 Hi, I'm Zor. Welcome to Unizor Education. Today I would like to talk about a special device which is called solenoid. It's very widely used in whatever we deal with related to electricity. It's just everywhere. Almost all the different devices have this type of device. Anyway, it's very important and it's also very important from educational point of view because the calculations related to magnetic field inside the solenoid are, well, I would say, exemplary as far as basically most of the calculations in physics are. It's a combination of good math and certain physical properties which we come up with from experiments or something like that. So it's very important to be familiar with a regular course of mathematics which includes calculus and simple integration because this particular lecture will involve vectors and calculus. That's what I'm always talking about. That at least these two parts of the math are very important. Now, this lecture is part of the course called Physics for Teens presented on Unizor.com and on the same website you can find the prerequisite course which is called Math for Teens. So I do recommend you to watch the lecture from the website because you might actually find it on YouTube where it's physically located but it's linked from the website and the website is important in the sense that every lecture has textual notes and it's basically like a textbook because everything whatever I'm talking is already there written down, maybe even better than the way I'm explaining this. So you can use it as a both as a textbook and as a source of visual representation. And again, the Math for Teens, the prerequisite course is very important. It's also on the same website and the site is completely free by the way. No advertisement. Alright, so let's go to our solenoids. So first of all, what is a solenoid and what do we want to do with it? Well, solenoid is a simple thing. It's basically a spiral of a wire. It's connected to some kind of a source of energy, a battery. We are talking about direct current and we know that around any electric wire, if there is a current in it, there is a magnetic field. So there is a magnetic field around each of these loops. Now in the previous lecture, we were talking about one single loop and magnetic field in the center of this loop and we will use the same technique here. But in our case, we would like to evaluate the magnetic field inside the solenoid. Now the way how mathematicians usually consider physical problems, they are abstracting out certain purely physical aspects. So number one, we are considering that our solenoid is infinite in both ends. That's number one. Well, obviously you can start calculating if it's a finite, but then the calculations might be a little bit more intense and it's not for the educational purpose. Because this lecture is just educational purpose. It would like to basically explain how things are done in some very general and as simple as possible cases. So we are dealing with infinite in lengths solenoid. Number two, we are assuming that the wire is very thin. So there are no like inside the wire, the direction of the current is always along the wire. Nothing goes sidewise. So it's a thin wire and so how can we define this solenoid? Well, there are two major characteristics. Number one is radius and number two is how many loops you have per unit of lengths. So it's density of loops because you can actually have it very loosely or you can have it very densely. And obviously the field inside should depend on the density of the loops. So r, radius and density, n. So it's n loops per meter. Let's say if we are talking about c, the system of physical units and r is also meters in this case. Alright, so these two are given. Now, it's very important to understand that in our case we will calculate the intensity of the magnetic field at the axis inside the solenoid. Well, that's simpler because it's the same distance from the surrounding wire and it allows to calculate things simpler. Now, it is possible to calculate the intensity of this field not on the center line of the solenoid. So this is the solenoid. So this is the center line, the axis. But let's say a certain distance from this wire. And I don't really know exactly the result. The textbooks, and I didn't do the calculations myself, textbooks usually are saying that you can consider the field inside the solenoid everywhere inside the solenoid as uniform. Well, being as it may, we are not really talking about calculation at any point but on the axis. Alright, now it's also important to understand since our solenoid is infinite in both ends, it's completely symmetrical as far as which point exactly on the axis I will choose to measure my magnetic field. Because in both ends from this field you will have an infinite solenoid. So I can choose any point basically. It should be the same. Because of infinity it should be the same. If it's not infinite then you might say that if it's finite length that somewhere at the edge it might actually be not exactly the same as in the middle. And probably it will not in real practical cases. But for simplicity we assume it's infinite because the calculations are easier basically that's it. I mean it's still involved but just easier. Now so what I will do is I will position this solenoid in such a way that it goes that its axis coincides with x axis in my system of coordinates. And a point where I will measure my or calculate rather my magnetic field intensity will be just the beginning the origin of coordinates. So this is the zero point. This is the x axis. Now somewhere there is z and y axis doesn't really matter. But I am stretching my solenoid along the x axis and I'm measuring at the origin of coordinate point x is equal to zero. I'm measuring my magnetic intensity. Okay now what's the plan? How can I really calculate the magnetic intensity of an entire solenoid? Well we will do it in three steps. Step number one we will choose one particular