Video Transcript
Part (a) of the diagram shows a bar magnet moving at a speed 𝑣 toward a stationary solenoid. This induces an electrical potential difference across the two ends of the solenoid. Part (b) of the diagram shows a stationary bar magnet with a solenoid moving toward it at a speed 𝑣. How is the potential difference induced in the solenoid in part (b) different from that in part (a)?
Okay, so, in this question, what we’re dealing with is a bar magnet and a solenoid, or a coil of wire. And, of course, in the case of a solenoid, the coil of wire has multiple turns on it. Now, in part a of the diagram, what we’ve got going on is that the bar magnet is being moved towards the coil of wire, the solenoid, whilst the solenoid remains stationary.
And we’ve been told that as a result of us moving the bar magnet towards the solenoid, an emf, or electromotive force, or in other words, a voltage, is induced across the two ends of the solenoid. Now, this phenomenon is electromagnetic induction. And in this phenomenon, an emf is generated across the ends of the solenoid because the magnetic field through the area of the solenoid is changing. And the reason that the magnetic field is changing is because we’re moving the magnet towards the solenoid.
Now, let’s recall that the emf generated is dependent on quite a few factors. And one of these factors is the rate of change of the magnetic field through the area of the solenoid, in other words, how quickly the magnetic field from the bar magnet is changing as it moves through the solenoid. We can even draw the magnetic field lines representing the magnetic field from the bar magnet. And we can see that as the bar magnet moves into the coil the magnetic field changes inside the area of the coil.
So, it’s this change in magnetic field in the coil and, more specifically, the rate of change of this magnetic field, that directly influences the magnitude, or size, of the potential difference generated across the solenoid. Now, what we’re being asked in the question here is that in part b, instead of us moving the bar magnet towards the solenoid, we keep the bar magnet in exactly the same place and move the solenoid toward the bar magnet instead. We need to find out how the emf generated across the solenoid differs in this case compared to the first case where we move the bar magnet toward the solenoid.
Now, the important thing to note is that we’re keeping everything else the same. The bar magnet itself is exactly the same. And so, the magnetic field it generates is exactly the same. The solenoid is the same solenoid as before. And so, the number of turns on that solenoid is not changing either. We’re not deforming the solenoid in any way, so the area through the coils is exactly the same as before as well. And we’re not rotating the bar magnet or the solenoid with respect to each other.
So, everything is the same as it was before. The only difference is that, in diagram b, the solenoid is moving toward the bar magnet, whereas, in diagram a, the bar magnet was moving toward the solenoid. Also, to note in both cases, the velocity of the moving object is 𝑣. It’s exactly the same in both cases. In diagram a, the bar magnet is moving at a velocity 𝑣, and in diagram b the solenoid is moving at a velocity 𝑣.
And so, let’s recall that we said earlier that electromagnetic induction is dependent on the rate of change of magnetic field through the area of the coil, where the area is the cross-sectional area that we can’t quite see because we’re looking at it side-on. Now, it doesn’t matter whether we remove the bar magnet toward the solenoid or the solenoid toward the bar magnet. As long as the relative speed between the two is exactly the same, the rate of change of magnetic field through the solenoid is going to be the same.
Because, look, in both cases, the two objects are moving toward each other, and both cases, they’re moving at the same speed toward each other. So, every second, the same amount of magnetic field from the bar magnet itself is entering the solenoid. And therefore, the emf induced across the solenoid is going to be exactly the same in both cases. And so, at this point, we’ve found the answer to our question. When asked how is the potential difference induced in the solenoid in part b different from that in part a, we can say that the induced potential difference is the same.
