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00:00hello my dear friends and students now I'm starting with electromagnetic induction fourth
00:09part of the series lecture in which we plan to do these seven topics which is clearly listed
00:17in upper part of the screen out of which I'm starting with self induction for this firstly
00:23I am to define the phenomenon due to which an EMF is induced in a coil due to varying current through
00:32the same coil is called self induction the flux to a coil of n turns is directly proportional to
00:49the current in in the coil pi is directly proportional to I and constant of proportionality
00:57would be L I where L is a proportionality constant known as self inductance public coefficient of
01:09self induction measured in Henry I must specify here in all these discussions concerned division
01:19device is inductor coil which is small solenoid here few more conceptual points related to self
01:28inductance Weber and Henry unit wise the same equation could be written like this with time
01:36d5e upon dt is proportional to e I by dt it implies that minus e is proportional to di by dt it implies
01:50that e is equal to minus l di by dt this is another interpretation and use EMF another way out to obtain
02:02e is equal to minus l di by dt interestingly if we analyze dimensional point of view from dimensional point
02:11of view volt is equal to Henry ampere per second another point for a coil of n turns
02:31n turns the magnetic flux is linked with each turn therefore the net flux will be n times of the flux associated with a single turn that is n5e is proportional to i and
02:53n5e is equal to l i let's understand theoretically what this phenomenon is by a simple diagram this is i increasing then battery is opposing this virtual EMF is arising in such a way that it should oppose this increasing current now second situation this time same current decreasing now EMF
03:21EMF induced will appear in such a way that it will try to increase this decreasing current so that's why this time virtual EMF is in opposite direction with respect to the previous situation these were the two simple situation of self induction definition for self inductance of a coil is numerically equal to the
03:50the amount of magnetic flux linked with the coil when unit current flows through the coil now i'll underline few keywords numerically equal to the amount of magnetic flux linked with the coil when unit current how y i is 1 ampere and pi is equal to l into 1 it implies that l is phi hence this definition
04:18the self inductance of a point is said to be 1 henry when a current change at the rate at the rate of to the coil induces and EMF of 1 volt in the coil and this analysis we have obtained from the formula is equal to minus l di by dt right
04:48so now here comes self inductance of a long solenoid so now here comes self inductance of a long solenoid that means we have to find or derive the expression for l self inductance of a long solenoid let's consider the diagram first this is battery emf here this is current here this is length l capital n is equal to n l this is total number of turns
05:18this is number of turns per unit length and this is length and this is length all the time solenoid behaves as a magnetic dipole so here also this is cross sectional area a l length already we have discussed now b due to this solenoid would be b that means magnetic field due to solenoid b is mu naught n i but we need to convert this small n into capital n
05:46capital n then flux flux for each turn flux for each turn flux means magnetic flux so phi b is b dot a but here that angle between both the vectors is zero so in general b into a
06:03b is mu naught n i by l into a but total flux it implies that total magnetic flux phi b total equals mu naught n i by l a into total number of turns because this whole expression was for single turn now total flux will be multiplied
06:31multiplied so it would be multiplied so it would be n so it would be n extra but phi total is also equal to l i therefore from two and three we'll be equating both the expressions mu naught n i by l a n is equal to l i i i i will cancel out
06:53therefore l is mu naught n square a pi l this is the expression for self-inductance of a long solenoid but there is one more point in this sometime it is asked in the form of tail question that any rod of permeability constant mu r is inserted in
07:21in between the solenoid then self-inductance of the system l dash this time mu r will be multiplied to the expression mu naught mu r n square a pi l therefore hence the topic is over then here comes energy stored in an inductor
07:49so energy stored in an inductor this topic is again very much magnetic counterpart of energy stored in a capacitor as energy stored inside the capacitor in the form of uniform electric field same way energy is stored in an inductor in the form of uniform magnetic field
08:15field field d lenses law and hence does work for the same that means in order to do opposition to the battery some work has to be done inside the inductor coil and this work done is stored in the form of inductor energy
08:39inductor energy that means inductor energy that means inductor energy stored inductor energy is in the form of uniform magnetic field considering same diagram of self-induction this very diagram i was considering theoretically this work done by the source stores form of magnetic energy in the coil
09:05the same way that means the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of the power of
09:35across that inductor coil minus for E minus L di by dt and outside this bracket i dt and
09:47dt dt and minus minus dw is equal to l i di it implies that w is l outside 0 to i naught i di and i
10:04square by 2 0 to i naught half l i naught square u b is half l i naught square and this is energy stored
10:20inside the uninductor coil and since l value of l depends upon b also so that's why it is in the
10:30form of uniform magnetic field here comes mutual induction this is topic number four in this video
10:39of emi04 that means three topics so far we have already covered now mutual induction as already
10:49you can guess by the word mutual in mutual induction we would have two coils primary and secondary and
10:59both will be mutually inducing each other and generating induced emf thereby induced current
11:09in them due to another so that's why it is defined as some keyword two coils by virtue of which each
11:19opposes any change in the strength of current flowing through the other by developing and induced emf
11:29now consider firstly i would like to draw the diagram of one primary and the other secondary coil we are to
11:40include a cell or battery in primary coil and because of that primary current in secondary coil induced current
11:49will be appearing this is primary now leveling thing this has a current plus one primary coil c1 this is
