applied to circuits, and we begin to see induced currents. This results in appliances and devices being powered without any batteries!
  • Consider a conductor of length ll moving at speed vv through a uniform magnetic field BB, while sitting on conducting wires, as shown below:
  • The Lorentz force on the charges inside the conductor results in a separation of charge that results in a potential difference across the conductor.
  • This potential difference is exactly equal to the induced emf in the circuit, and is sometimes called motional emf because it arises due to the motion of the conductor:
ε=Blv\boxed{\varepsilon=Blv}
  • If there are no other circuit elements present, we can use Ohm's Law (V=IRV=IR) to determine the induced current in the circuit, noting that the induced emf ε\varepsilon and circuit voltage VV are equal:
I=BlvR\boxed{I=\cfrac{Blv}{R}}

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Example: Conductor on Rails


A circuit is located in a constant magnetic field of strength 2.4 T, shown below. The conducting 10 centimeter bar moves to the right with a constant speed of 20 m/s and the resistor R in the circuit is 20 Ω.

a) What is the induced emf in the circuit?
b) What is the induced current?

Part a)

The induced emf can be found with the usual formula:
ε=Blvε=(2.4 T)(0.1 m)(20m/s)ε=4.8 V\begin{aligned} \varepsilon&=Blv \\ \varepsilon&=(2.4~T)(0.1~m)(20m/s) \\ \varepsilon&= 4.8 ~V\\ \end{aligned}
Part b)

The current is found using Ohm's Law:
I=εR=4.8 V20 Ω=0.24 A\begin{aligned} I&=\frac{\varepsilon}{R}\\ &=\frac{4.8~V}{20~\Omega}\\ &=0.24~A \end{aligned}

Practice: Motional EMF


In the diagram below, the blue rod has a resistance of 5 Ω5\ \Omegaand it is moving with a speed of 5 ms5\ \frac{m}{s}. The magnetic field is constant at B=1 TB=1\ T. The length of the rod is 10 centimeters.

a) Find the induced emf and current in the circuit.
b) What force magnitude is required on the blue rod to maintain the constant speed of the bar?
c) How much power is dissipated in the blue bar?

Part a) - induced emf