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### AuthorTopic: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]  (Read 713 times) Tweet Share

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#### Jefferson

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##### Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« on: August 14, 2019, 06:09:59 pm »
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Can someone explain to me why the graph looks the way it does (and not flipped the other way, e.g positive first peak)
Also, why is the (inflexion?) point where it changes polarity not instantaneous and is dragged out over a longer time?

Thanks

#### myopic_owl22

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #1 on: August 14, 2019, 06:42:35 pm »
+2
Hello,
Here's what I'm thinking - mightn't be spot on but here we go...

Can someone explain to me why the graph looks the way it does (and not flipped the other way, e.g positive first peak)

Imagine the solenoid as a battery (similar to generators), in that a current is being produced. So, like in a generator, (or in the little bit between two battery terminals) the conventional current goes from negative to positive. Other things outside of this will have current flowing in the normal, conventional way from positive to negative. These 'other things' could be light globes, etc. In this case, it's a voltmeter.
Now looking at the current that the magnet induces in the solenoid. As the south pole moves towards the top of the solenoid, a south pole is induced to oppose the change (Lenz's Law). Using the right-hand grip rule, the conventional current moves from the top to the bottom of the solenoid. In between the terminals of the voltmeter/ datalogger is what we're looking for, as that's when the current leaves the 'battery' and enters the circuit. Hence -ve to +ve. The first blip is in the negative direction to reflect this. The second shows the current moving from the bottom of the solenoid to the top, so from the datalogger's POV, current is going from +ve to -ve.

Also, why is the (inflexion?) point where it changes polarity not instantaneous and is dragged out over a longer time?
From what I'm guessing, we're supposed to assume that the solenoid of considerable length. In which case, there will be some time when the magnet is completely inside the solenoid ( and still falling down). It won't be inducing any current during this time, so the voltmeter won't pick it up. I'm not sure on the specifics of why this is the case, though. Perhaps it's because while it's inside, a south pole wants to be generated at the top as the north pole at the top of the magnet leaves, it - but at the same time a south pole wants to be generated at the bottom of the solenoid to repel the magnet's south pole as it accelerates towards it (therefore while there is an EMF, no net current is produced)? Please don't take my word for it, though.

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#### DrDusk

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #2 on: August 14, 2019, 07:09:21 pm »
+3
From what I'm guessing, we're supposed to assume that the solenoid of considerable length. In which case, there will be some time when the magnet is completely inside the solenoid ( and still falling down). It won't be inducing any current during this time, so the voltmeter won't pick it up. I'm not sure on the specifics of why this is the case, though. Perhaps it's because while it's inside, a south pole wants to be generated at the top as the north pole at the top of the magnet leaves, it - but at the same time a south pole wants to be generated at the bottom of the solenoid to repel the magnet's south pole as it accelerates towards it (therefore while there is an EMF, no net current is produced)? Please don't take my word for it, though.
This is a bad explanation. An EMF is produced when there is a change in Magnetic FLUX. Think of it like this. When the magnetic is entering the solenoid, the Area that the magnetic field encompasses is constantly becoming larger. This is easy to see, as constantly more an more parts of the magnetic field are entering the Solenoid.

However when it is fully inside the solenoid(specifically at the centre), the area of the Solenoid that the Magnetic field encompasses above the middle will be the same as below. So there is no NET change in Magnetic Flux, thus the NET change in EMF is also zero.

Also the reason the second Peak is narrower is because the Magnet has a greater velocity when exiting, as it is constantly being accelerated by gravity, and thus the time it takes to exist the Solenoid will be shorter than the time it takes to enter. So we have

$V = -N\frac{\Delta \phi}{\Delta t}$

According to this to produce the same change in Magnetic flux, it takes a shorter amount of time, this skews the graph into becoming thinner at the second peak.
« Last Edit: August 14, 2019, 08:31:48 pm by DrDusk »
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#### vox nihili

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #3 on: August 14, 2019, 09:57:54 pm »
+2
This is a bad explanation. An EMF is produced when there is a change in Magnetic FLUX. Think of it like this. When the magnetic is entering the solenoid, the Area that the magnetic field encompasses is constantly becoming larger. This is easy to see, as constantly more an more parts of the magnetic field are entering the Solenoid.

However when it is fully inside the solenoid(specifically at the centre), the area of the Solenoid that the Magnetic field encompasses above the middle will be the same as below. So there is no NET change in Magnetic Flux, thus the NET change in EMF is also zero.

Also the reason the second Peak is narrower is because the Magnet has a greater velocity when exiting, as it is constantly being accelerated by gravity, and thus the time it takes to exist the Solenoid will be shorter than the time it takes to enter. So we have

$V = -N\frac{\Delta \phi}{\Delta t}$

According to this to produce the same change in Magnetic flux, it takes a shorter amount of time, this skews the graph into becoming thinner at the second peak.

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#### DrDusk

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #4 on: August 14, 2019, 10:03:51 pm »
0
A gentle reminder that there are more tactful ways of correcting people and that using said ways is generally a good plan.

I'm sorry but I don't get what you are trying to tell me. Did I sound rude or something?

Sorry if that's the case!
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#### myopic_owl22

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #5 on: August 16, 2019, 05:30:29 pm »
+2
Yeah, I wasn't quite sure about that second question - your explanation is so so much better! Hopefully OP finds it too
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#### DrDusk

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #6 on: August 18, 2019, 03:25:06 am »
+2
Yeah, I wasn't quite sure about that second question - your explanation is so so much better! Hopefully OP finds it too
No worries. Apologies if I sounded rude, that definitely was not my intention
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#### Jefferson

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #7 on: August 25, 2019, 07:28:02 pm »
0
This is a bad explanation. An EMF is produced when there is a change in Magnetic FLUX. Think of it like this. When the magnetic is entering the solenoid, the Area that the magnetic field encompasses is constantly becoming larger. This is easy to see, as constantly more an more parts of the magnetic field are entering the Solenoid.

However when it is fully inside the solenoid(specifically at the centre), the area of the Solenoid that the Magnetic field encompasses above the middle will be the same as below. So there is no NET change in Magnetic Flux, thus the NET change in EMF is also zero.

Also the reason the second Peak is narrower is because the Magnet has a greater velocity when exiting, as it is constantly being accelerated by gravity, and thus the time it takes to exist the Solenoid will be shorter than the time it takes to enter. So we have

$V = -N\frac{\Delta \phi}{\Delta t}$

According to this to produce the same change in Magnetic flux, it takes a shorter amount of time, this skews the graph into becoming thinner at the second peak.

My second question was that why is there a "horizontal point of inflexion" in the graph. Why is it that the induced voltage is 0 for the entire duration of when the magnet is inside the solenoid (middle section of graph).
I was expecting it to be a "normal point of inflexion", where it just cuts the horizontal axis (exact moment where magnet is in the centre) and then goes up. Not sure if that makes sense.

#### S200

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##### Re: Dropping Bar Magnet Through Solenoid (Induced EMF) [Nesa Sample Question]
« Reply #8 on: August 25, 2019, 07:35:16 pm »
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There is only induced current when the magnet is either entering or exiting the soleoid...

in the middle space, the induced voltage is zero...
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