this week we started talking about chapter 27 magnetism yesterday we introduced magnets and magnetic fields and magnetic field lines we said that the source of magnetic field could be traced to currents currents generate magnetic fields and currents actually interact with magnetic fields if you have a wire with current of length l the force on the wire by an external magnetic field b is given with this form and we said actually we could trace this even further and then relate the interaction between the current and magnetic field to interaction between a moving charge and a magnetic field we just showed that and then we set actually a force on it moving charge is given by this f is equal to q v cross p and this was important and we saw that f was perpendicular to be at any moment and if your velocity is a purely um pure velocity that's orthogonal or like perpendicular to the force you had this nice circular motion coming from magnetic field but we also consider another example if you have like two projections of velocity one is perpendicular to magnetic field the other one is along the magnetic fields the one along the magnetic field does not affect the force right because v parallel cross b is zero so this particle this charge will move with constant velocity along this direction but in the perpendicular direction this component of velocity interacts with b as a result you have this nice spiral trajectory and when we combine electric and magnetic fields we had something like this f was equal to q times e which was the earlier electric force we have studied extensively and then q times v cross b and we call this lawrence force or lorenz equation we introduce a torque on a closed loop of currents and then from that we defined a so-called magnetic dipole moment which we showed with this meal and we defined this like the magnitude of the current in that loop if there's like a loop here and there's like a current that's that i times area vector an area vector was magnitude of area times normal of this area we related this to like electric field for example in the case of magnetic dipole moment the torque was equal to mu cross b in the case of electric field the dipole that torque on an electric dipole was equal to torque is equal to p cross e similarly potential energy of a dipole in a magnetic field is minus mu dot p whereas was very similar in the case of electric dipoles which was minus p dot e we saw that even atomic hydrogen can be taught f of like a current in a loop and we calculate its magnetic moments to be this number let me introduce whole effects so what is all of it if you have pass a current through a rectangular region of conductor and then if you apply a magnetic field which is in this case into the page this crosses these moving charges will experience the force coming from the magnetic field and move down if they are negative as a result there will be charge accumulation here and opposite charge recognition in the opposite side and there was going to be an electric field from top to bottom so it's important that this electric field or like if there's electric field there's a potential difference this potential difference was perpendicular to the direction of the current so current is in that way let's say call is x and the voltage difference which we called hole voltage was along y direction and this gave us an experiment to distinguish between positive and negative charge carriers for example we said suppose as the people in early years of electricity are magnetism taught suppose we think of plus sign charge as the charge carrier and do the same experiment we found that the whole voltage actually changes the sign so initially this voltage was plus if the charge carrier was positive this whole voltage was minus so this gave us a way to distinguish between positive and negative charges experimentally all right so let's just start so this is example 13 a long copper strip of 1.8 centimeters wide and one millimeter thick okay i think this is like 1.8 and the thickness which is into the page is one millimeter thick it's placed in a 1.2 tesla and now you know that this is actually a rather large magnetic field when a steady current of 15 ampere passes through it the whole emf is measured to be 1.02 micro volts and we'd like to determine the drift velocity of electrons and the density of free conducting electrons number per unit volume in the copper so let me go ahead and start so we set then the steady current is established that means everything is subtle and you know the whole equilibrium is reached what this means is the force on each charge coming from magnetic field which is charge qv cross b it's going to be minus e v which is the derivative because it's the velocity they move it cross b this should be equal to electric force coming from this whole voltage difference or whole electric field which is going to be minus e which is the charge times e i think you can work through the vectors and cross product but what you'll find i'll just write the magnitudes maybe i can just write it here because everything is perpendicular let me just delete these vectors because it's going to be like this this is like a scalar product now you can cancel ease and ease here and now i need to relate this to voltage difference because because problem gives us voltage difference can basically multiply both sides with this d okay if this is electric field electric field times the distance is your voltage because this is constant electric field so let's multiply with this d and multiply the other side with d good is that clear now i can solve this for v d drift velocity okay this kind of looks like b d is equal to now this one is hole voltage v h let's call it divided by i think it's called b times d now let's put the numbers this is like 1.02 times 10 to the minus 6 volt divided by b is equal to 1.2 tesla times d is equal to 1.8 times 10 to minus 2 meter okay if we do this calculation is