loop. Now this is an x axis. I will choose one particular loop at a distance x from the origin of coordinates and I will try to calculate the magnetic field at this point. Now the previous lecture calculated at the center of this loop if you remember that was a previous lecture. Now in this lecture we will calculate a certain distance x from the center. And this is still a known radius, right? So that's my step number one. Step number two. Step number two I will have certain infinitesimal piece dx on my axis of solenoid and I will consider all the wires, all the loops which are inside this dx. And I will use obviously the density for this. And I will basically multiply whatever the magnetic field from one loop I will multiply by this number of loops to find out what is the contribution of a dx length of an entire solenoid contribution into the magnetic field of this at this point. And finally the third step is I will integrate the whole thing by dx by x from minus infinity to plus infinity because the solenoid is infinite in both ends. So again one loop first then I have all the loops which fit into the infinitesimal part of the solenoid along its axis and then I will integrate. So let's just do it one by one. Number one again let's go back to our primary more I would say most involved. If you have one loop and I would like to find out this is zero, this is x and this is r obviously. This is r. Let's put it inside. So how can I calculate the magnetic field from this loop? I will basically repeat the same logic as I was using when I was calculating with the center but it will be a little bit more involved. Not by much. Some geometry might be involved and some trigonometry and then integration. Okay so what I will do first is I will choose, yes before that let me just say that obviously all these pieces of this loop are in exactly the same position relative to point where I would like to measure calculate my magnetic field intensity because it's completely symmetrical right? This is the loop. Now the axis is perpendicular to it and so the distance from this point to any point on the on the circle is the same. Now so what I will do is I will choose one particular infinitesimal piece of the lengths ds differential of s and I will calculate what is the magnetic field intensity produced by this little piece only and then I will add all these magnetic field intensities. Well actually it's integration but it's a very simple integration because the magnetic field intensity at this point from this will be exactly the same. So the way how you calculate the result of the direct current in this point result of magnetic field intensity in this is something which we have discussed many times before. Now the basic formula is that intensity B is equal to zero and we did discuss this before times I. I is obviously the current it should be proportional to the current right? It's also proportional to the lengths of this segment and inversely proportional to square of a distance right? So d is a distance. Now more exact formula if you remember was I have to add 4 pi here. Now speaking about distance well when I was calculating the center my distance was the radius. Now when I'm calculating on the axis which is perpendicular to the circle but on the distance x from the center it would be what? It's a Pythagorean theorem right? This is x and this is r so the square of a distance would be r square plus x square. So this is something which we have started with but then if you remember if this is my current I this is my ds and if I'm measuring here which is different from here. Now this is perpendicular and this formula is good for the perpendicular. If I'm at the angle here then this segment is seen a little bit smaller right? So I have to multiply by a sign of this angle. Now let's see if I have this situation here. If this my segment is on a circle and the point is on the perpendicular to the plane then since this is perpendicular to the whole plane where this circle is located where this loop is located my segment belongs to the plane right? So my O x is always perpendicular to this particular O O A. My O A is always perpendicular to to the segment. It's probably easier to see if you draw a cone. If this is a cone this is our loop. This is point O. Now no matter where I take this particular segment this line would always be perpendicular okay? Because this segment is supposed to be a tangential right? So tangential is perpendicular to radius and this is perpendicular to the radius so it would be a right triangle if you wish but in any case it's kind of obvious that this particular line is perpendicular to any tangential line here. So we have the perpendicularity of this segment of this segment to the point and they don't have to add any kind of a sign of any angle here because this is always perpendicular. So this line which goes from the top of the cone to the circle at the base is always perpendicular to to the tangential line to a circle. That's kind of a very easy problem in space geometry and the game if you will go to the math 14 course here there is the whole chapter on space geometry which I do recommend you to know. So in this case this is something which we are using from the geometry which actually allows us not to put any kind of a sign of anything because this segment is always perpendicular. If it's here it's perpendicular if it's here it's always perpendicular to the distance to the vector to my point wherever I'm measuring this. Okay so this is a contribution of one particular segment. Okay now what should I do next? Next I would like to know what's the magnetic field at this point from an entire loop. So what I have to do is I have to basically integrate it along the whole circle. Well but since it's it does not really depend on anything I is constant, r is constant, x is constant so the only thing integration is gs so it's like integrative integral of gs which is actually the total circumference of of the loop which is 2 pi