Transcript for:
Magnetism

in this video we're going to talk about magnetism perhaps you're familiar with bar magnets and you know that when you place the north pole of a barge magnet with another north pole these two they repel they're gonna push apart however let's say if you were to face the north pole of one bar magnet with the south pole of another they won't repel these two will feel a force of attraction the north floor is attracted to the south pole but if you put the north pole with another north pole of another magnet they will repel or if you put the south pole of one magnet with the south for another they will also repel every barbed magnet has its own magnetic field which emanates away from the north pole and it travels towards the south pole likewise this one has a magnetic field emanating away from the north pole notice that the magnetic field cancels in the middle those two they repel each other and in the case of the other example the magnetic field leaves the north pole but it enters the south pole so in the case of attraction notice that the magnetic field between the north and south pole of the two barmatics notice that they're in the same direction and so the additive you have that force of attraction so what causes magnetic fields magnetic fields are created by moving electric charge and this example can be illustrated if you have a wire whenever there's an electric current that flows through a wire it creates its own magnetic field and the magnetic field created by this wire it looks like this it's circular on the left side the magnetic field is leaving the page so it's represented by a dot or circle and on the right side it enters the page which is symbolized by an x you need to be familiar with that by the way you can use the right hand rule to figure this out if you take your hand and if you curl it around a pen with your thumb facing the direction of the current the way your fingers curl around the pen is the way the magnetic field travels around the wire try that so here's a picture that describes it as you can see you want the thumb facing the direction of the current which is upward and as you curl your hand around the wire notice the way your hand curls around it on the left side you can see how it's coming out of the page and then on the right side your hand curls into the page and so the way your hand curls around that wire is the way the magnetic field created by the moving charge travels around the wire it's out of the page on the left side and it's going into the page on the right side now there's an equation that allows you to calculate the strength of the magnetic field created by such a wire and here's the equation b is equal to u0 times i divided by 2 pi r so let's say if you want to calculate the magnetic field some distance point a away from the wire r is the distance between the wire and point a b is the strength of the magnetic field and b is measured in units of tesla or capital t u 0 or mu 0 it's equal to four pi times ten to the minus seven this is known as the permeability of free space and the units are tesla times meters per amp now notice that the current and the magnetic field are directly related if you increase the magnitude of the current the strength of the magnetic field generated by this wire will increase as well and the reason for that is because the current is on the top of the fraction whenever you increase the numerator of a fraction the value of the entire fraction will increase now r is on the bottom so that means that r is inversely related to b if you increase the distance between the wire and the magnetic and uh point of interest i should say the magnetic field at that point will be weaker as you move away from the wire the strength of the magnetic field weakens by the way the number of magnetic lines that you see in a picture is proportional to the strength of the magnetic field so for example let's say if the magnetic field in this region looks like this versus two lines as opposed to three the magnetic field on the left is stronger than the one on the right if you have more lines that are closer together the strength of the magnetic field is stronger but make sure you know this relationship so anytime you increase the electric current in a wire the strength of the magnetic field will increase and as you move away from the wire the strength of the magnetic field will decrease let's work on this problem a vertical wire carries a current of 45 amps do self calculate the magnitude and the direction of the magnetic field two centimeters to the right of the wire so go ahead and try this problem so let's say this is the wire and the current is due south so it's going down and that means that using the right hand rule the magnetic field is going to enter the page on the right side and it's going to be out of the page i mean it enters the page on the left side but on the right side it comes out of the page so when it enters the page put it x and when it leaves the page put a circle a closed circle now our goal is to find the magnetic field two centimeters to the right of the wire so we already have the direction of the magnetic field it's out of the page all we need to do now is calculate the magnitude so we can use this equation b is equal to mu zero times i divided by two pi r mu zero is equal to four pi times 10 to the minus 7 and the current is 45 amps r is the distance between the wire and the point of interest in this case r is two centimeters but we need to convert that to meters so we got to divide by a hundred one meter is equal to 100 centimeters so that's going to be point .02 meters so all you got to do is just type this in and you should get 4.5 times 10 to the negative 4 tesla so that's going to be the strength of the magnetic field 2 centimeters away from the wire number 2 a wire carries a current of 10 amps at what distance from the wire will a magnetic field of 8 times 10 to the minus 4 teslas be produced so we got to find r in this case we can use the same equation we don't have to worry about the direction so we don't really need to draw a picture so let's solve for r let's multiply both sides by r so on the right side it's going to cancel so b times r is equal to mu 0 times i divided by 2 pi now let's multiply both sides of the equation by 1 over b if we do that on the left side b will cancel so the distance