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Electromagnetic Induction

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Electromagnetic Induction

Students will learn how to determine the flux and how to calculate the induced voltage. In addition, Students will learn Lenz law and how to use it to determine the direction of the induced current in a loop of wire.

Key Equations

 \Phi = N \vec{B} \cdot \vec{A} &&\text{ Electromagnetic Flux}

 emf = -\frac{\Delta \Phi}{\Delta t}&& \text{Faraday's Law of Induction}

Guidance

To understand induction, we need to introduce the concept of electromagnetic flux . If you have a closed, looped wire of area A (measured in  \mathrm{m^2} ) and N loops, and you pass a magnetic field B through, the magnetic flux \Phi is given by the formula below. Again, the relative direction of the loops and the field matter; this relationship is preserved by creating an 'area vector': a vector whose magnitude is equal to the area of the loop and whose direction is perpendicular to the plane of the loop. The directions' influence can then be conveniently captured through a dot product:  \Phi = N \vec{B} \cdot \vec{A} &&\text{ Electromagnetic Flux}

The units of magnetic flux are \text{T} \times \mathrm{m^2} , also known as Webers \text{(Wb)} .

In the example above, there are four loops of wire (N = 4) and each has area \pi r^2 (horizontally hashed). The magnetic field is pointing at an angle  \theta to the area vector. If the magnetic field has magnitude  B , the flux through the loops will equal  4 \cos \theta B \pi r^2 . Think of the magnetic flux as the part of the “bundle” of magnetic field lines “held” by the loop that points along the area vector.

If the magnetic flux through a loop or loops changes, electrons in the wire will feel a force, and this will generate a current . The induced voltage (also called electromotive force, or emf ) that they feel is equal to the change in flux \triangle \Phi divided by the amount of time \triangle t that change took. This relationship is called Faraday’s Law of Induction:  emf = -\frac{\Delta \Phi}{\Delta t}&& \text{Faraday's Law of Induction}

The direction of the induced current is determined as follows: the current will flow so as to generate a magnetic field that opposes the change in flux. This is called Lenz’s Law. Note that the electromotive force described above is not actually a force, since it is measured in Volts and acts like an induced potential difference. It was originally called that since it caused charged particles to move --- hence electromotive --- and the name stuck (it's somewhat analogous to calling an increase in a particle's gravitational potential energy difference a gravitomotive force). For practical (Ohm's Law, etc) purposes it can be treated like the voltage from a battery.

Since only a changing flux can produce an induced potential difference, one or more of the variables in equation [5] must be changing if the ammeter in the picture above is to register any current. Specifically, the following can all induce a current in the loops of wire:

  • Changing the direction or magnitude of the magnetic field.
  • Changing the loops' orientation or area.
  • Moving the loops out of the region with the magnetic field.

Example 1

You are dragging a circular loop of wire of radius .25 m across a table at a speed of 2 m/s. There is a 2 m long region of the table where there is a constant magnetic field of magnitude 5 T pointed out of the table. As you drag the loop across the table, what will be the induced Emf (a) as the loop enters the field (b) while it is in the field and (c) as it exits the field.

Solution

(a): As the loop enters the field, the flux will start at zero and begin to increase until the loop is entirely inside the field. The flux will increase from 0 Tm 2 to some maximum value in the time it takes for the loop to move into the field. We can find this maximum value using the dimensions of the loop and the strength of the magnetic field. The dot product will be equal to one since the area and magnetic field vectors are parallel.

\Phi_f&=NBA\\\Phi_f&=1*5\:\text{T} * \pi (.25\:\text{m})^2\\\Phi_f&=.98\:\text{Tm}^2\\

Since we also know the radius of the loop and the speed at which it is being pulled, we also can find out how long it will take for the loop to move within the magnetic field.

d&=vt\\t&=\frac{d}{v}\\t&=\frac{2*.25\:\text{m}}{2\:\text{m/s}}\\t&=.25\:\text{s}\\

Now we can find the induced Emf in the loop.

