Class 12 Physics in 30 Days|Day 18|Complete Crash Course|Board 2025|Most Important Topics + PYQs#pyq

追加: 2026-04-28

[00:00:11]So chapter 5 chapter 6 PQ.

[00:00:18]So first one going to be somewhat interesting. Draw a labelled diagram of an AC generator.

[00:00:28]Right. drawn labeled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil. And then a circular coil of uh cross-sectional area 200 cm square and then 20 turns is rotated about the vertical diameter with the angular speed of 50 rad/s in a uniform magnetic field 3 into 10 ^ 2 - 2 Tesla. Calculate the maximum value of the current complete five answer.

[00:01:08]Five mark answer.

[00:01:17]2017 2020 2023.

[00:01:30]2017 18 212 maybe we may expect in the we may expect first you going to draw the AC generator concept diagram derive the expression for the instant instantaneous value of the emf induced in the coil in the AC generator diagram you're going to follow your diagram Imagine diagram Answer sorry omega you're going to say as maximum emf maximum emf sin omega diagram followed with principle.

[00:03:25]A circular coil of cross-sectional area.

[00:03:28]They have given the area very clearly.

[00:03:30]So first write the given data. Area 200 into 10 ^ -4 m² area number of turns 20 is rotated about the vertical diameter.

[00:03:48]Right? Rotated angular speed. It is omega angular speed. F angular frequency angular frequency chapter velocity 2 which is equal to 1 by 2 pi roo<unk> that becomes your angular angular frequency with the angular speed omega which is 50 radian per second in a uniform magnetic field of 3 into 10 ^us 2 tesla maximum value of current so I s omega First you need to find your emf. EMF sin omega t voltage by resistance voltage which is your maximum resistance resistance resistance right resistance So you keep r as it is n b a omega upon r you're going to substitute only the given term EM not a current. Maximum current.

[00:06:35]No need to mention the other terms.

[00:06:38]Simple 20 into 10^ -2 A value 200 omega.

[00:06:49]Yeah. Right. Angular. Yeah. Correct.

[00:06:52]Omega angular speed which is 50 rad per second.

[00:06:58]50 rad per second.

[00:07:02]resistance value.

[00:07:08]You just multiply this term alone.

[00:07:21]Two plus one five up 6 into 10 ^ 5 by R square you have to convert into sorry you have to convert into me 200 suppose term like cm you can write 10 minus - 2.

[00:07:53]So this becomes 10 ^ -4 into 200 correctus rightus 20.

[00:08:12]Yeah, right. Thank you so much ma. This is your maximum current. Right.

[00:08:28]Okay. Fine. Fine. So be very cautious when you substitute the value.

[00:08:35]Check whether it is correct or not.

[00:08:41]The complete five mark very important repeat I have mentioned it in the first case itself you have this one a.5 m metal rod long metal rod.5 the three mark. Okay. The three mark completes the circuit as shown. Yeah, of course. The area of the circuit is perpendicular to the magnetic field area.

[00:09:47]So this says the angle between them is 90° perpendicular and that's they said it is perpendicular the resistance of the total circuit resistance they have given most of the things they have mentioned very clearly calculate the EMF induced in the rod.

[00:10:23]for motional emf.

[00:10:38]Emotional EMF magnetic field the.15 you can convert into like this.

[00:11:13]You will feel light.

[00:11:20]15 into 10 power.5 into 10 ^ -1 in easy right and then they are asking they they need velocity v the vi 2 m/s 2 just multiply 10^ us 15 into 10 ^ -2 right 15 into 10 ^ -2 be very careful first segment ment.

[00:12:19]Calculate the force needed to move the rod in the direction as indicated with a constant speed.

[00:12:44]magnetic field.

[00:12:57]So force B I L sin theta 90° B I going to do the simple resistance R which is 3 home.

[00:13:31]Since we talking about the force, we have another beautiful formula which is B² L² V upon R force, right?

[00:13:54]Right.

[00:14:00]just we have substituted right we didn't include anything resistance I'm going with this right I'm going with this format Correct.

[00:14:32]Just substitute the B value, length value, velocity, and then the resistance R, right? Resistance R. So just you going to substitute 15 into 10 ^us 2 square length you have length so five into 10 ^us the square right you have the next term which is V upon resistance of the circuit is You get the term like five time 15US 25 into 10 ^ -2 and then this normal term. So the term is 5 into 15 into 10 ^ -4 into 25 into 10 ^ -2 into 2 upon normal going to calculate normal calculation force.

[00:16:00]So this is your value right? So final value make sure that you're mentioning the newton.

[00:16:26]So you cannot go with this formula. You may have confusion.

[00:16:40]Vit replace right= emf by rv.

[00:17:08]So this is the way you have to solve. So first try to think why it is needed.

