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Hi, myself Sufal Kumar,
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Hi, myself Sufal Kumar,
Physics Faculty.
Dear friends, viewers & students, my channel is about Physics Education. I am to concentrate mainly on JEE, NEET, CBSE & ISC in near future.
#sufalphysicsforum
#jee
#neet
#physics
#cbse
#iitjee
#igcse
#cbse12thexam
Pls like, share & subscribe, if you like my educational video....
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LearningTranscript
00:00This is a second part that is second video in the series of electromagnetic induction
00:08lecture in which I plan to do these topics. First, Fleming's right hand load, methods
00:16of generating EMF, emotional EMF, emotional EMF in rotation, energy conservation and
00:24lastly relation between amount of charge induced and change in flux. So let's begin with Fleming's
00:32right hand rule. A lot of students think while discussing Fleming's right hand rule that what's
00:40the difference between Fleming's left hand rule and right hand rule. Fleming's right hand
00:51rule is concerned with induced current only. That means whenever electromagnetic induction
00:59is to be talked about, then only Fleming's right hand rule is needed. Otherwise, in general
01:07situation of magnetic field and magnetic force, always Fleming's left hand rule is
01:14needed or discussed.
01:21Fleming's left hand rule was FBI. That means force, magnetic field and current for thumb,
01:32index and middle finger. Same way this time for Fleming's right hand rule. Again, same way
01:39this time right hand rule. We are to think on the line of same pattern VBI, velocity of straight
01:53conductor, B magnetic field and I is induced current such that these three physical quantities are
02:06mutually perpendicular to each other. Then velocity of straight conductor should be along thumb.
02:15B magnetic field should be along index finger and third middle finger along which induced current should
02:26be located. And when out of these three, two directions are given and third is to be found. Then,
02:36only Fleming's right hand rule is to be applied. I must mention it is thumb index middle and you
02:47should remember VBI instead of FBI. So this was Fleming's right hand rule.
02:54Let's see. Whenever we would be needing, we will apply that. Here comes our second topic was
03:07methods of generating EMF. Now I would be considering a solenoid. Around this solenoid I am to consider
03:19one circular coil of N turns whose radius is B and radius of this solenoid is A. So let us consider this solenoid
03:37having time varying current. Time varying current I is equal to I0 sin omega t. Time varying that means current
03:47is varying with respect to time. Number of turns per unit length N. Number of turns per unit length N.
03:59And radius A. Let me indicate again with some other color. Radius A and this is N number of turns. And already we know magnetic field due to solenoid is
04:17is mu naught N I. Mu naught N. I is replaced by I0 sin omega t. And phi is N B A. In general flux is B dot A. But considering total number of turns. So phi is N B A.
04:39By that phi is N B mu naught N I naught sin omega t. And A is phi A square. Remember I am telling you in advance wherever mutual induction is considered.
04:58That means one here. That means one here in this diagram. Both if this solenoid has a current time varying current. Then because of this green circular coil having N turns will have induced current because of this current. That means it is a case of mutual induction. Later on I am going to clarify in third video in detail.
05:27In detail what is mutual induction. But I didn't plan to discuss mutual induction in the current video. For the time being you just take mutual induction is the phenomenon in which primary and secondary coil both mutually induce each other. Okay. So here.
05:49Again and for mutual induction any and every time smaller radius is considered. Whether this induced current or this green coil induced current. In both the induced current we need to consider smaller radius. Because empty space between the this. That means this is empty space.
06:18This has got no magnetic field lines. Hence this larger radius is irrelevant for mutual induction. That's why we have considered small radius only. Therefore induced EMF induced EMF.
06:42induced EMF E is minus D phi by DT. Now I am to differentiate this whole expression. And while differentiating. We will be taking out all the constants. Out. Only variable are to be differentiated. Minus D DT of N.
06:49Mu naught. Mu naught. Mu naught. Mu naught. Mu naught. N. I naught. Then pi A square. And all these constants will be taken out. Minus N. Mu naught. Mu naught. N.
06:56Mu naught. Mu naught. Mu naught. N. I naught. Then pi A square. And here variable coefficient is omega.
07:03So that will also come out. So that will also come out. And differential of sine is cos. So this much.
07:10large expression is.
07:19large expression is.
07:23and here variable coefficient is omega so that will also come out and differential of sine
07:30is cos so this much large the expression is.