12:02secondary c2 secondary map there would be here it should be i1 here it should be i2 all leveling is done
12:13i1 consider two coils c1 and c2 placed as shown by varying current i1 and c1 we change the flux
12:28change in flux y2 through c2 due to change in i1 induces an emf in c2 this emf is
12:44known as mutually induced emf and the process is known as
12:58mutual induction i'm willing to underline few keywords now mathematically mathematically phi 2
13:09is proportional to i1 is proportional to i1 that means flux for second coil will be directly proportional
13:18to current in first coil that means primary one one notation is being here used only for primary here
13:28and two and two second here is secondary coil it implies that phi 2 is equal to mi1 where m is called
13:39m is called mutual inductance of a pair of coils here c1 is primary and as you can see primary is c1 coil
13:41and secondary is c1 coil and secondary is c2 coil now emf induced in secondary coil emf induced in secondary coil
13:57e2 is d5 dt di1 upon dt
14:14point to be noted where notation 2 is used and where one is used mutual inductance is
14:29maximum when the coils are wound up on the same axis and it is minimum when the axis of coils are
14:43normal to each other maximum and the same axis minimum normal that means when both the coils are like this
14:56then mutually both the coils will be influencing each other in least manner but suppose
15:05if these two are placed like this then they both there of them will be influencing mutually more strongly
15:15most strongly i mean and i had to underline this this point again m capital m is also measured
15:26in henry i'll be coming on to this defining capital m also but before that i'm willing to share one simple diagram
15:38exhibiting mutual induction among primary and secondary coils so here clearly seen how primary coil
15:48and secondary coil both the coils are sharing flux and henry already i have talked about now defining mutual
15:58inductance from the formula e2 is equal to minus n di1 upon dt if di1 upon dt is equal to 1
16:14then e2 is minus m so the mutual inductance of two coils is said to be 1 henry when a current current changes at
16:25rate of 1 ampere per second
16:29in one point in one coil induces and emf of 1 volt other coil this point i am to write and coefficient of mutual
16:42inductance i is 1 ampere then m is equal to if i1 is 1 ampere then m is equal to y2 is correct
16:55coefficient of and of mutual induction mutual induction of two coils is numerically equal to the
17:07amount of magnetic flux length one coil unit current neighboring coil this is also known as mutual inductance
17:20both are same thing coefficient of mutual induction or mutual inductance and last point under this topic of
17:29mutual induction mutual inductance depends upon factors affecting capital n number one geometry of two coils
17:41number two distance between two coils and number three relative placement two coils orientation of the now i'm switching over to next topic mutual inductance of two long solenoids
18:00here we are given two solenoids s1 and s2 and both are shown
18:07in the form of coaxial solenoids that means both the solenoids axis are common in first solenoid current is
18:21i1 and in second coil second solenoid s2 current is i2 here in the given situation as we can see l1 is equal to l2
18:34equals l but a1 is less than 8 it implies that r1 is less than r2 see this is r1 and r2 is shown like this
18:47but both the solenoid has a length l that is same now flux for first coil phi1 equals m12 i2 that means flux for
19:02first solenoid will depend upon second solenoids current that is i2 also phi1 is equal to b2
19:14a1 for each turn and this is very very important derivation for 12 boats it implies that this was for each turn
19:25each turn but now total flux for total flux purpose total number of turns has to be multiplied that is
19:37n1 but for first solenoid magnetic field will be second because of second magnetic field since
19:46we are into mutual induction we are into mutual induction so second solenoids magnetic field will be causing flux in
19:56first solenoid but number of turns will be khud ka self of b2 a1 and n1 b2 will be mu naught n2 i2
20:07because already we knew b is mu naught n2 i2 because already we knew b is mu naught n i and also
20:17also n1 is equal to n1 l for a1 pi r1 square n2 l this is equation 2 therefore from 1 and 2 m1
20:33i2 i2 i2 is equal to n2 i2 i2 is equal to n2 i2 pi r1 square n2 l and here we go i2 and i2 cancel
20:46it implies that m12 is mu naught n1 n2 pi r1 square l this is equation number three now same process
21:01we'll be doing for second solenoid s2 we'll be doing for second solenoid s2 also similarly for s2 phi 2 is
21:12m21 and let me clarify m12 was mutual inductance of s1 with respect to s2 same way inductance of
21:29s2 with respect to m21 i1 this is equation number four also phi 2 is b1 a2 n2 total flux for s2 again let
21:50us first clarify flux for second coil second solenoid is due to s1 current and also flux for
22:01second solenoid s2 is due to first solenoid s1's magnetic field because it is mutual induction now i'm to
22:12to expand again b1 b1 a2 n2 phi 2 equals mu naught n1 i1 pi r2 square here n2 l this is equation number
22:285 now from 4 and 5 now from 4 and 5 we can write m21 pi 1 mu naught n1 i1 pi r2 square n2 l i1 pi 1 cancel
22:46it implies that m21 is mu naught n1 n2 l l that time also i wrote l pi r2 square now coming back to our
23:00original diagram here now empty space between here empty space between these two solenoids here in this
23:12region there lies no magnetic field lines for s2 also r2 is irrelevant for second solenoids flux also we
23:26would be considering r1 only although finally we have r2 but we'll be ignoring this r2 that means exchanging
23:40r2 for r1 now what do we mention empty space both the solenoids has no magnetic field therefore only
23:54inner radius is to be considered for second solenoid s2 also therefore m21 is equal to mu naught n1
24:09n2 l pi r1 square now this has become equation number six now coming back to equation number three as you can see
24:23same expression is obtained mu naught n1 n2 pi r1 square l mu naught n1 n2 l pi r2 square we have replaced by r1
24:38square it implies that m12 it implies that m12 is equal to m1 from equation 3 and 6 m12 equals m21 equals mu naught n1 n2 l pi r1 square
24:59and that is equal to m and that is equal to m and that is equal to m and if we transform this expression into
25:07slightly simpler form therefore m is mu naught for small n1 n1 by l for small n2 n2 by l and that is 1 l is already there
25:24the pi r1 square it implies that m is mu naught n1 n1 n2 a by l where a is smaller cross sectional area
25:41which is relevant for the purpose so both these expressions are useful for numerical purpose hence