is going to be mu zero times i divided by two pi times b so it's 4 pi times 10 to the minus 7 times the current of 10 amps divided by 2 pi times the strength of the magnetic field so we can cancel pi in fact four pi divided by two pi is just two so it's going to be two times ten to the minus seven times ten divided by eight times ten to minus four and so this is equal to two point five times ten to the minus three and the units is meters so if we want to we can convert it to millimeters and to do that you need to divide actually multiply by a thousand there's a thousand millimeters per meter if you multiply by a thousand this will give you 2.5 millimeters and so that's the answer that's how far away from the wire that you have to be to measure a magnetic field of eight times ten to minus four by the way if you ever were to place a compass near a wire whenever there's an electric current flowing through that wire it will cause the compass to deflect you should try it now let's say if we have a current carrion wire what's going to happen if we place this wire inside a magnetic field let's say the magnetic field is directed east and the current is moving north a magnetic field exerts no force on a stationary charge however if the electric charge is moving then the magnetic field will exert a force specifically a magnetic force so whenever you have a wire with an electric current that means you have moving charges in the wire the magnetic field will exert a force on the wire you can calculate the strength of the magnetic force using this equation f is equal to i lb sine theta so the strength of the magnetic force is proportional to the current if you increase the current the magnetic force will increase the magnetic force is also proportional to the strength of the magnetic field if you increase the magnetic field the magnetic force will increase as well and it's also proportional to the length of the wire now depends on the angle as well so here's one example where the current and the magnetic field are perpendicular and here's another example when the current and the magnetic field are at an angle and here's one that's parallel now theta is the angle between the current and the magnetic field when they are perpendicular sine 90 is equal to one and one basically represents a hundred percent so the maximum force occurs when the current and the magnetic field are perpendicular to each other now when it's at an angle it's going to be between anywhere from zero to 100 percent of its maximum value so then you can use this equation when they're parallel the angle is equal to zero degrees sine zero is equal to zero therefore the magnetic field exerts no magnetic force on a moving charge that moves parallel or even anti-parallel to magnetic field so for the third example there's no magnetic force acting on the current they have to be at an angle with respect to each other they can be parallel now what about the direction let's go back to our last example so let's say if the current is due north and magnetic field is directed east in what direction is the force now the force has to be perpendicular to the current anti-magnetic field so if the current is in the north-south direction and if the magnetic field is in the west east direction then the force is either into the page or out of the page that is along the z axis so how can we figure this out well we have to use the right hand rule so take your right hand and you want to extend it you want your thumb to be in the direction of the current and you want your other four fingers to be in the direction of the magnetic field so this represents b and this represents the current i so using your right hand look at where your right hand opens towards if you direct it the way it's presented here it's going to go into the page the force comes out of the palm of your right hand and so it's going to be directed uh into the page and that's how you can figure it out let's try another example so let's say if we have a wire and a current is directed east and the magnetic field is directed into the page and what direction is the force going to be so if the current is in the east-west direction and well the current's really west to east and the magnetic field is in the z direction that is between out of the page and into the page then the force has to be in a north south direction these three variables have to be perpendicular to each other so what you want to do this time you want to direct your four fingers into the page and you want your thumb directed east that is in the direction of the current but you want your four fingers to be into the page and the force comes out of the palm of your right hand so the force will be directed north if you do that it's kind of hard to draw the direction of the hand on this video but hopefully you can visualize it number three a 2.5 meter long wire carries a current of 5 amps in the presence of a magnetic field with a strength of 2 x 10-3 teslas calculate the magnitude of the magnetic force on the wire using the picture shown below so feel free to try that the equation that we need is f is equal to i lb sine theta now let's talk about the angle so the current is directed east and the magnetic field is directed 30 degrees relative to the horizontal so theta is always going to be the angle between the magnetic field and the current so you can also use this angle as well because that angle is between b and i 180 minus 30 is 150 and it turns out that sine of 150 and sine of 30 they're both equal to one half so it doesn't matter so whether you choose this angle which is between i and b or if you use this angle the answer will be the same so just something to know so now let's go ahead and calculate f so it's going to be the current which is 5 times the length of the wire which is uh 2.5 meters times the strength for the magnetic field which is uh 2 times 10 to the minus 3. multiplied by sine of thirty so the magnetic force is very small it's .0125 newtons and so that's going to be the force exerted on this current carrying wire number four a current of 35 amps flows due west in a wire that experiences a magnetic force of 0.75 newtons per meter what is the strength of the magnetic field which is directed to south so here's the wire and the current is directed west and the magnetic field is directed itself our goal is to find the