Electromagnetic Induction | Induced EMF Due To a Falling Magnet | A-level Physics
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- Опубликовано: 16 окт 2024
- This video discusses the induced emf captured by a data logger as a bar magnet falls through a coil. This is a classic example of EM induction, and we discuss the following questions about the emf (ɛ) versus time graph:
1. Why do the peak and trough develop?
2. Why is ɛ = 0V between 0.2 and 0.3s?
3. Why is the magnitude of the trough bigger than that of the peak?
4. Does the magnet accelerate at g throughout?
5. Is there anything interesting about the 'area under' the curve?
/ @forestlearn
for the question i'm doing, the trough is first and then the peak, basically the opposite of the graph in this video. how is that so? the magnet in the question i'm doing is also being dropped into the coil like the one in the video so i don't understand why the graph is different?
Thanks for your question - it's a good one! Just to be clear, I think you mean that your graph is a reflection in the horizontal axis of the graph in the video, right? There's actually nothing fundamentally different about the two graphs - they both describe exactly the same physics. In the video, we're assuming that the 'normal line' (which is used to calculate magnetic flux) is pointing up. As a result, as the magnet falls into the coil, the flux linkage changes from zero to a negative value - i.e. the change in magnetic flux is negative. From Faraday's law, this means the induced emf would be positive (as in the video graph). [You may want to check out our video on magnetic flux if the preceding explanation re flux isn't clear.] Had we chosen the 'normal line' to be pointing down though (and we're free to choose as we please), the change in flux would be positive and the induced emf would be negative - as is the case with your graph. Importantly, there's no difference in the underlying physics - both choices/graphs 'encode' exactly the same induced currents and magnetic poles in the coil discussed in the video. I could've made the video with your graph - the actual physics discussed would've been identical. Hope this makes some sense (sorry it's a bit long!), don't worry too much if it doesn't quite all click, it's a subtlety that goes beyond A-level. Happy to answer any follow ups you might have!
@@ForestLearn makes so much sense! thank you so much :)
@@dearlantsov That's great to hear, it's my pleasure :)
amazing video and thanks! How can i work out the results if I connect the falling bar magnet through a spring so it falls and rises continuously?
Glad it was helpful! Sounds like an interesting investigation - do the experiment and then try to explain the results using your theoretical understanding :)
I am checking out your channel after reading the article you published on chatphysics. I would love to see more phd graduates show this much dedication to increasing standards in teaching A-level physics.
Thanks :)
Thanks for the video! I have one question:
When exactly do the peaks happen? Is it when the middle of the magnet enters the coil, when the bottom of the magnet enters the coil or something else entirely?
Thanks for watching :) The peak and trough occur when the magnet is fairly close to the coil, but not inside the coil (see the magnet positions at 3:00 in the vid, for example). At these positions, the rate of change of flux linkage is greatest, corresponding to max values of induced emf. As I discuss for the second question in the vid, when the middle of the magnet passes through the coil, the induced emf vanishes for an instant. Hope this helps - let me know if you have any further questions.
Im doing my a levels and this is by far the best explanation of this topic. Many thanks for the playlist :).
Thanks for the kind words :) Will be adding to the playlist soon! All the best with your A-level exams!
I can't find the other videos for 'type 1 and type 3.' This video is here is type 2.
Type 3 doesn't exist yet (unfortunately), this vid introduces type 1: ruclips.net/video/ZkO_fBTbtec/видео.html
@@ForestLearn It doesn't matter. This topic didn't come up in paper 2. Well apart from one question on transformers in the multiple choice.
Amazing video!
Answered every question that i was unsure of!
Thanks, great to hear that :)
Very Informative, great video!
Thanks :)
is there any function that fits the curve and brings any data about the magnetic field passing through?
Thanks for your question! I'm unaware of a function that fits the curve - in principle, you could simulate what's going on here computationally (i.e. code the equations of motion etc) and arrive at the induced emf graph. You could then numerically (computationally) integrate the induced emf with respect to time to arrive at how the flux linkage changes with time. Hope this helps!
@@ForestLearn Thanks, luckily someone already did the math in physics forum, if anyone needs it , tittle of the post is "how to model a magnet falling through solenoid"
@@sirlimonada2 Many thanks for sharing this!
would the graph change at all if the south end was at the bottom? would that result in a reflection in the time axis? Thanks for the video!
That's correct! Apologies for the late response.
nice video！ But i still have some questions
1, Why does the gradient of the graph increase, instead of the straight line?
2. I don't understand why the magnetic field is uniform at the middle ?
Thanks for your questions!