[00:17:27]Sorry. Coming to this one. Two coils C1 and C2 are placed close to each other.

[00:17:36]The magnetic flux 52 linked with the coil C2. Right. of the coil varies with the internal sorry varies with the current I1 flowing in coil C1 as shown in the diagram. I think uh we missed the diagram. We will try to sort it out.

[00:18:03]Obviously there is a magnetic field and the magnetic field find the mutual inductance of the arrangement inductance First term. First term.

[00:18:35]First just previous directly proportional to I and proport proportance and mutual inductance.

[00:19:13]We have studied with the emf dt this can also be written as m d upon dt.

[00:19:28]So first segment. Okay.

[00:19:39]I think I missed the diagram.

[00:19:48]So you'll be finding your mutual inductance. Okay, mutual inductance.

[00:19:54]Find the rate of change of current. Find the rate of change of current that will induce an emf of 100.

[00:20:12]Correct.

[00:20:15]They have given correct diagram. they have given something. So just in in the KV just take a screenshot or just go with that diagram number of terms.

[00:20:48]Okay.

[00:20:51]given the next term.

[00:20:57]Okay.

[00:21:02]Self inductance.

[00:21:04]Selfind.

[00:21:11]Okay. We'll move the next question.

[00:21:20]Coil, right? A coil of number of turns.

[00:21:26]A coil of number of turns is rotated at a constant angular velocity omega. Right? Is rotated at a constant angular velocity omega. Omega in a uniform magnetic field.

[00:21:56]A angular velocity omega magnetic field B connected to a resistance R.

[00:22:02]Okay. First obtain an expression for the maximum emf induced in the coil. First part maximum We can easily sort it out. Number of angular velocity it rotates in per second which mean it is rotating number of you can visualize. You have to visualize this.

[00:22:48]It can be any shape but closed loop which has n number of turns uniform magnetic field.

[00:23:03]You can write simply your first thing EMF E sin omega t but you have to express B which is B A cos theta generator generator right. So different generator generators omega omega you're going to follow the same equation but in a random Obtain the expression for the power dissipated power dissipation equation.

[00:24:31]We cannot go with this equation. Next we can go with this which is nothing but emfist you can write everything sin omega t by r Maximum current problem school students problem spend at least two minutes time to read.

[00:25:27]First 12 numbers calculation is not so tough for you guys since you are in CBC.

[00:25:43]Take one minute time to read the question.

[00:25:56]Spend some quality time to read the answer. Sorry. Uh question because most of the students problem.

[00:26:23]So keep in mind power. So finally okay you can complete with this if needed.

[00:26:42]You can substitute this term n² b² omega square a omega a² / 2 rating sin² omega t.

[00:27:11]Okay.

[00:27:23]So always take some time to read the other things also. Okay.

[00:27:36]Fine.

[00:27:38]The magnetic field through a circular loop of wire 12 cm, right? A magnetic field through a circular loop. Circular loop 12 cm.

[00:27:55]Okay. uh radius radius and then resistance of that circular loop on the 8.5 ohm resistance changes with time as shown in the figure. The magnetic field is perpendicular to the plane of the loop. The magnetic field perpendicular to the plane of the loop.

[00:28:23]We'll try to sort it out.

[00:28:27]We'll try to figure out how the diagram will be the magnetic field through a cilistance 8.5.

[00:28:47]Okay.

[00:28:50]The magnetic field is perpendicular to the plate. First calculate the current calculate the current in the loop and plot it as a function of time a function of time. So given radius resistance.

[00:29:18]So which makes an angle 90° in terms of which mean the function of time simple they have given the term It is very easy for us to find the area r² r² you can mention as it is right 12 into 10 ^us2 right so 12 into 10 ^us2 mention next you're going to find only the induced closed loop again sme problem motion emf which mean towards the left towards right B.

[00:31:17]Next.

[00:31:26]First of D5B by DT first number of they didn't mention anything. So we can consider your to be one and then magnetic field A cos theta if I mention it is perpendicular perpendicular magnetic field is perpendicular to the plane like table. Okay.

[00:32:42]and the magnetic field.

[00:32:45]Right? This becomes 0°.

[00:32:59]So that is Right? Zero emf which is equal to minus of which is and then you have your change in ba by dt right by dt.

[00:33:21]So is nothing but so we have d * of b upon dt by r right by r according to this one. So the term is B because it is rotating magnetic field. They didn't say anything current induced in the loop as a function of time. But you need to draw a graph.

[00:34:09]Graph a function of time. There is a change in time. Change in time.

[00:34:14]what we can say since we don't have any value.