strength of the magnetic field we need to solve for b so let's write the equation f is equal to ilb now because the current and the magnetic field are at right angles to each other because it's 90 degrees sine 90 is one so we don't need the sine portion of this equation now we're given the force per meter so that's f divided by l if we divide both sides by l we're going to get this equation f over l is equal to the current multiplied by the magnetic field we have the force per meter that's 0.75 that value takes care of two of these variables we have the current which is 35 amps so we got to solve for b so it's just going to be 0.75 divided by 35 and so b is equal to 0.0214 tesla so anytime you have the force per unit left or newtons per meter make sure you understand that it's f divided by l the entire thing is f over l so you might see that expression in this chapter in a few problems now what is the direction of the magnetic force we know the current flows west the magnetic field is south so the force is either into the page or out of the page so what you want to do is you want to direct your four fingers south and your thumb west and the force should come out of the page so let's see if i can draw that so you want your thumb facing this way and the four before fingers facing this way using your right hand if you do that i put into the page this should be out of the page the force should come out of the palm of your hand so make sure the current is aligned with your thumb the magnetic field is aligned with your four fingers and then the force should come out of the palm behind that's out of the page now let's say if we have a rectangular metal loop with a current that flows in the metal loop clockwise so in this section the current's going up here it's going down here it's directed towards the right and here is directed towards left and only a portion of this loop is inside a magnetic field that is only the bottom portion now let's say the magnetic field is directed into the page so i'm going to put x everywhere where will this rectangular loop move if initially it's at rest will it move towards the right will it begin moving towards the left up or down what would you say so let's start with this portion of the wire or of the the metal loop let's see what the magnetic force on that portion is directed so what you want to do is you want to place your thumb facing south you want the four fingers of your hand to be going into the page and the magnetic force should be directed east so let's see if i can draw that so you want your four fingers going into the page you want your thumb going south so make sure the magnetic field or your four fingers is going into the page and your thumb is in the direction of the current and the force should come out of your hand and that is out of the palm of your right hand and it should be directed east if you do it correctly so make sure you try that and make sure you can master this right hand rule now for the other side the left side everything is the same except the current because the current is in the opposite direction the force has to be in the opposite direction now these two forces are equal in magnitude and because they're opposite direction they will cancel out so the loop is not going to move towards the left or towards the right these forces balance each other out now in the top part of the loop there's no magnetic field in that region so therefore there's no force the net force is going to be based on this portion of the loop because it's not balanced by this portion of the loop if the entire loop was in the magnetic field all the forces will cancel but since it's not this one will create a net force so now let's draw another picture point your thumb towards the left and make sure your four fingers are going into the page if you do it correctly the force should be direct itself so what you want to do is you want your thumb directed uh west and you want your four fingers going into the page so if you do it correctly the magnetic force should be coming south out of the palm of your right hand now let's move on to another topic we talked about how to calculate the magnetic force on a current carrying wire but what about the magnetic force on a single point charge because any move in charge will create a magnetic field so let's go ahead and come up with an equation let's start with this equation f is equal to ilb sine beta current is defined as the amount of electric charge that flows per unit time and capital q is the total amount of charge it can be due to many charged particles so capital q is going to be equal to lower lowercase q which is the magnitude of each charged particle times n which is the number of charged particles and that will give you the total charge divided by t now distance is equal to the speed multiplied by the time and length could be thought of as distance they're both measured in meters so we can replace out with vt so let's replace i with q n over t and let's replace l with v t so we can cancel t now if you want to find the magnetic force on a single point charge that means there's only one charged particle so n is one so when n is one we can get rid of it so this leaves us with f is equal to bqv sine theta so that's how you can calculate the magnetic force on a moving charge particle when it's inside a magnetic field so let's say if we have a proton and let's say it's moving towards the right and also the magnetic field is directed towards the right if these two are parallel sine zero is equal to zero so there's going to be no magnetic force they have to be perpendicular now if the proton is moving at an angle relative to the magnetic field and then you can use the equation f is equal to b q v sine theta where theta is the angle between the magnetic field and the velocity vector now let's say if the proton it's moving perpendicular to the magnetic field that is they're at right angles or 90 degrees relative to each other sine 90 is one so the magnetic force will have its maximum value at this point and it's equal to simply bqv now let's say if we have a proton and it's moving towards the right and the magnetic force or rather the magnetic field is directed north what is the direction of the magnetic force it has to be in the z direction if the velocity is in the x direction and if the magnetic field is in the y