equal to 0.47 times okay the exponent is minus 6 plus 2 10 to the minus 4 [Music] meter per second and now the other one is asking for what the density of three conducting electrons number per unit volume in the copper how can i do this any suggestions j the current density is going to be minus e okay okay minus e the charge density you know volume density times v drift and if i multiply both sides with the area and the area is like okay here this guy has like a width and it has like a thickness so area is this area right and the current is passing through here through this area if you like or another area here any area is fine so i which is j times a which is minus e times n times the drift velocity and now let's put the area this is 1.8 times 10 to minus 2 times the other one is what the other one is one millimeter one point zero times ten to minus three this is meters squared now so we know this i is equal to fifteen and you'd like to find n everything else is known okay number density is okay let's write this 15 appear divided by charge of an electron 1.602 times 10 to the minus 19 coulomb v drift we just found times 0.47 times 10 to the minus 4 and the last one is 1.8 times 1 is 1.8 times 10 to the minus five so let's do this calculation eleven point maybe almost one times let's sum minus 19 minus 4 minus 9 minus this is plus 28 electrons per meter cube right a lot of electrons huge number of electrons that's why they are conductors that make sense to me all right next question this is the first size problem a current carrying circular loop of wire radius r current i is partially immersed in a magnetic field of constant magnitude b naught directed other page as shown in figure and you would like to determine the net force on the loop due to the field in terms of theta 0 not the theta 0 points to the dashed line above which is b is equal to okay so there's like a magnetic field like here and this loop is partially inside this magnetic field and we would like to find the net force on the loop we can solve this as if there is a magnetic field everywhere yes i like that suggestion let's do that my question is like suppose there is magnetic field everywhere what is the net force on this entire loop like suppose zero right nice yes then rather than just so trying to find this like really big chunk i can just try to solve this little chunk and then if i put a minus in front of this chunk that's going to be the force this guy right because the force on this guy should be the minus of this guy so that the total force is zero let's do this let's set up the geometry so i want to take a point here like a very tiny segment in which the wire is almost almost straight so that so let's determine the okay the current is like this way if i do the right hand rule it starts with from this and this this is going to be the force i determine the force with the right hand rule okay and let's call this angle theta and i want to also introduce this angle with the perpendicular this is also theta now let's write the force it's df is equal to the current which is constant along the loop times dl dl is what dl is r d theta and b which is b naught is this all we know that's not all right we also need to project this because for every point here there's another point here and the thing about these pair of points their projections they are they their horizontal projection projections cancel whereas they are perpendicular their vertical components add up so i need to project this guy to this axis and then integrate so for that i put cosine theta okay now i can find the force by integrating f is equal to let's take out i r b naught and integral d theta cosine theta and the limits are minus theta 0 to theta 0. so this integral is like simple this is like sine theta from minus theta 0 to theta 0 which is going to be equal to 2 times sine theta 0 then force is equal to i r b naught there's like you know rule two sine theta up so like along this direction but this is like to remind you like this is the force on little segments so we have to take the okay then the force we are looking for which is actually this segment is missing that missing part is the opposite of this force here then force on the bigger segment is minus of that or we can write it the same thing i r e naught sine theta and the direction is down okay the force is like this way an electron enters a uniform magnetic field b is equal to 0.28 tesla at an angle of 45 degrees to b and would like to determine the radius r and the pitch which is like the distance between two peaks here of electrons helical path assuming its speed is 3.0 times 10 to 6 meter per second okay this is the figure we said if velocity makes an angle with magnetic field we just decompose it let's let me decompose it like the perp and the parallel okay so parallel component does not affect the force no effect on force but we'll see it's going to be responsible for motion along this direction right so thanks to this force this guy is moving and doing this spiral rather than a circle like we saw earlier so we can find the circle the radius of circle how do we do this let's write what we know about circular motion it is going to be m v square over r should be equal to force which is equal to e v okay this is of course i have to be careful because we have two components maybe i can be more explicit and v perpendicular squared over r is equal to charge the perpendicular times b and i take the cross product already because they are perpendicular this and this component and this component of magnetic field they're perpendicular so let's cancel these these and these so what is given is v is given m is given we are trying to find r e is given b is given okay no t right luckily we don't have to do as much work so then i just saw this equation for r which is going to be equal to m v perp divided by e times b now i think i can calculate this 9 point 11 times 10 to minus 31 this is kilogram times