[00:34:20]So you have your magnet change in time current with respect to time we can simply write dB dt db dt area doesn't change magnet db by dt Since current induced in the loop and plot as the function of time because I don't have anything to refer also for magnetic field okay current current in terms of function of time since we don't know anything resistance AU R² I I'll move the next page I equal to minus A by R DB by dt previous you can substute the value resistance resistance yeah they have given 8.5 dB by dt 8.5 dB upon dt I think for sure there won't be any changes so r² by 8.5 you can get the term in terms of current with respect to time. Okay. DB by DT. I think this is the final answer once in the class refer this also mostly right. So since with respect to time we will keep this as a constant right because I think this would be the final term I think but as you said we can conclude the answer last step maybe right? Yeah, I think nothing will be there because we will cross check once which mean in terms of rad changes with the time correct as shown in the figure. Magnet they are 0°.

[00:37:36]So by dt db by dt resistance db by dt this would be the final one graph you can get the final answer just we we need to plot graph they they won't mention many magnetic field what we did is perfectly done. Nothing wrong in that. Okay. Once.

[00:38:25]Yeah. Next we are in this part. A circular copper disc of radius 10 cm rotates. I think previous similar a circular copper disc of radius rotates angular speed about an axis through its center and perpendicular to the disk.

[00:38:51]Okay, perpendicular to the disk. A uniform magnetic field of 2 Tesla acts perpendicular to the disk. Okay.

[00:39:03]Perpendicular to the disc.

[00:39:11]Calculate right? Calculate the potential difference of potential difference developed between the axis and the rim. In the rimf we have studied one thing the cycle spokes dx whatever it is equal to b in the r omega in the Rate.

[00:39:56]So your value becomes like R² by R² by the term.

[00:40:06]I think we will follow this first. We will read magnetic field. Yeah. Exactly.

[00:40:20]EMF even text which is nothing but 1 by 2 B omega R² omega B straight substitution you can find the value 1 by into 2 into 10 ^ -1 omega 20 or you can write 3.14 or 20 22 by 7 whichever you like you like 10^us so have square okay so when you calculate you get your emf do do that. Okay. Second part.

[00:41:21]What is the induced current? If the resistance of the disc is 10 ohm induced induced in the coilp resistance resistance divided by 10 you get the final value you get the final value spend time to read the question first sir please 02 0.628 >> 0.628.

[00:42:28]Nice. Okay.

[00:42:29]>> Next 0.02 >> 0 to p amp. Ah okay. Okay. Got it. 0.06.

[00:42:43]Okay. You can also convert this if you wish.

[00:42:49]02.

[00:42:59]>> Okay, >> you got it.

[00:43:01]>> Got it.

[00:43:02]>> So if you you can substitute you can end with this also. This is your current.

[00:43:08]Okay. You can also convert into mill if you wish.

[00:43:12]Next mill.

[00:43:16]Yeah. Next. The magnetic field in a certain region is given by five ICAP and then minus 2 kcap Tesla which mean they have the XY X is plane okay how much magnetic flux passes through passes through a loop of area in this If the loop lies in X Y magnet.

[00:43:59]Okay.

[00:44:01]Magnetic field.

[00:44:08]So how what can be done here?

[00:44:11]First try to sort it out.

[00:44:15]You have your two component.

[00:44:17]Uh imagine this is X Y sorry passes. Yeah. How much magnetic flux passes through the loop of area in the in this region if it lies in XY in the plane in the plane but magnetic field obviously in the Okay.

[00:45:06]XY sorry X plane 5 I - 2K the cap follow and then vector in the comp I'll be substituting here.3 or 3 into 10 ^ - 1 whichever way you dot product A cos theta B vector dot A vector Max 90° that becomes zero vector value 2 into.3 becomes Okay. Minus and then since it is a flux you can write your webber.

[00:46:53]Okay. So this is your thing.

[00:46:57]Okay.

[00:47:03]Yeah. Next one. A wire of length.1 right moves with a speed of 20 m/s in a magnetic field.

[00:47:23]So velocity magnetic field you will get the answer.

[00:47:35]Okay, next we are moving to this one.

[00:47:39]A coil of wire area 100 cm square - 4 m² okay is placed with it plane making an angle 60° magnetic field and the magnetic field the magnetic flux linked with the coil.

[00:48:09]First one you can write directly a 100 into 10 power into cos 60 circular circular they didn't mention anything right wire enclosing an area obviously can be a rectangle square circle but plane making an angle 60° making an angle 60° magnet area vector in the surface is the surface.

[00:49:17]Yeah. Magnetic field.

[00:49:20]Right.

[00:49:26]Are vector magnetic field. Correct. Are vector magnetic field.

[00:49:32]The remaining term 60° magnetic field magnet Right. Area vector magnet.

[00:50:06]If I do so, remaining vector vector 30° right formula 30° Maybe it becomes <unk>3 by 2 I think 3 correct I think you will have only minus 3 up roo<unk>3 by 2<unk> 3x 2 - 3 Weber flex Tesla square magnetic field square or T m² if the magnetic field is reduced to zero reduced to zero in 10 ^ milliseconds.