direction the magnetic force have to be in the z direction and you can use the right hand rule think of velocity as the current so to speak that's where the charged particles are moving so you can use the the same right hand rule as we've been doing you want your four fingers to be in the direction of magnetic field and you want your thumb to be in a direction of the velocity so if you orient your right hand this way the force should come out of the palm of your right hand and so it should come out of the page so i'm going to put a circle for that now for a proton the magnetic force is out of the page but for an electron the magnetic force will be in the opposite direction that is it's going to be going into the page so for any negatively charged particle simply reverse the direction of magnetic force if you need to find it for a positively charged particle it's going to work in the exact same way as the right hand rule directs it number five a proton moves east with a speed of four times ten to the sixth meters per second in a magnetic field of two times ten to minus four teslas directed into the page what is the magnitude of the magnetic force acting on a proton so if we need to find the magnitude we don't have to worry about the direction all we need to know is that the velocity and the magnetic field are perpendicular to each other the proton is moving east in the x direction the magnetic field is directed into the page that is in a negative z direction so therefore the magnetic force has to be in the y direction so all we need to do is find the magnitude so we just got to use the equation f is equal to b q v sine theta now because the velocity and the magnetic field are perpendicular that is the velocities in the x direction the magnetic field is in the z direction the angle has to be 90 degrees and sine 90 is one so f is simply equal to b q v b is the magnetic field in tesla that's two times ten to the minus four teslas q is the charge of just one proton the charge of a proton is 1.6 times 10 to the negative 19 coulombs and that's something you just got to know and the speed of the proton is 4 times 10 to the 6 meters per second so we just gotta multiply these three numbers and so you should get 1.28 times 10 to the minus 16 newtons so that's the magnetic force acting on the proton now let's talk about a proton so let's say if we have a proton and it's moving towards the right and the magnetic field is directed everywhere into the page so let's say there's an x everywhere what's going to happen where is the magnetic force so if you direct your thumb towards the right and your four fingers into the page the magnetic force will be directed north whenever force and velocity are perpendicular to each other the object will turn so it's going to go this way if force and velocity are in the same direction the object will speed up if force and velocity are in the opposite direction the object slows down but if they're perpendicular the object will turn and eventually the particle is going to be moving in a direction of the force but now that the velocity is directed north the force is going to change if you use the right hand rule again if you direct your four fingers into the page the velocity north the force will be directed west and so what's going to happen is this particle the proton is going to move in a circle i haven't drawn a knife circle because i ran out of space but you get the picture when it's at the top the force is going to be directed south as it moves to the left and as the proton moves south the force will be directed east so notice that for a moving charge particle the magnetic force behaves as a centripetal force or a sensor seeking force now what if we have an electron if we had an electron the situation will be opposite as the electron is moving in the same direction as the proton it's going to fill a force in the opposite direction so the proton felt a force that was directed north the electron will feel a force directed in the opposite direction at itself so as the proton moves in the counterclockwise direction the electron will move in a clockwise direction so they will move in an opposite direction now how can we calculate the radius of curvature that a proton or electron might travel in a circle how can we figure out the radius if you ever get a question like this what you need to do is set the centripetal force equal to the magnetic force let's call fb the magnetic force f status and triple force the centripetal force is provided by the magnetic force in this particular example the centripetal force is equal to mv squared divided by r which is the radius of the circle the magnetic force is equal to bqv now because the magnetic field and the velocity are perpendicular we don't have to worry about the sine portion of this equation now we're going to do is multiply both sides by 1 over v so on the right side v will cancel on the left side because we have a v squared one of them will remain so mv divided by r is equal to bq so multiplying both sides by r we have this equation so make sure you uh write down this equation mv is equal to bqr because in this format you can solve for anything so let's say if we want to find the radius of the circle all we need to do is divide both sides by bq so the radius of the circle is equal to the mass of the charged particle times the velocity divided by the magnetic field times the charge number six a proton moves with a speed of five times ten to six meters per second in a plane perpendicular to a magnetic field of 2.5 tesla calculate the radius of its circular path so here's the proton and if it's moving perpendicular to a magnetic field it's going to move in a circle our goal is to calculate the radius of that circle so r is equal to based on the equation that we had before it's mv divided by bq so what is the mass of a proton now the problem doesn't give it to you which means you can either look it up online or you can look it up in the reference section of your textbook the mass of a proton is about 1.673 times 10 to the minus 27 kilograms the speed which is given that's 5 times 10 to 6 meters per second the strength of the magnetic field is 2.5 tesla and the charge of a proton which is the same as that of an electron but the opposite sign it's 1.6 times 10 to the negative 19 coulombs so if we type these numbers into the