v perp is what 3.0 times 10 to the 6 times okay this is going to be if this is 45 it's going to be sine 45 degrees and i'm going to divide this whole thing with charge which is 1.602 times 10 to minus 19 times the magnetic field which is point 28 it's like what this is equal to this is equal to 43 times an exponent is going to be equal to minus 31 plus six plus 19 is equal to 10 to the minus 6 meters and the second one is what is the um what was the question and the pitch p which is like this guy okay and now okay this is the same time that this particle completes one right one okay maybe i can tell it this way from here to here you get one full circle but due to this translational motion this particle is shifted a little bit then that's the definition of pitch so for this i want to write it like this okay v perp is equal to 2 pi r over t then t is equal to 2 pi r over v perp now i can put this back for the p p is equal to v parallel times t which is going to be equal to 2 pi r times v parallel divided by v perp but since this is like 45 these two guys are going to be equal so this is going to be basically 2 times pi times r which is equal to 2 pi of this number is what 270 270 times 10 to minus 6 maybe can write it like two point seven times ten to minus four meter good the net force on the current loop whose face is perpendicular to enough magnifier is zero since contributions to net force from opposite sides of blue cancel however if the field varies in magnitude from one side of the loop to the other then there can be a net force on the loop consider its curl up with sides whose length is a located with one side that is x is equal to b in x y plane a magnetic field is directed along z other page and magnetic field is like directed on z with magnitude that varies according to this equation if the current in the loop circulates counterclockwise that is the magnetic dipole moment of the loop is along z axis okay this is like this find the expression for the net force on the loop so basically the current is like this way this way this way and this way so we can calculate the magnetic field here so b is equal to we just plug in x is equal to b here and this is going to give us 0. so the b magnetic field here is going to be uh what b 0 1 minus a plus b over b which is minus p0 a over b good so that's actually into the page so it seems like since magnetic field is varying so after this point it actually changes sign and we need to calculate okay since p is zero the force on this side is zero first of all so f is equal to zero on this side right so this side and this side do i need to calculate these forces no right exactly they cancel so i just whatever they are they are symmetrically placed so i don't have to worry about this the only part i need to carry is this little chunk and for that one the magnetic field is given like this so let's calculate it f is equal to i l is a and b is equal to minus b0 a over b and this is going to be equal to minus i p 0 a square over b and the direction is like let's use right hand rule i start with my fingers along here since this is out of page there i curl it like that is that correct yes then the force is like that direction to the left okay it's like oh i see it's into the page sorry so i curl and then into the key my fingers and then curl it into the page so this is going to be yeah that's correct thank you but this is f so then this is i mean there is an overall sign but to the left a uniform conducting rod of length d and mass m sits atop of a fulcrum which is placed the distance d over four from the roads left hand end and is immersed in a uniform magnetic field of magnitude b directed into the page an object whose mass is m is eight times greater than the rod's mass is hung from the road rod's left hand and our task is to find the current the direction and magnitude that should flow through the road in order for it to be balanced or to be addressed horizontally on the fulcrum and there's like a knot there are like flexible connecting wires which exert negligible force on the okay there are wires but we don't care those there's some currents rest of or maybe like this way and we are trying to find the current okay so let's let me go ahead and start so for the okay maybe i think your favorite thing is to draw free by diagram so i'll do this real quick so there is the force first mg right and also at the same position but let me shift it a little bit to make it visible and there's also f magnetic fields otherwise i mean there would be no way to balance these two maybe if m is so heavy so lights you had to lift it that way but it seems like m is eight times heavier so i suppose it's going to be down and we need to calculate the distances so this is d over two this should be like d over four this is d over two then this little chunk is d over four all right and now i can write the equation the equation is going to be m times g times d over 4 should be equal to f magnetic field plus m g times d over four so d over force they're gone and let's also put m is equal to eight times m so [Music] f m in this case is going to be 7 times mg and it should be down as we anticipated earlier and the question is what is the current that can give rise to this kind of force from the magnetic field and that should be equal to i okay l cross b now b is other page into the page so let's try a few things let's try with this one and if the current is this way i curl my hands and it's into the page the force turns out to be this way so that doesn't work so then there's the only other one is this way i curl my edge i put my fingers along the wire curl it into the page and it's going to be down yes then that works so the current should be this way and its magnitude should be okay now i know the direction so i can forget about cross product i times n is what d db and this one we found had to be equal to 7 mg so current should be equal to 7 mg divided by i think d times b