r. So whenever I'm integrating this I will get this. Well can I do it? Actually not exactly and here is the point. Now the vector of magnetic field is always perpendicular to both direction of the current which is basically direction of the this segment and the line which connects my source of the electricity and the point where I'm calculating. So remember if you have vector of i is this vector of point to towards here my magnetic always my magnetic intensity is always perpendicular to both. What does it mean in this particular case? It would be more convenient if I will look at this not in a 3g way but if I will look perpendicularly to the loop. If I will look perpendicularly to the loop what would happen? So my loop will be like a line. So my loop is this line is axis. Now how is my electricity direction? Let's say at this point this is A. If this is my loop my electricity is going perpendicular to the board. So this is my vector of electricity. So this is direction of the Gs. On the bottom is also this. Now as the loop goes this way for instance it would be this direction. So this is direction of the but in this case on the top it will be perpendicular to the board. That's very important it's easier to do this. Now all A is on the board. Now so what is the direction of my magnetic field intensity at this point? It should be perpendicular to this so it should be somewhere on this plane and it should be perpendicular to this but since this is perpendicular to the board then my line should be really within the board. So this is real direction of the B as a vector. It's perpendicular to this line and it's perpendicular to the DS because DS is perpendicular to the whole board which means perpendicular to any line on the board. Again back to the space geometry. So this picture now which is on this board is really a two-dimensional exact two-dimensional representation and this is exact direction of the B vector magnetic intensity. It's perpendicular to this. Now if you will take for instance point here it will be perpendicular to this. Now what happens here the B vector would be in this particular kind of position here here here here and now I have to add them up altogether. I cannot add them up just based on their magnitude because vector like this for instance will always be represented as a sum of two vectors. One goes along the axis which is B prime and another is B perpendicular to the axis. Now this vector magnetic intensity which is produced by this segment will also have this component and this component and what's interesting is that all components along the axis will be the same for all the different piece but the components which is perpendicular to axis for instance this will be cancelled with the opposite one and if you take any point on the loop and the opposite point again their projections on the perpendicular to the axis would cancel each other and only those projections on the axis will be in the same direction and that's what I have to add. So instead of adding the vector with magnitude of vector B I have to actually add only their projections on the axis because these projections are always the same for any point on the loop. So what is exactly this particular projection? Well look at it this way. This is perpendicular to this. Now this is perpendicular to this so this angle is equal to this angle and I need a sine of this angle right? I need to multiply my magnitude by sine of this angle to get projection on the axis and I know the sine of this angle because this is x this is r and this is square root of r square plus x square right? So my sine is equal r divided by square root of so I have to really multiply this by r and divided by square root so this is the first degree and square root of 1 half so it will be 3 half here and this is x component that's what we are looking for not just B because we cannot really add vectors which are directed differently but we will take projections on the x axis that all these vectors can be added together and there another component will always be cancelled and will always be if I will summarize them it will always be zero. So these projections are going in all different directions like this and again this will be cancelled with this. Now these projections are always going along the x axis and I can add them together so if I will add this together this is actually not the B the entire Bx this is only differential which is produced by a specific sinc and now I basically have to integrate it which means I have to multiply it instead of ds I have to multiply it by the entire 2 pi r that gives me r square and this is the final magnetic field produced by the one particular loop at point zero if loop is on a distance x okay fine by the way the picture in my on my website is slightly better than whatever I have just drawn okay so this is something which is important this is from one loop only now what I will do next is so let's go back to my solenoid so what I will do next is from x to x plus dx all the loops which are on this particular segment of the x axis how many of them well if density is n so for every unit of lengths I have n loops now the if length is dx then the number of loops is n times dx right so in this case b of let's put it off x would be equal to the same formula just multiply by n and dx and I will have differential also here i well obviously 2 pi will cancel with 2 pi here so I will have r square and dx that's number of loops per dx from 4 pi 2 pi I have still 2 r square plus x square to the power of three seconds and what I have to do next is I have to integrate it by x from x is equal to minus infinity to x is equal to plus infinity and that would be my answer that would be the real b so b is equal to integral from minus infinity to plus infinity mu i r square and dx divided by 2 r square plus x square 3 2 don't get scared it's not a big deal now but I would like to point out that your mass knowledge should be sufficient not to get scared by these