calculator we should get point zero two zero nine meters which is equal to two point zero nine centimeters so that's the radius of the path that the protons gonna travel in and so that's how you could find it now what about part b what is the energy of the proton and electron volts how can we find the answer to that question well first we need to find the energy in joules a moving object has kinetic energy any object in motion contains kinetic energy so we've got to find the kinetic energy of the proton which is one-half mv squared so we know the mass of a proton it's 1.673 times 10 to the negative 27 and we also have the speed 5 times 10 to the 6 meters per second squared so this is equal to 2.09 times 10 to the negative 14 joules now once you have the energy in joules you can convert it to electron volts electron volts is basically another unit of energy it's very useful for small particles like protons and electrons one electron volt is equal to 1.6 times 10 to the negative 19 joules we'll talk about why later but that's what it's equal to and so you just got to convert it to electron volts and this will give you the answer which is a 130 000 and 703 electron volts so that's it for this problem so why is it that one electron volt is equal to 1.6 times 10 to the negative 19 joules why is that the case electric potential which is measured in volts is equal to the electric potential energy which is measured in joules divided by the charge so the unit volt one volt is equal to one joule per one column so therefore an electron has a charge of 1.6 times 10 to the negative 19 coulombs it's negative but let's ignore the negative sign that's the charge of an electron and an electron that has one volt or one ev that's one electron volt an electron with one volt will have an energy of 1.6 times 10 to the negative 19 joules volt is basically the ratio between joules and coulombs so if you have a charged particle that has one joule and one clue its voltage is one volt or the electric potential is one volt voltage is really work per unit charge electric potential is energy per unit charge now if we have a charged particle that has an energy of 1.6 times 10 to negative 19 joules and a charge of that many coulombs then these two will cancel and the electric potential will be 1 volt which is the case of an electron an electron has a charge of 1.6 times 10 to negative 19 coulombs and if that electron has an energy of one electron volt its voltage is one volt which means its energy is equal to this number and that's why one electron volt is 1.6 times 10 to the negative 19 joules just in case you're wondering now let's say if we have two wires parallel to each other and let's say that there's a current in wire one and in wire two and these two currents are in the same direction will these two wires attract each other or will they repel it turns out that these two wires will attract each other if they have a current in the same direction now if there's two wires with the opposite current then the situation is different they will not feel a force of attraction rather they will repel each other so why does that happen why do we have a force of attraction in the first case but in the second scenario these two repel what's going on here wire one creates a magnetic field because it has a moving charge it has a current and that magnetic field exerts a force on wire two now let's focus on wire one the current is going north if you use the first right hand rule that we talked about earlier in the video where you curl your hand around the wire and your thumb is in the direction of the current the magnetic force created by wire one will be out of the page on the left side but it's going to be into the page on the right side so wire 2 is on the right side of wire 1 therefore wire 2 sees a magnetic field that's going into the page now using the second right hand rule that we talked about what you want to do is you want your fingers to point into the page but you want your thumb pointing north in the direction of the current so here's your thumb it follows i2 the magnetic field which is b1 that's created by i1 you want that to be in the page and what's going to happen is the force is going to come out of the palm of your hand and it turns out that force is directed towards wire one so it's a force of attraction and so anytime you have two wires with the current going in the same direction it's going to create an attractive force the two wires will be tracked to each other if the current is in the opposite direction then the two wires will repel each other now how can we calculate the magnitude of the force between the two wires so we said that the first wire wire one creates a magnetic field that causes the second wire to be attracted to the first if they're moving in the same direction and also the second wire creates a magnetic field that exerts a force on the first wire causing the first wire to move towards the second if the currents are in the same direction so let's start with this equation wire one creates a magnetic field b1 which is u0 times i 1 over 2 pi r where r is the distance between the two wires so wire 1 which produces a current one will generate a magnetic field that is at wire two and so r is the distance between the two wires now we need to use the other equation that is the force that acts on a current carrying wire that's f is equal to ilb sine theta but we're not going to be worried about the angle in this problem so the force on wire two which we'll call f2 is equal to the current on that wire times the length times the magnetic field created by wire one so what we're going to do now is replace b1 with mu 0 i 1 over 2 pi r so it's going to be i 2 times l times mu 0 i 1 over 2 pi r so that's how you can calculate the force between the two wires f2 and f1 they have the same magnitude number seven what is the magnitude and direction of the force between two parallel wires that are 30 meters long and two centimeters apart each carrying a current of 50 amps in the same direction let's draw a picture so we have two wires and they have the same current so i1 and i2 both equal 50 amps now we have the length of the wire which is l and the left of both wires is 30 meters and the distance between them which is r that's two centimeters which is equivalent to point zero two meters now for these type of problems