Emf&=\frac{\Delta \Phi}{\Delta t}\\Emf&=\frac{.98\:\text{Tm}^2 - 0\:\text{Tm}^2}{.25\:\text{s}}\\Emf&=3.92\:\text{V}\\

(b): There will be no inuduced Emf in the loop once the entire loop is inside the magnetic field because the magnetic flux will not be changing.

(c): As the loop exits the magnetic field, the induced Emf will have the same magntiude as when it entered the field except that this time it will be negative because the flux is decreasing.

Emf&=-3.92\:\text{V}\\

Watch this Explanation

Simulation

Faraday's Electromagnetic Lab (PhET Simulation)

Inductance Problem Set

  1. A speaker consists of a diaphragm (a flat plate), which is attached to a magnet. A coil of wire surrounds the magnet. How can an electrical current be transformed into sound? Why is a coil better than a single loop? If you want to make music, what should you do to the current?
  2. A bolt of lightening strikes the ground 200 \;\mathrm{m} away from a 100- turn coil (see above). If the current in the lightening bolt falls from 6.0 \times 10^6 \;\mathrm{A} to 0.0 \;\mathrm{A} in 10 \;\mathrm{ms} , what is the average voltage ,  \varepsilon , induced in the coil? What is the direction of the induced current in the coil? (Is it clockwise or counterclockwise?) Assume that the distance to the center of the coil determines the average magnetic induction at the coil’s position. Treat the lightning bolt as a vertical wire with the current flowing toward the ground.
  3. A coil of wire with 10 loops and a radius of 0.2 \;\mathrm{m} is sitting on the lab bench with an electro-magnet facing into the loop. For the purposes of your sketch, assume the magnetic field from the electromagnet is pointing out of the page. In 0.035 \;\mathrm{s} , the magnetic field drops from 0.42 \;\mathrm{T} to 0 \;\mathrm{T} .
    1. What is the voltage induced in the coil of wire?
    2. Sketch the direction of the current flowing in the loop as the magnetic field is turned off. (Answer as if you are looking down at the loop).

  4. A wire has 2 \;\mathrm{A} of current flowing in the upward direction.
    1. What is the value of the magnetic field 2 \;\mathrm{cm} away from the wire?
    2. Sketch the direction of the magnetic field lines in the picture to the right.
    3. If we turn on a magnetic field of 1.4 \;\mathrm{T} , pointing to the right, what is the value and direction of the force per meter acting on the wire of current?
    4. Instead of turning on a magnetic field, we decide to add a loop of wire (with radius 1 \;\mathrm{cm} ) with its center 2 \;\mathrm{cm} from the original wire. If we then increase the current in the straight wire by 3 \;\mathrm{A} per second, what is the direction of the induced current flow in the loop of wire?
  5. A rectangular loop of wire  8.0 \;\mathrm{m} long and 1.0 \;\mathrm{m} wide has a resistor of 5.0 \ \Omega on the 1.0 side and moves out of a 0.40 \;\mathrm{T} magnetic field at a speed of 2.0 \;\mathrm{m/s} in the direction of the 8.0 \;\mathrm{m} side.
    1. Determine the induced voltage in the loop.
    2. Determine the direction of current.
    3. What would be the net force needed to keep the loop at a steady velocity?
    4. What is the electric field across the .50 \;\mathrm{m} long resistor?
    5. What is the power dissipated in the resistor?
  6. A small rectangular loop of wire 2.00 \;\mathrm{m} by 3.00 \;\mathrm{m} moves with a velocity of 80.0 \;\mathrm{m/s} in a non-uniform field that diminishes in the direction of motion uniformly by .0400 \;\mathrm{T/m} . Calculate the induced emf in the loop. What would be the direction of current?

Answers to Selected Problems

  1. .
  2. 1.2 \times 105 \;\mathrm{V} , counterclockwise
  3. a. 15 \;\mathrm{V} b. Counter-clockwise
  4. a. 2 \times 10^{-5} \;\mathrm{T} b. Into the page c. 2.8 \;\mathrm{N/m} d. CW
  5. a. 0 .8 \;\mathrm{V} b. CCW c. .064 \;\mathrm{N} d. .16 \;\mathrm{N/C} e. .13 \;\mathrm{w}
  6. 19.2 \;\mathrm{V}

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