How can you produce an electric current in a copper ring, using just a magnet?

Easy. Just keep moving the magnet in and out of the ring. In the diagram above, an anticlockwise current (looking leftward) is produced when the magnet is moving towards the ring. A clockwise current (looking leftward) is produced by the magnet is moving away from the ring. But no current is produced in the ring when the magnet is stationary (regardless of the position of the magnet).

It is also observed that the faster the magnet moves, the larger the current. Also, the larger the area of the ring, the larger the current.

So what’s happening? Michael Faraday had it all figured out in 1831. He created this concept of magnetic flux linkage of a coil (introduced in the previous section). If the flux linkage is constant, nothing happens. But whenever the flux linkage changes, then an emf e is induced in the coil. Furthermore, the magnitude of the induced emf is proportional to the rate of change of magnetic flux linkage. All these is encapsulated in Faraday’s Law of Induction, which can be expressed mathematically as $\displaystyle \varepsilon =-\frac{{d\Phi }}{{dt}}$

Let’s go back and see if we can explain the magnet and coil phenomena now.

Recall that the magnetic field of a bar magnet is not a uniform one. It is stronger near the poles and weaker further away. So when the distance between the magnet and the coil changes, so does the flux linkage of the coil. As Faraday pointed out, a changing flux linkage (of the coil) is always accompanied by an induced emf (in the coil). Since the coil provides a complete conducting path, an induced current is produced whenever the magnet moves.

Next, we realize that the faster the magnet moves, the faster the rate of change of flux linkage and thus the stronger the induced emf. This explains why the induced current is stronger when the magnet moves faster.

It is crucial to understand that change is required for induction. If the magnet does not move, it does not matter how strong or how near the coil the magnet is. In other words, if $\displaystyle \frac{{d\Phi }}{{dt}}=0$, the induced emf is zero, regardless of the value of Φ.

Now, why is the negative sign doing in $\displaystyle \varepsilon =-\frac{{d\Phi }}{{dt}}$? Ah, that brings us to Lenz’s Law.

Demonstrations

Magnet in-and-out of Solenoid 1

Magnet in-and-out of Solenoid 2

Bungee Magnet and Coil

Concept Test

3010

Comics