there's only two answers for the direction either the force is attractive or they will repel because the currents are in the same direction we have a force of attraction and so that answers the direction of the force it's simply attraction now all we need to do is find the magnitude so let's use the formula f is equal to mu zero times i1 times i2 times l which is the length of the wires divided by 2 pi times r so it's 4 pi times 10 to the minus 7 times 50 times 50 which we can write as 50 squared times l which is 30 meters divided by 2 pi times the radius of 0.02 meters now 4 pi divided by 2 pi is going to be 2. so it's 2 times 10 to the minus 7 times 50 squared times 30 divided by .02 so the force between these two wires is 0.75 newtons and a direction is a force of attraction these two forces will be pointed towards each other now the next thing that we need to talk about is ampere's law ampere's law describes the relationship between the current and magnetic field produced by that current perhaps you've seen this equation the sum of all the magnetic fields that is parallel to any uh segments that the magnetic field passes through that's going to be equal to mu zero times the current enclosed by the path that the magnetic field makes and has to be a closed path a good example is using a wire so let's say if we have a current that passes through this wire now this current will create a magnetic field that travels around the wire in a circular path so if we take the magnetic field and multiply it by the path that is parallel to it let's call it delta l we can use that to get an equation that will give us the magnetic field created by this wire so the path that the magnetic field travels is basically the path of a circle so delta l is really two pi r if we add up all the small segments so if we add up this segment plus that segment plus that segment we're going to get the circumference of a circle which is 2 pi r so it's going to be b times 2 pi r which equals mu is not over i i mean times i so the magnetic field if you divide both sides by two pi r we get this familiar equation mu zero times i over two pi r now it's important to understand that the current in this equation is the current that is enclosed by this loop so that loop has an area and the current that passes through that area that's this current is the current enclosed by that circular loop and so that's how you can use ampere's law to get the equation of a magnetic field by a current carrying wire now let's use ampere's law to come up with an equation for a solenoid a solenoid is basically a device with many loops of wire and the reason why it's advantageous to create a solenoid anytime you create a loop of wire whenever you have a current the magnetic field that is at the center of the wire is very strong and for every loop that you add you increase the strength of the magnetic field inside the wire and so solenoids are very useful for creating powerful magnetic fields outside of the loop the magnetic field is weak so the magnetic field will travel in a circular pattern so to calculate the or to derive the formula for a solenoid we need to create a path of the magnetic field so let's draw a rectangular path so let's say this is a b c and d so what we need to do is add up the magnetic field that is parallel to each segment so that's going to be the magnetic field times the left of segment ac plus the magnetic field times the length of segment cd plus the magnetic field times the length of segment db plus the magnetic field times the length of segment ba now let's focus on segment bd segment bd is outside of the solenoid and the magnetic field is very weak outside of the solenoid so we can say that the contribution for uh bl and segment bd is very small so it's negligible now ba and dc they're perpendicular to the magnetic field that is inside the solenoid that magnetic field is much stronger than the magnetic field on the outside so it's because it's perpendicular to the magnetic field that's inside the solenoid its contribution is going to be negligible so we can eliminate ba and dc so therefore all we have is the segment that is parallel to the magnetic field that is inside the solenoid that's ac segment ac is the most important segment because one it's parallel to the magnetic field that is inside the solenoid and that magnetic field is the strongest one compared to the ones that are outside of it and so bl in segment ac will have the greatest contribution towards the sum of all the magnetic fields that is parallel to each segment so now using ampere's law which is basically this equation we can now replace this term with b times l where l is simply the length of the solenoid the length of the solenoid being segment ac but we'll just call it l so bl is equal to mu zero times i but this is four if we only have one loop let's say if you have a wire with a current of ten if you have one loop then the enclosed current is 10 amps but it turns out that if you add another loop the current enclosed by the magnetic field it's going to be twice as much even though 10 amps is flowing through the wire the enclosed current is now 20. and if you add another loop the enclosed current is 30 even though 10 amps is still flowing through the same wire so therefore we need to add n to this equation because the enclosed current increases so the enclosed current is basically the current that flows in one loop times the number of loops so now all we need to do is divide both sides by l so b is equal to mu zero times n times i divided by l lower case n is equal to capital n over l capital n represents the number of terms or loops l is the length in meters so lower case n is the number of loops or turns per meter so we're going to replace capital n over l with lowercase n so now we have the equation of a solenoid b is equal to mu zero times n times i so the magnetic field produced by a solenoid is proportional to the current that passes through as the current increases the strength of the magnetic field will increase the second way to increase the magnetic field is to increase the number of turns if you can increase the amount of turns per unit left or per meter the strength for the magnetic field will greatly increase so you want to increase the number of turns you want to increase the current but you also want to decrease the length if you can decrease the length the magnetic field will increase as well so the magnetic field is directly related to the current it's related to the number of turns and it's inversely related to length number eight a solenoid has a length of 15 centimeters and a total of 800 turns of wire calculate the strength of the magnetic field at its center if the solenoid carries the current of 5 amps so first let's calculate lower case n which is capital n over l so there's 800 turns and the length of the wire is 15 centimeters but we need to convert that to meters so let's divide by 100. 