One of the Hardest A Level Physics Questions Ever

Hinzugefügt: 2026-06-02

[00:00:00]In 2024, there was an A-level physics question that was part of a paper which caused lots of difficulties. The grade boundaries were extremely low, and that also produced a lot of interesting comments on YouTube. I think the electromagnetism question was one of the hardest electromagnetism and in general A-level physics questions ever given.

[00:00:19]So, what was it? It starts off with an interesting situation. So, imagine that we have a rod which is going into the page, so a little bit like this. And then let's say that the ground was, I think, 8 m. And then we also had a magnetic field which was going in this direction.

[00:00:40]Like that.

[00:00:42]The angle here was 68°.

[00:00:46]It was then asking us to calculate the amount of EMF induced across the ends of the rod as it's falling. We're also given the length of the actual rod.

[00:00:57]Let's say that L was, I think it was 2 m. And that's 2 m into the page. So, already we're starting to see that the problem is relatively hard to see in an exam situation because it's a little bit tricky to imagine. The EMF and the rod are going into the page. We're looking at it from the side. Okay. Well, how are we going to calculate this? We know that the magnitude, and I'll do the magnitude only, of the induced EMF is going to be given by delta NBA over delta t. The component of the field will be the same, but during the motion, the rod will sweep out an area. This area is kind of interesting to imagine.

[00:01:39]Let's think of this from the side. We're going to have a rod which kind of falls like this.

[00:01:46]And it sweep sweeps out an area which is going to be 2 m by 8 m. Again, I'm looking at the rod kind of falling like this. Okay, so we know what our area is going to be. What about the component of the B field? Well, we need the component which actually cuts through that area.

[00:02:07]So, therefore, I'm going to take the cosine component. This here is B. That's a terrible straight line.

[00:02:16]So, I'm going to say that I'm going to take B cosine of theta. Theta in this case will just 68, so might as well just write 68°.

[00:02:23]Okay, so N will just be one in this case. We are going to have B. Oh, we're also given the magnetic field, by the way. B was just 4.9 * 10 ^ -5 T, which is essentially the magnetic field due to the, well, in the surface of Earth in most places. Okay. So, this here is just going to be 1 * B, which is 4.9 * 10 ^ -5.

[00:02:53]Then, I'm going to take the perpendicular component of B, which is the cosine component.

[00:02:58]Cosine of 68.

[00:03:01]That's B cos theta. My area, well, the area swept will just be 2 * 8, but I still don't know my delta T.

[00:03:13]Okay, I can probably figure this out because, just on the side here, we have an object which falls down under gravity. And if the object falls down under gravity, we can probably just use Let's do this over here. Uh S is equal to UT + 1/2 AT squared. There's no initial velocity.

[00:03:35]We know that S is H. We know A is 9.81, so we can just say that T will be the square root of 2S over A. So, what is that going to be? 2 * 8.0 / by acceleration, which is 9.81.

[00:03:55]This here will just give us around 1.277 seconds.

[00:04:03]Okay, we have the time. So, this can go into here. So, it's going to be 1.2 77. And now, we can just calculate our induced EMF, which up to, let's say, two significant figures should be just 2.3 * 10 to the -4 V.

[00:04:22]We can make sense of this answer because this answer is not very, very big. And after all, it's the EMF induced on a rod that's falling due to the magnetic field on Earth. Okay, the second part of this question is much more interesting.

[00:04:38]Let's clean the board, and then we can solve it.

[00:04:40]So, in the second part of this question, we actually put the rod in a non-conducting support, a little bit like this. Now, the rod is able to essentially rotate around this hinge.

[00:04:58]So, it can do this and that.

[00:05:02]Essentially, we're going to be kind of following a circular curve this way. I know this is not circular, but please bear with me.

[00:05:10]And kind of like this this way. What the question was asking was what are some differences to the EMF induced if the rod falls to the left and if it falls to the right.

[00:05:25]Tricky.

[00:05:26]Now, let's consider the geometry of the situation. First of all, if we're falling to the left, we're going to reach this point right over here.

[00:05:35]Notice something which is highly interesting. If that's the case, our velocity of the rod V will be along the field line. Let me just summarize it.

[00:05:49]So, there will be a point here where V is parallel to the field line. And that point is right over here. When they're parallel, the rod will not be cutting through the field lines. So, that means that the induced EMF here must be zero.

[00:06:04]If If it was some other value, and if it reached zero, that means that in this region the induced EMF was decreasing.

[00:06:11]It's passed through zero, so that means it changes sign. And then in this region, the magnitude of E is increasing yet again. It's also reminding me of this portion of the curve. So, imagine the cosine graph. So, it was, let's say, somewhere here. It reaches through zero, and then it starts increasing in magnitude yet again. Prior to that, the EMF was decreasing. So, let's just say that initially it decreases. It's parallel to be at this exact point is zero, and that point, let's call it uh point one or something. After point one, the EMF is increasing yet again with the opposite sign. Tricky to spot. Now, if the falls in the other direction, this is going to be interesting, because it's impossible for the geometry of the situation to give us that the rod is parallel to the field lines yet again, or the velocity is parallel to the field lines. In fact, we're going to reach a region maybe somewhere here where it's perpendicular to the field lines. If you imagine it kind of like this, approximately.

[00:07:23]So, what we're going to get is we're going to get a maximum.

[00:07:26]So, if we're falling to the right, we're going to get this region. Let's call that point two.

[00:07:33]So, at point two, the speed of the rod, and in the exam I would word this uh very carefully, is perpendicular to the field lines.

[00:07:44]Because of that, because it's perpendicular to the field lines, there will be a maximum flux. It will be cutting for the maximum amount of field lines. After it's reached the max, the magnitude of E will be decreasing. It will still be with the same sign because we've not passed through zero. Just before that though, if this here was a maximum, that means that in this region, the EMF must have increased.

[00:08:08]Here, it must have decreased. Here, at two, uh is maximum. So, just before that, we can also say that the EMF was increasing. In other words, it was probably, if you think about a cosine graph, uh it was probably somewhere within like, let's say, this region. It increases, it reaches a maximum, and decreases, but it remains in the same direction.

[00:08:33]Now, these are just notes. I've not worded this properly, and this is why I think that in an exam situation, as part of uh what is overall really hard paper, this makes this one of the hardest A-level physics questions of of all time. Do let me know how did you find it and what you think of it. And if you're revising for A-level physics tomorrow, first of all, good luck, but you also need to check out my explanation on binding energy, and I'm going to leave this video right over here. Good luck.