What is Solenoid? - Definition, Uses & Examples

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  • 0:00 What is a Solenoid?
  • 0:30 How Do Electromagnet…
  • 1:20 Uses of Electromagnet…
  • 2:45 Lesson Summary
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Lesson Transcript
David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

Expert Contributor
Elaine Chan

Dr. Chan has a Ph.D. from the U. of California, Berkeley. She has done research and teaching in mathematics and physical sciences.

In this lesson, you will learn what a solenoid is, how it relates to electromagnets, and how it is used in a few real-life situations. A short quiz will follow.

What is a Solenoid?

Solenoids and electromagnets technically are not the same thing, but people talk as if they are. A solenoid is just a coil of wire, but when you run a current through it, you create an electromagnet. Since this is by far the most useful application of a coil of wire, it's not surprising that when you say the word 'solenoid,' people tend to assume you mean electromagnet. Electromagnets are particularly useful because, unlike regular magnets, they can be switched on and off, and strengthened by increasing the current flowing through them.

How Do Electromagnet Solenoids Work?

When a lazy charge sits on its couch, doing nothing, it is surrounded by an electric field. This makes sense, because it's an electric charge, after all. But once that charge gets some motivation and goes for a run around the block, suddenly it produces a magnetic field. This might strike you as odd, and you wouldn't be alone! As physicists figured out later, both fields are part of the same force of nature: electromagnetism.

Because of this, we can create a magnet by simply running a current through a wire. When we run a current through a solenoid, however, we get a super strong magnet because the magnetic field is concentrated inside the coil. This can be incredibly useful in our everyday lives.

Uses of Electromagnet Solenoids

Electromagnetic solenoids find uses all over the world. They're in hotel door locks, water-pressure valves in air conditioning systems, MRI machines, hard disk drives, speakers, microphones, power plants, and cars. You can hardly swing a bat without hitting a solenoid.

Speakers and microphones, for example, both contain solenoids. In fact, a speaker and microphone are pretty much exactly the same thing in reverse of each other. A speaker takes electrical signals and runs it through a solenoid to create motion; that motion drives the speaker and creates a sound. A microphone does the opposite; your voice pushes the solenoid back and forth, and that motion of the solenoid creates an electrical signal that can be used to create the sound elsewhere. Without solenoids, we wouldn't be able to record or reproduce sound at all.

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Additional Activities

Calculate the Magnetic Field of a Solenoid

Ampere's Law applied to a solenoid gives the magnetic field as (mu-zero)(n)(I), where mu-zero is a constant, n is the number of loops per unit length, and I is the current. The magnetic field is uniform in the direction of the axis of the solenoid, using the right-hand rule. This is strictly true only for an infinite solenoid but is a good approximation in practice when not close to the ends of the solenoid.

Problem 1:

Calculate the magnetic field inside, near the center of a thin 10 cm long solenoid of 400 turns of wire carrying 2 A of current.

Answer 1:

The number of turns per unit length n= 400 / 0.1m = 4000 per meter

mu-zero is a constant equal to 4*pi/10000000 with units of Tesla meter per ampere

I is the current in amperes (A)

B = mu-zero (n)(I)= (12.57 / 10000000 )(4000) (2) = 0.01 T (Teslas)

Wire Wrapped Around Solenoid

The magnetic flux passing through a loop of area A is equal to the product of the area and the component of the magnetic field perpendicular to the area (for a uniform field). If the flux changes in time, there is an electromotive force produced that is equal to the negative of the time rate of change of the flux, multiplied by the number of loops.

Problem 2:

A wire is wrapped with N2 turns around a solenoid with N1 turns, length L, cross-sectional area A, and a current changing at rate dI/dt. What is the electromotive force in the wire that is wrapped around the solenoid?

Answer 2:

emf = -N2 (mu-zero N1 A/ L) dI/dt

mu-zero is a constant equal to 4*pi/10000000 with units of Tesla meter per ampere

A is the area in meters squared (not amps)

L is the length of the solenoid in meters

dI/dt is the rate of change of current with time in amps per second

The units on the right-hand side are Tesla meters squared per second and this is equal to volts.

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