15 divided by 100 is 0.15 or 0.15 meters so 800 divided by point fifteen that's equal to five thousand three hundred and thirty three turns per meter or loops per meter so now to calculate the strength of the magnetic field it's mu zero times n times i so it's four pi times ten to minus seven times the number of turns per meter which is five 5333 times the current which is 5 amps so the strength for the magnetic field at the center it's about .0335 tesla and so that's it for this problem now what's going to happen if we have a current carrying loop inside a magnetic field in a magnetic field the loop is going to rotate it's going to produce a torque so let's say on the left side there's a current and on the right side there's a current so the current travels clockwise in this loop and is also a magnetic field that is directed east now what type of force will we have on the left side of the wire or the metal loop so using the right hand rule place your thumb going up and your four fingers in the direction of the magnetic field and if you do that notice that the palm of your hand opens into the page and so that's where the force is going to be so the magnetic field will exert a force on the left side of the loop going into the page therefore on the right side of the loop the current is reversed so it must be out of the page so therefore this loop is going to turn it's going to turn this way and then into the page on the other side now let's talk about how to derive an equation for the torque of this current carrying loop f is equal to ilb it's ilb sine theta but the force is perpendicular both to the current and magnetic field so sine 90 is one so the force acting on the right side and on the left side of the loop it's going to be ilb there's no force on the top section in the bottom section of the loop the reason for that is because the current is parallel to the magnetic field and whenever the current is parallel to the magnetic field there's going to be no magnetic force it has to be perpendicular to it so there's only a force on the left side and on the right side not on the top or bottom section of the loop so since we're focused on this side that's where the magnetic force will be exerted on l represents the length of that section so that's this l b is the magnetic field the width of the loop let's call it w let's put that here now we need to calculate the torque this force will create a torque and this force will create another torque that's additive so the net torque is going to be t1 plus t2 and just to refresh you on how to calculate torque torque is the product of the force times the level arm so it's f times r and if the force is at an angle is going to be fr sine theta so t1 and t2 is going to be f1 times r1 times sine theta plus f2 times r2 sine theta so what exactly is r in this example we know that r is the distance between where the force is applied and the axis of rotation that's where the object moves around so in this particular problem the axis of rotation is here so which means that r is half of w so r is w divided by two so now we can replace f with ilb and we can replace r with w over two and then t2 is going to be the same thing it's the same current times the same length they both have the same life they're both exposed to the same magnetic field and r is the same r1 and r2 is the same so we no longer need this picture so what i'm going to do is i'm going to factor out ilb and sine theta so t is ilb sine theta w over two plus w over two half plus half is a whole so t is equal to i l w b sine theta now for the rectangular loop that we have which has a length l and a with w the area of the loop is basically the length times the width so a is l times w so let's replace lw with a so the torque of a single loop is the current multiplied by the area times the strength of the magnetic field times sine of theta now what if we have many loops if you have two loops the torque is going to be twice as strong three loops three times strong so we need to add n to this equation for the number of loops so it's n i a b sine theta by the way the quantity n i a is known as the magnetic dipole moment represented by capital m so it's equal to the number of loops times the current times the area that's the magnetic dipole moment but this is the equation that we need to calculate the torque it's n i a b sine theta now let's say if this is the face of the loop let's draw the normal line perpendicular to the face of the loop so this is the area of the loop which we could describe it as a and the red line is the normal line that's perpendicular to a and sometimes the magnetic field it's not going to be parallel or perpendicular to the surface it can be an angle the angle theta is between the magnetic field and the normal line which is perpendicular to the surface of the coil so make sure you understand that so theta is between b and the normal line now instead of drawing a side view of the current carrying loop let's analyze it by drawing the top view so here's this is the top view of it and here's the rest of the loop if you wish to see it this way and this part is perpendicular to the plane of the loop let's say the magnetic field is at a angle of 90 degrees when it's at an angle of 90 degrees you're going to get the maximum torque possible so make sure you understand that i'm going to redraw like this so let's say b is directed east we're going to have one force going up and the other force going down so this part is perpendicular to the area or to the the surface of the plane of the loop i should say and here's the magnetic field so it's at an angle of 90 degrees and so you're going to get the greatest torque in a situation that's when the magnetic field is perpendicular to the normal line which means it's parallel to