07:40Now second method theta as a function of time this is also quite popular for this
07:49I'll have to show into the plane magnetic fields now it is b into the plane now circular coil this is
08:02a a is the radius and omega is the angular velocity about this axis of rotation situation
08:15you must have understood omega is the angular velocity rotating the coil about the axis of rotation
08:30then phi is nba cos theta because it is phi is in general b dot a so here nba cos theta
08:42and then later on theta is equal to omega t is to be considered as s is equal to bt it implies that
08:51phi is nb for a will be considering pi a square yeah and thereafter differential of cos omega t is minus sign
09:12therefore e is minus d phi by dt is equal to minus nb phi a square d dt of cos omega t minus nb phi a square
09:31minus omega sin omega t minus minus minus cancel e is equal to nb phi a square sin omega t this is what
09:45our target was now coming back to our original list third topic
09:52next topic we are supposed to do is emotional emf that means emf induced in a coil due to motion of
10:06something in fact that something is straight linear conductor of length l next topic is emotional emf in
10:16which we are to study the appearance of induced emf due to motion of straight conductor that means here
10:26in the current situation straight conduct conductor will be moved through the velocity v such that magnetic field
10:38b length l of the rod and v velocity of the rod should be mutually perpendicular
10:49then only emotional emf will be appearing otherwise there won't be any emotional emf right let's see the
11:00situation again i'll explain this thing this is known as a rail because it is very much similar to railway track
11:09so this is the rod the rod is at a distance l is the length this is v velocity of straight conductor
11:17the emf generated by moving a conductor in a magnetic field that is by changing the flux in the circuit
11:30which is called emotional emf that means when this rod is moved in this
11:39leftward direction with speed v then magnetic flux is changing how why because in the process of
11:49movement of the rod this whole thing is area area is decreasing in the process and hence flux is
12:00decreasing and if flux decreases that means emf will be induced across the ends of the rod these two
12:09are the ends and thereby induced current will be appearing here along the rod and then thereafter
12:19current will be continuous in this closed loop but the problem arises how to find the direction of induced
12:28current that for that purpose we will be applying fleming's right hand rule and then let us apply right
12:36hand rule let's see what are fleming's right hand rules said initially we had a discussion on this
12:46we have a discussion on this v v v b i velocity of straight conductor thumb magnetic field index and induced current middle
12:59now i'll be considering c this is into the plane of my laptop into the plane and this thumb is pointing
13:12towards the motion direction towards the motion direction of the rod which is leftward and direction of induced current
13:21rightly across the pq end this is induced current and here v is constant speed now while discussing mathematical
13:34analysis of the situation we will be considering constant speed there is no loss of energy due to friction
13:46that means we are considering ideal situation then by using faraday's law by using faraday's law
13:55phi b is b l x y phi b that means magnetic flux magnetic flux is b a b is already given
14:08and area is length into l into x length of the rod into x is the separation between this length towards the end of the
14:21the rail rail here here x is changing with respect to time faraday's therefore e is minus d phi b upon dt
14:39minus d dt blx already we had that is equal to it imply that e is
14:50b l v so this is emotional emf and point to be noted here for appearance of this particular type of induced emf
15:02always d l and v should be mutually perpendicular otherwise there would be no motionally now same expression
15:12we are to analyze and conclude by using lawrence force lawrence force mainly this would be needed for
15:24tackling j mains related questions and even advanced level also here induced charges also taken into account
15:33the charge q in the conductor will also move frame velocity as the charge will be moving in uniform
15:47magnetic field then definitely it will be experiencing magnetic force q vb charge q of force
15:49it will be experiencing magnetic force q vb charge q of force f equals q vb
16:01d two words now for this purpose now for this purpose we are to apply flaming's left hand rule because
16:10now magnetic force is taken into account due to moving charge
16:16so let me let me apply magnetic field into the plane of my laptop screen and force direction
16:29direction and positive charge is moving and rightly my thumb is pointing downward in air
16:41that means if we consider my laptop screen then it is p2q this is magnetic force i guess
16:51you must notice you must notice this magnetic force and this was determined this direction was determined
16:58by applying fleming's left hand rule work done in moving a charge from p to w is equal to
17:10force into displacement and force is q vb but emf is the work done
17:20per unit charge therefore e is w by q w is q vb into l upon q therefore e is blv
17:40so just now we proved the expression for motional emf
