 So let us try to derive it now so let's assume that this is the bar magnet draw a cylindrical shape all of you I hope you have copy and pen with you you're not watching it like a movie with a popcorn in your hand okay so this just to be you know being a just to make it simpler the diagram I am taking it as just a cylinder the entire solenoid so this entire solenoid if I treat it like a cylinder okay the first let me draw the basic things so these are the two axis okay so I'll say that this is axial and this one is equatorial okay and the dimensions of this solenoid the radius is given as a okay and the length of this solenoid is also given the length of the solenoid is L okay so L is a solenoid length and the current is flowing in the solenoid that has a value of I capital I is the current okay you are measuring all the distances from this point let us say this is O okay and number of turns per unit length in this solenoid is given as small n okay so given all these things I want to find out forget about magnetic dipole moment right now okay just try to find out magnetic field at this point P okay which is at a distance of small r from this line okay and just like in case of electric dipole we have assumed that what we have assumed distance between the charges we have said that distance between the charges whenever we find electric field okay is very less compared to where we are finding the electric field into a dipole okay so distance of the point where you are finding the electric field that distance is very large compared to the distance between the charges when you were talking about the electric field due to a dipole okay similarly when we are finding electric field due to this solenoid and we are treating this entire solenoid as a magnetic dipole then our assumption here is okay this is too well sorry for that so this is too well just like in in case of electric dipole you have taken distance as to a here I'm taking distance between these two poles let us say this is North Pole and this is South Pole distance between these two pole is too well okay so it's an analog of electric dipole where too well is the distance between the pole and the tool is very less compared to where we are finding the magnetic field alright now just spend some time and see can you get an expression of magnetic field in terms of whatever is written here forget about a magnetic dipole moment the hint is you can take rings like this you know you can take you can imagine a ring like that the width of this ring is dx and this ring is at a distance of x okay just spend some time in it and try to see whether you can get something you know you may not be able to get the final expression but at least try to analyze how you can you know get the equation another hint is a question to you do you remember the magnetic field due to a ring along its axis so you have to use the formula for magnetic field due to a ring along its axis and you will be able to find the magnetic field due to the entire solenoid okay I'll just write down here magnetic field due to a ring along the axis so let us say this is the ring okay this is the ring here current going through is i number of turns on this ring is n okay I am trying to get the magnetic field at a distance r or let us say at a distance you know x from its center at this point which is at a distance x okay the radius let us say is a what will be the magnetic field you guys remember magnetic field in this particular situation would have been mu naught n i a square divided by two times of a square plus x square raised to power 3 by 2 you guys remember this yes or no please message okay so now can you find out magnetic field due to this ring just because of this ring forget about the solenoid you just have to find magnetic field due to this ring which is of width dx and the radius is small a okay get the magnetic field due to this ring here at a point p how far is this point from the ring this point p is at a distance of r minus x isn't it this is x and r is from the center so distance of the ring from this point is r minus x so get the value of magnetic field due to this ring only first of all the number of turns on the ring that will be equal to what n into dx fine and this x was what this x was distance from the center of the ring to the point where you find the magnetic field okay so that x over here is actually r minus x r minus x is distance from the center of this ring to this point p okay so we can write magnetic field due to the ring will be equal to mu naught n into dx i a square divided by two times of r minus x whole square plus a square raised to power 3 by 2 fine so this is the magnetic field due to this ring only alright now in order to get the total magnetic field I need to integrate this right total magnetic field will be the integration of this magnetic field isn't it and I know that direction of magnetic field along the axis will be straight so I don't need to worry about any components here because all the magnetic field due to all the rings okay where ever you consider ring you can consider ring here also there also you can consider anywhere you consider along the axis the magnetic field will be always along a one straight line so you just have to integrate it and the limits of integration yes same here limits of integration you are integrating what you are integrating x right and x is a distance taken from this line right so you are going from this point you're going from this to that isn't it so your first going left and then going right so the limits will be minus l to plus l fine so that's what the limit is now we are going to use an approximation which is not very obvious over here as in the usage how it is getting used when you try to integrate this it becomes very tedious okay but then if I use the approximation straight away what is this approximation that r is very large okay r is very large compared to l okay and I know that x is smaller than l right so if x is smaller than l then r will be always much greater than x fine and if r is much greater than x can I say that r minus x is approximately equal to r only yes or no so I have effectively removed this variable from integration because it is very small it will hardly affect it will hardly affect anything so I'll say that magnetic field B is integral minus l to plus l mu naught n i a square into dx divided by two times of r square plus a square to the power 3 by 2 all of you comfortable with this any doubt till now please message any doubt till now this is very important derivation okay so now that this expression has this expression has come I can you know further make it more simple you know simpler I can say that r is so large that it is much greater than the radius of the solenoid also okay so if I say that then even this I can drop off even this is gone so what I will get is magnetic field B is equal to now I will take all the constants outside mu naught n i a square divided by two r cube so I will get integral minus l to plus l dx so it becomes straight forward fine so this you will get as mu naught n i a square into two l when you put the limit minus l to plus l you'll get two l divided by two r cube okay so this is what you get as magnetic field due to the entire solenoid at a distance r from the center of the solenoid along the axis fine so I say that this is the magnetic field which is axial okay now you guys remember what we were trying to do we were trying to find magnetic field as a function of magnetic dipole moment m okay so can you modify this expression further and get the magnetic field which is this in terms of magnetic dipole moment quickly do that can you first type in what is a total magnetic dipole moment of this solenoid B's sorry m is equal to what total magnetic dipole moment of this solenoid is what all of you type in magnetic dipole moment through any loop is current into area of the loop right now here we have total number of turns n into i into pi a square now how will you get total number of turns capital n is what capital n is number of turns per unit length which is small n into total length of the solenoid which is two l this is capital n this into i into pi a square this is what is the magnetic dipole moment m okay now please modify this and do it quickly now it's not small n i a it is capital n i a small n is number of turns per unit length once you get the answer please message that you have got it or just put the thumbs up that you have got it once you got as a function of m okay poor we got it ritwick got it syme here got it with caps on okay a lot of people are getting it right so let me write down the expression here so what you'll get is this mu naught divided by 4 pi multiplied by 2 m by r cube all of you got this only m is n i a square into pi into 2 l so denominator and numerator pi will come okay so this entire thing this thing is capital m so i have multiplied and divided by 2 also so that i get nicely this mu naught by 4 pi outside which is like typical a magnetic field constant getting it so we have just now derived magnetic field along the axis due to a magnetic dipole moment fine now you guys remember electric field due to a dipole moment along the axis how much it was it was 1 by 4 pi epsilon naught 2 p by r cube so you can see how similar these expressions are okay the direction of magnetic dipole moment is the direction of the area curl your hands along the direction of current you'll get the direction of magnetic dipole moment okay so this is axial now if i correlate with electric dipole moment field and i know that along the equatorial plane if i know that along the equatorial plane electric field if you guys remember was 1 by 4 pi epsilon naught minus p by r cube so the magnetic field will be what magnetic field should be mu naught by 4 pi into minus m by r cube okay so the formulas look analogous fine now we will take up a few more i mean we'll just discuss a couple of more theoretical aspects and then we'll start solving you know numericals so just a couple of more things