the face of the coil now let's show another picture when it's at an angle so here is the normal line which is perpendicular to the face of the coil and let's draw the magnetic field which is directed ease so this time the angle theta is less than 90. here theta is equal to 90. now f1 and f2 they're still directed on north and south but they're much less than the original values so the torque is less and let's see what happens when the angle is zero in this case the normal line is parallel to the magnetic field so therefore the angle between the two is zero and when this happens the magnetic field is perpendicular to the face of the coil it passes through the coil and so there's going to be no torque whenever you have this situation now let's understand why there's no torque if you look at the two forces f1 and f2 notice that they're parallel to the level arm anytime you have a force that's parallel to the lever arm it cannot create a torque the only way a torque can be created if it's perpendicular let's say this is the top view of a door here is the level arm r the only way for you to push a door is to apply a force perpendicular to the level arm and the door is going to turn if you try to push a door from this side it's not going to move it won't turn and that's what's happening here the axis of rotation is here and the way the forces are oriented since they're parallel to the level arm which is r there's gonna be no torque there's no rotation and so that's why whenever the angle is zero sine zero is zero so the torque is going to be zero so the maximum torque occurs whenever the angle is 90 that is the angle between the magnetic field and the normal line so whenever the magnetic field is parallel to the face of the coil you're going to have maximum torque whenever it's parallel to the face of the coil it's perpendicular to the normal line the angle is 90. and whenever the magnetic field passes through the coil that is when it's parallel to the normal line there's going to be no torque the angle is zero now notice what happens let's just erase a few things i think i'll keep that if we look at the first picture on the left f1 will create a torque that will cause it to rotate and the clockwise direction and f2 will also create a torque that will cause the system to rotate in the clockwise direction now once it rotates it's going to move towards a picture 2. it's going to look like this and as you can see f1 will still create a torque that will cause the system to rotate clockwise and the same is true for f2 eventually the system is going to reach this point and f1 and f2 will still be directed north and south which we see that in the third diagram but once we reach the third diagram where the angle is zero then the system is at equilibrium the two forces they're opposite and they're equal so they cancel out there's no net force in fact there's no net force for all pictures in the first one these two forces they balance each other out however there is a net torque for the first and second picture but in the third diagram the net torque is zero this force creates no torque and the same is true for the second one so the net torque and then that force is zero for the third diagram which means it's at equilibrium so basically the loop moves from this position from an angle of 90 and the magnetic field causes it to rotate to an angle of zero and then it stops now let's work on some problems number nine a circular coil of wire has a radius of 30 centimeters and contains 50 loops the current is 8 amps and the coil is placed in a magnetic field of five tesla what is the maximum torque exerted on the coil by the magnetic field so if we draw a picture we're gonna have uh a coil of wire that is circular and has many loops 50 loops and so it's going to look something like that and the radius is 30 centimeters to calculate the torque it's equal to n i a b sine theta now if we wish to find the maximum torque then the angle is going to be 90 and sine 90 is equal to 1. so n is the number of loops and is 50. the current is 8 amps to find the area it's going to be the area of a circle which is pi r squared so it's pi times the radius squared the radius is 30 centimeters but we need to convert that to meters 30 centimeters is 0.3 meters so it's pi times 0.3 squared so that's the area the magnetic field is 5 tesla and sine 90 is 1. so let's start with the area 0.3 square times pi which is 9 pi over 100 that's 0.2827 if we multiply that by 5 times 8 times 50 you should get 565.5 tesla so don't forget to square 0.3 and so that's it for this problem actually the unit is not tesla that's for the magnetic field the unit for torque is newtons times meters so this is it number 10 a rectangular coil contains 200 loops and has a current of 15 amps what is the magnitude of the magnetic field required to produce a maximum torque of 1200 newtons times meters so let's draw a picture and so here's a rectangular loop and there's 200 loops but i don't want to draw it 200 times the area is going to be the width multiplied by the length and so we have the dimensions so the torque is going to be equal to the number of loops times the current times the area times the magnetic field and since we're looking for or since we're using the maximum torque the maximum torque occurs at an angle of 90 degrees so this is going to be sine 90. now this problem the area of a rectangular coil is going to be the length times the width and sine 90 is 1. so the maximum torque is let's put this on the left it's 1200 newtons times meters n is equal to 200 loops the current is 15 amps and then the area the length times the width we need to convert 40 centimeters to meters so that's 0.4 meters times 0.5 meters and the magnetic field we're looking for b so we got to solve for it so let's multiply 200 times 15 times 0.4 times 0.5 so that's equal to six hundred and so now all we need to do is divide both sides by six hundred so these two cancel twelve hundred over six hundred we can cancel the two zeros so it's twelve divided by six and that's two so the magnetic field is two tesla and so that's the answer so that's it for this video um if you like this video feel free to subscribe you can check out my channel for more physics videos if you look out for my physics video playlist if you want to find them all so thanks for watching and have a great day