17:45by second method alternatively by using lawrence force again here also we obtain e is equal to blv only
17:57now how to determine the direction of positive polarity of motional emf one method was already i told
18:07you that is to apply fleming's right hand rule because by applying fleming's right hand rule we are able
18:18to determine the direction of induced current so indirectly we are able to find the polarity also of
18:27this induced emf this induced emf generated across these two ends if this is the direction then obviously
18:36positive polarity will be here and negative is here and alternatively polarity is direction of
18:46positive polarity direction of v cross b now i'll explain how for example this is rod of length
18:58this is b into the plane this is b into the plane now i'm to apply b cross b simply cross product i'm applying
19:08cross b the first vector to second vector anti-clockwise this way positive polarity
19:17y is sine theta z b direction of emf g is blv sine theta another example second example this time rod
19:33b and x is b and x is equal to n b is y is cross 90 degree z is b here e is equal to zero since any two of
19:50b l v are parallel what do we learn then b l v out of these three any two is parallel then no motional emf
20:04so that is the point right general formula became for emotional emf e is v cross b dot l now last point
20:20under this heading would be for stationary conductor in case of stationary conductor the force on its charges
20:30will be already you must be knowing f is equal to q v cross b plus e which is already dot e
20:43v is equal to zero because we are talking about the stationary conductor the force must be due to e therefore to
20:53explain to explain the induction of current of current we must assume that a time varying magnetic field
21:06generates an electric field and time varying magnetic field generates an electric field this kind of
21:15point will come across while studying electromagnetic waves but the but the electric field due to static charges
21:27and that due to varying magnetic field have different properties and time varying magnetic field that
21:40means db upon dt hence we have discussed this point also and here under this heading we are to consider
21:50some rod of length l has to be rotated in perpendicular uniform magnetic field in this kind of rotation flux lines
22:04are being cut so that's why induced emf will be appearing across these two ends of the rotating rod such
22:14that one end is fixed at the center of this virtual circle circle and the other end is rotating forming the
22:25circumference of the circumference of the virtual circle this thing only we are two and one standard formula
22:33e is equal to half v omega l square r to be is to be used all the time just formal proof we'll be doing so in two ways
22:46that is two methods we'll be following now discussion is very small although this is the rod of
22:55length r which is constituting the radius this is the rod of length r in the process of rotating like this radius of this virtual circle is length of the rod r
23:15r the charges in the rod experiences of force and move towards r the charges in the rod experiences of force and move towards both at a certain
23:27value of emf there is no more flow of electrons and steady state is reese considering
23:38considering elemental length dx at distance x from o as the rod moves in moves perpendicular
23:52to the field to the field the magnitude of emf across dx will be d e is equal to
24:05d v dx now i am to consider in the same diagram suppose here dx at a distance x omega is the
24:19angular velocity with which the rod is being rotated about o firstly i am to consider this dx element because of this dx
24:32element we'll have very small induced emf de so i am intended to write the expression for de first
24:42then afterwards we'll be integrating that de and limit would be zero to capital r so here it is whatever
24:53thing i explained this is de is equal to b v dx on the line of b v l and since v is equal to r omega
25:06so here v is equal to in the given picture small r is capital r and then dx is r here dx is the
25:21for a small element for a small element this is the radius so v is equal to x omega it implies that de is
25:31b x omega dx it implies that e is equal to b omega x dx limit from zero to capital r
25:50x dx
25:54x square upon two zero to r
26:01it implies that e is equal to b omega r square minus zero square by two therefore
26:12e is equal to half p omega r square and it is very very important and alternative method also i
26:24i'm willing to use that is again very very simple alternatively area swept by the rod area swept
26:34by the rod in covering theta of angular displacement this area would be area swept by the rod in covering
26:51theta angle at the center angular velocity omega a is pi r square into theta y
27:042 pi that is equal to theta r square pi 2 this is just by using the concept of ratio and proportion
27:16therefore phi is pa theta r square by 2 it implies that mod of e is d phi by dt that is whatever thing
27:30we obtain now we have to differentiate it imply that e is all will come out b r square by 2
27:40and d theta by dt and that is from the basic fundamental of rotational dynamics e is p r square by 2
27:52it is omega therefore e is half b omega r square hence it is done omega is d theta by dt as v is ds by dt
28:09meaning that you have to Kahili
28:14you
28:17you
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