1. The gradient of the graph (emf vs time) corresponds to the second derivative of the flux linkage versus time graph. If you don't know what this means, I'd say don't worry! Basically, how exactly the flux linkage changes with time determines what the emf vs time graph looks like, including changes of the gradient. This isn't something you'd be asked about for A-level/high school physics - at least I hope not :)
2. As I mentioned in the video, you may find it helpful to check out the first worked example in the following video where I discuss in more detail the uniform magnetic field inside a bar magnet: ruclips.net/video/vMPSzMwee9s/видео.html
Let me know if you have any further questions!
Finally found the best explanation for this graph 😎🔥
Thanks :)
Great video! How do you know the direction of flow of the induced emf?
Thanks for watching and the kind words :) Induced emf doesn't flow, so best to avoid saying that. We can certainly talk about the direction of the induced current tho, and this is determined by whether the induced emf is positive or negative (the 'direction' of the induced emf). Check out the video between 1:33 to 3:14 where I discuss all this at length, and let me know if you still have any questions.
Why does the direction of rotation of the induced current switch halfway through the tube
Thanks for your question - you may want to re-watch the video around the 2:12 mark. It's related to why the poles of the coil switch, which can be explained by Lenz's law (check out the Lenz's law revisited video for more explanation on this: ruclips.net/video/y2c36W9QhzE/видео.html ). Let me know if you have further questions.
Very informative, very well explained. Thanks!
Thanks for watching, glad you found it useful :)
Sorry this might not be totally related to the video but I have a question. Is there a way to theoretically do this without experimental data?
Like i wanna find out induced emf theoretically .
Thanks for the question! One could simulate what's going on here computationally (i.e. code the equations of motion etc) and arrive at the induced emf graph. Probably if you search for this you'll find examples of this already. I'm not sure about deriving it analytically (pen and paper) though, seems rather tricky to do that ....
@@ForestLearn thanks for the reply . I really wanna solve this analytically , I'll look into solving it using a program if i can't 👍
Lenz Law is 2nd Law of Thermodynamics
It signify the continious mechanical energy required to get and electical energy
Why is the induced emf initially positive? When the magnet falls through the coil, isn't the change in magnetic flux linkage positive due to more field lines? A positive value in Faraday's Law would make the induced emf a negative value?
Thanks for your question! In the video, we're assuming that the 'normal line' (which is used to calculate magnetic flux) is pointing up. As a result, as the magnet falls into the coil, the flux linkage changes from zero to a negative value - i.e. the change in magnetic flux is negative. From Faraday's law, this means the induced emf would be positive. (There was a similar question to yours asked that I've pinned in the comments, you might want to check out my reply to that as well.) Let me know if anything is unclear and and if you still have any questions.
Thankuu, heared about this trickier topic, but u made it super easy 🎉 plz do upload it's pps😢
Thanks, happy to hear that :) What do you mean by 'upload it's pps' ?
Hey! I got a question, I did this experience in experimental physics, but my graph of induced voltage vs time has a trough and then a peak, where the peak is greater than the trough. Why is that?
Thanks for your question! At least one other person has asked the same question. See my answer to dearlantsov, which is pinned in the comments. Hope that helps, let me know if you still have further questions.
@@ForestLearn Thank you, I read your previous answer and it did help! However, the following I found curious: I carried out this experience while studying induced voltage vs position. On Day 1, my experimental graphs showed a peak and then a trough. On Day 2 I repeated the experiment and my graphs showed a trough and then a peak! I hadn’t changed anything, the magnet, the pole with which it fell, the coil, all were the same. I was baffled. But a couple of hours ago, I found a lecture on EM induction by MIT professor Walter Lewin. At the end he explains that it is the non-conservative electric field due to the changing magnetic flux that causes two voltmeters connected to the same points to read different results. Also, the graphs from the two voltmeters were the inverse of the other!
I think this explains my data: in day 1 I may have connected the voltmeter differently than in day 2. I’m not sure though… What do you think?
@@danielavargas3091 Yes, most likely - if you have a voltmeter connected across a component and the reading is 2V, switching the leads of the voltmeter will result in a reading of - 2V . That's basically how the two graphs are related to each other: you times one graph by minus one and you get the other graph (a reflection in the horizontal axis).
Hello i am confused as to why the trouph is greater than the peak... Why is the magnet faster through the second portion of its journey? To my understanding ,Initially the magnet is travelling at g and then it decelerates due to opposing magnetic field from the induced current. Its acceleration decreases until it reaches a constant velocity. Eventually after passing the midway point the acceleration of the magnet starts increasing as it exists the coil, increasing to the initial g.
How come the magnet accelerates inside the coil? Shouldnt it be moving a constant velocity.