calculations whatever i'm just making this is the goal and if you are uncomfortable go back to calculus course here or math proteins or or anywhere else whatever and you have to really make yourself comfortable if you would like to really know physics you have to be very comfortable with mathematics okay now what about this guy well first of all let me just take all the multipliers out so I will so I will have mu zero is a constant i is a constant r square is a constant n is a constant divided by 2 is a constant and then I have integral from minus to plus infinity dx divided by r square plus x to the power of 3 second now how this particular integral should be dealt with well first of all the obvious representation x should be equal to r times let's say t which means dx is equal to r dt and why did they do it well well if x is minus infinity to plus infinity t will also be minus infinity to plus infinity because it's just multiplication by r so I will have the same mu i r square n divided by 2 integral of minus to plus infinity and now we are integrating by t so instead of dx I will put r dt and instead of x here I will have r square plus x square would be r square t square to the power of three seconds right well obviously now r square can be taken out from here r square to the power of three seconds would be r to the power of two times three divided by two so it's r cube so the whole result would be mu zero i r square and divided by two now this r cube obviously should go out I think I made some mistake by r square right and integral dt divided by oh I have another r here that's why so it's cube I see that something is wrong cube r square and r is r cube and r square to the power of three seconds is also cube I know that r should actually disappear that's why I was concerned and here I have one plus t square to the power of three seconds right now you see how important this is our final formula does not depend on the radius of solenoid that's what's important okay now what about this integral well the easiest thing I can say right now is I can just give you an answer because it's really not very difficult again you have to be relatively comfortable to to take the integrals like this but basically I think the story is that the t divided by one plus t square to the power of let me see I think it's three seconds but I'm not sure once one square sorry I think it's squared yes I think this is the function which if you will get its derivative will give you this and we can check it out I just don't want right now to do it because it will take some time but when I was actually trying to find out this integral now this is from minus infinity to plus infinity first you have to do is you have to get the function derivative of which is equal to this and there are some methods which we used to get this function and I think this is the function which I have found now if this is an indefinite integral therefore if derivative of this is equal to this so this is the indefinite integral of this well plus c obviously and I have to really put the limits right now this is the formula Newton Leibniz formula if you have the indefinite integral so the definite integral should be just indefinite substituting the limits okay now if we substitute the limits obviously constant disappears so we don't really need it if you substitute infinity you will have this okay t divided by one plus t square square root if you will now if it's a plus infinity so let's divide everything by t it would be one divided by this divided by t or square root divided by t square would be one divided by t square plus one right and obviously if t is equal to infinity the whole thing is equal to one now how about minus infinity well if minus infinity then minus t is positive so I will divide it to minus t and that would be what if I divide by minus t which is positive in both cases I will have minus one here now minus t is positive so I will just put it under the square root so it will be the same thing and again it would be infinity so the whole thing would be minus one so the whole result would be one minus minus one which is what two so the whole thing is equal to two and that two will cancel out this two and the whole answer would be mu zero i m you see how simple the formula final formula is that's worse actually all these calculations the complicated calculation sometimes give you a simple formula which which signifies that maybe we did something right okay all right so what's important is this is the magnetic field at the center line of the solenoid at its axis and textbooks usually are saying that everywhere else inside this infinite solenoid the magnetic field is exactly the same I did not check it and I think there is some deviation from this but probably very small ones so I'm not making any judgment right now you can just trust the textbook textbooks I did not verify it with calculations other than that that's a very interesting formula it does not depend on the radius of the solenoid so if solenoid is big or small as long as it's infinite the magnetic field inside at least along the axis will be the same but now my feeling is that if really the radius is very big then somewhere near the wire itself it should be different than let's say in the center because in the center if a very big radius in the center well it looks like it's further from the from the edges but at the same time it's infinite length so again it needs some really detail and involves calculations which I did not do and let's just trust the textbooks that they're saying that okay the the field is uniform and does not depend whether it's on the axis of the solenoid or slightly above or below it and it does not depend on the radius all right so that's basically it for today thank you very much I do suggest you to read all the notes for this lecture on the unizord.com you go to electromagnetism and magnetic properties of the current and that's where you find this lecture thanks very much and good luck