[00:51:43]Okay.

[00:51:50]Magnetic field reduced to zero.

[00:52:04]because there is a change in magnetic field which results in em.

[00:52:10]So you can take that as one dy by dt final minus <unk>3x 2 into 10 ^ -3 upon time change in time 10 ^us 3 cancel so you will have <unk>3x2 MF cannot be in negative value. So you can write root3 value 1.732.

[00:52:48]Yeah, most welcome to do that.

[00:52:51]Right. 1 732 divided by two. Most welcome. Okay, you can do that.

[00:52:58]Roo<unk>3 by 2 V happy.

[00:53:04]Next we'll move to the next problem.

[00:53:08]Yes. So we are moving to now fifth chapter. Sixth chapter we'll move to fifth chapter.

[00:53:20]A bar magnet exactly in the sixth chapter. Fifth chapter problems similar you no need to worry about it. A bar magnet of magnetic moment lies along with the directions of the uniform magnetic field magnetic field of 22 Tesla or 22 into 10 ^ -2 what is the amount of work done work done an external talk to turn the magnet so has to align its magnetic moment. Okay.

[00:54:12]The bar magnetic field it is nothing but your potential.

[00:54:24]Second chapter.

[00:54:26]Second you have your cos right because you would have mentioned very carefully final final theta initial correct similar magnet Imagine you hanged using a thread force into distance which is equal Force distance. Final equation. Minus P E cos theta torque potential ma if it is 90° right. Initially if it is right so that becomes your sorry correct.

[00:56:08]So you have only this term minus PB 2 L P 2 sorry 2 A Q is nothing but your potential equation equation. So MB initial cos final minus cos theta final initial cos theta final minus become plus theta that becomes your minus Okay.

[00:57:06]What is amount of work done of torque to turn the magnet as so as to align its magnetic moment normal to the field direction opposite to the field direction.

[00:57:26]One is stable.

[00:57:29]Unstable. Stable 0°. unstable 180° maximum work done normal to the field.

[00:57:48]But normal to the field initial it is zero right? Initial it is zero. Final unstable opposite 180°.

[00:58:17]Okay.

[00:58:19]stable unstable conditional.

[00:58:29]Yeah.

[00:58:31]If it is stable conditional it is 90°. Okay.

[00:58:39]Maxute.

[00:58:44]You get your work done.

[00:58:50]You get your work done. Wet B.

[00:59:04]0 90 director will be multiplying this term alone you'll get the answer two times I think this is okay Anything with us - 1 - 1 -1 becomes + one two times of segment. What is the torque on the magnet in case conditions stable unstable 90° maxite unstable integrate Right.

[01:00:23]90° max is zero sign of z.

[01:00:35]We are moving to the next one. Next problem. A short bar magnet placed with its axis inclined at 30°. Okay.

[01:00:51]to the external magnetic field of 800 G 10^ -4 - Tesla acting horizontally experiences a torque of either just to make it simple one 16 into 10 ^us 3 newton per meter maybe sometimes difficult first magnetic movement of the 30° magnetic field obviously rotation sin theta correct b sin theta torque value 16 into 10 ^ - B 800 into 10 ^ -4 okay sin 30° 30° 1 by 2 the normal you divide you get the final value of magnetic moment work done by the external moving the magnet of the most stable 0° to the most unstable 180 Stable. Unstable.

[01:02:22]Cos 180°.

[01:02:36]+ 2 * you get the final answer.

[01:02:49]Next, a coil, right? A coil of n turns.

[01:02:57]The coil we don't know how many turns but it has n turns a radius R carries a current of I unwound and rewound to make another coil.

[01:03:28]by for example.

[01:03:54]magnetic moment. It has a magnetic moment magnetic moment. The magnetic moment you can write Q the Qgen because it carries a current and then number of you can write number of times I they're asking the ratio of both area is nothing but pi r² square.

[01:05:01]N right cancel R. Sorry.

[01:05:10]R2 R1 duplicate duplicate R square 1 by 2 four.

[01:05:34]So this is your final term.

[01:05:46]I think. Yeah. MC current sensitivity sensitivity.

[01:05:55]Number of increase voltage sensit by K motor. Resistance number of increase of number of two times of radius number of number of increase of n_sub_2 by m1 is nothing but 1 by 2 ratio 1 is to two.

[01:07:12]So you have to be very cautious.

[01:07:21]number of number of so right. So next I think problem as a homework right a small magnetic needle. So you have your XY plane just give a try. So try when it is share the answer to me and I'll cross check and then the right one will be posted in the group. Okay.

[01:07:54]Uh wait I'll check whether this is the last or not magnet placed with the axis similar to the previous condition which is equal to B sin theta problem take as assignment. Thank you so much for joining the session. Thank you.