Hey, thanks for your question! You may want to check out the vid between 5:00 and 6:00 again. The magnet does not decelerate (slow down) at any point - for this to be the case, the upward magnetic force would have to be greater than the downward weight of the magnet. It also doesn't reach constant velocity - for this to be the case, the two forces would be equal in magnitude. When moving through the middle of the coil, the acceleration is g for an instant - for a moment, there is no induced emf or current and the magnetic force is zero (look back at the vid for discussion of this).
What is true though is that the acceleration is less than g during various intervals of the magnet's journey (the magnetic force is less than W=mg in magnitude, meaning a resultant force downwards, meaning the magnet picks up speed). This means that by the time the magnet exits the coil, it's moving faster than when it entered, which explains why the trough is bigger than the peak.
Hope this helps, let me know if you have any further questions!
@@ForestLearn oh wow i was thinking this concept fundamentally wrong. Thank you for clearing that up, your doing gods work pal, or no-ones work if your not religious. Any how thanks so much 😂
@@clement_jacob My pleasure, and thanks so much for the awfully kind words - it means a lot to me :)
Why would the magnet take a longer time to reach the ground than if it had been dropped outside of the coil? Is it because the coil and the magnetic field and the induced emf slow it down?
The coil's magnetic field repels the magnet as it enters, and attracts the magnet as it exits. As a result, the acceleration of the magnet is less than g (acceleration due to gravity) on entry and exit, meaning it would take longer to reach the ground compared to something dropped outside the coil. See the discussion at 5:00 . Hope this helps, let me know if you still have any questions.
oh crap right tysm for your response i got it now!
Great!
I have a question but its not related exaclty to this video, when a queation asks work out the Average emf on a rotating coil in a time frame lets say 0.1 seconds.
And say the coil rotates 90 degrees in 0.1 seconds.
Would you have to divide the total emf by 2 to get the Average Emf? Because the emf goes from a maximum to a minimum in the time frame.
If you need some context on this question i can link the question.
I would appreciate the help sir.
You need to work out the change in flux linkage (due to the coil rotating 90 degrees) and divide that change by 0.1s , which gives you the average induced emf over the 0.1s (this is Faraday's law of induction). You probably mean that the flux linkage goes from max to min in 0.1s . Let me know if this helps and if you need further clarification.
@@ForestLearnWhat i dont understand is, how is that the average emf? Isnt that just the peak emf?
@@clement_jacob The induced emf at any moment is given by the gradient of the flux linkage vs time graph at that moment. The peak emf will therefore be given by the gradient of the steepest part of the flux linkage vs time graph. You may find the discussion of graphs in the latter part of this video useful: ruclips.net/video/kO73UYmJiW0/видео.html
Let me know if you need any further help.
@@ForestLearn I understand thank you very much :)
@@clement_jacob Glad to hear that!
Best explanation 🥰
Thanks for the kind words :)
What would happen if the falling magnet wasn’t a bar magnet but a horseshoe magnet?
Interesting! Judging by the magnetic field lines around such a magnet, I'd expect there to be no (or very little) induced emf, assuming it's dropped 'poles first' (poles closest to the ground). Can you see why?
@@ForestLearn Oh, maybe because both poles induce a current in the coil in the opposite direction of the other, cancelling each other?
What about an Alnico button magnet, for instance? I’ve been researching and I found its magnetic field resembles that of the horseshoe’s rather than the bar’s, so I’d expect a similar result for the induced emf. Or would there be any difference?
@@beccacross7880 Apologies for late reply. No, it's not correct to talk about opposite currents cancelling out. There is however a cancellation of magnetic flux in the following sense: the field lines pointing down away from the north pole would result in negative flux (once the magnet is close to the coil and the field lines are 'poking through' the coil). However, the field lines pointing up toward the south pole would result in positive flux, of same magnitude as negative flux, therefore cancelling it. With no net magnetic flux, there would thus be no change in flux (remains zero) and therefore no induced emf. Hope this makes sense, if anything needs clarifying feel free to ask. You might find it useful looking at a picture of the field lines around a horseshoe magnet to follow the explanation I've given.
@@beccacross7880 I see no reason why your expectation wouldn't hold :)
when u raise a magnet weighing 100 grams upto a hieght of 100 meters, u put 100 joules of energy into it, when u drop that 100 grams magnet through the coil, the magnet fall takes 2 seconds and the coil generates 50 joules of energy.
is there a way to generate more energy from the falling magnet of same weight from same hieght?
yes, just increase the winding on the coil,
now the coil has 4 times winding than it was before, now raising the magnet to same hieght takes the same amount of energy (=100 joules), but dropping the same magnet through the coil of bigger winding now generates more energy, and the fall of magnet takes more time (let's say 8 seconds) because of more back emf because of more current in coil.
can you make and show graph and calculations of this situation please?