ElectroMagnetic Induction 03 : Motional EMF 1 II e = Bvl Derivation and Best Numericals JEE/NEET

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Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App. Download the App from Google Playstore ( bit.ly/2SHIPW6 ) Physicswallah Instagram Handle : instagram.com/physicswallah/ Physicswallah Facebook Page: facebook.com/physicswallah Physicswallah Twitter Account : twitter.com/PhysicswallahAP?s=20 Physicswallah App on Google Play Store : bit.ly/2SHIPW6 Physicswallah Website: physicswallahalakhpandey.com/ LAKSHYA Batch(2020-21) Join the Batch on Physicswallah App bit.ly/2SHIPW6 Registration Open!!!! What will you get in the Lakshya Batch? 1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey. 2)Live Classes and recorded Video Lectures (New, different from those on KZbin) 3)PDF Notes of each class. 4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level. 5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture. 6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021) 7)In case you missed a live class, you can see its recording. 8)You can view the videos any number of times. 9)Each chapter will be discussed in detail with all concepts and numericals 10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams. ****Test Series for XI & XII**** We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year. The test series follows very logical sequence of Basic to Advance questions.& Evaluation of Test and Solution to all the questions at the end of the test. ========================================================= For PDF Notes and best Assignments visit @ physicswallahalakhpandey.com/ Follow us on: Instagram instagram.com/physicswallah/ Facebook: facebook.com/physicswallah =========================================================== 12 Chap 6 II ElectroMagnetic Induction 01 : Magnetic Flux II Faraday's Law & Lenz's Law JEE/NEET kzbin.info/www/bejne/qafHZmV9fs-perc ElectroMagnetic Induction 02 : Induced EMF in Rotating Coils - Numericals on Faraday's Law JEE /NEET kzbin.info/www/bejne/o2WZnJamf9uDrqM ElectroMagnetic Induction 03 : Motional EMF 1 II e = Bvl Derivation and Best Numericals JEE/NEET kzbin.info/www/bejne/foLXm4uXbL-hoKc ElectroMagnetic Induction 04 : Motional EMF 2 II Conceptual RAIL Problems in Motional EMF JEE/NEET kzbin.info/www/bejne/panZioKgq9N0l5I ElectroMagnetic Induction 05 : Motional EMF 3 II Motional EMF in Rotating Rod JEE /NEET kzbin.info/www/bejne/Y6SahGt_btR-jas ElectroMagnetic Induction 06 II Self Inductance & Inductor II Self Inductance of a Solenoid JEE/NEET kzbin.info/www/bejne/fZexgGSIir59qpY ElectroMagnetic Induction 07 II Growth and Decay of Current in L-R Circuit JEE/NEET kzbin.info/www/bejne/o5i2lZKNlMSBitE ElectroMagnetic Induction 08 II Mutual Inductance : Mutual Inductance of co-axial Solenoids JEE/NEET kzbin.info/www/bejne/b2Svmaudardgjc0 ElectroMagnetic Induction 09 II A.C Generator - Working of A.C Generator and a Famous Story JEE/NEET kzbin.info/www/bejne/j5-1Z4yfesiIpa8

[@SD-pz6dj]

sirr...i have a doubt...emf nikaal ne wala 4th question(semicirle) m circumference ka tangent use nhi krskte? as length vector? asa karke mera ans e= -VBl aaya...kya gaalat krrhi m🤔

[@paramveerdosi]

Sir aap nhi hote to hum 12th kse nikalte

[@aashishsanjayambast322]

In last question and last part shouldn't L = L ROOT2 PLZ REPLY

[@yogeshmishra7699]

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[@vishwamagrawal7253]

Sir iska pdf mil nhi rha he

[@PhysicsWallah]

Download PDF Notes of this Lecture @ physicswallahalakhpandey.com/class-xii/physics-xii/06-electromagnetic-induction/

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Correct

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Soooo True......!

[@whomptilizes]

For those who are struggling to solve the questions relating v, B, l, e.... You can simply take the components of v/l/B such that those components are mutually perpendicular to one another.

[@GauravYadav-ry3nu]

Acha....

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Sahi ayaega

[@AanchalSingh-ch9te]

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[@Cuppy-e4r]

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[@shivamgoel2788]

Dear Sir, in the last question the displacement length would be l= root 2 x r and hence the answer will come out to be bvr..commenting so that other students dont get confused 😄

[@abhishekshivpuri2473]

Ys u are right....

[@nothingbad8019]

ty fam

[@yashsaxena7642]

No l would be r/root2

[@nobitasharma6272]

it's -ve

[@Shivam....]

Also in last 3 no ans=bvl as bv2r

[@rosellia_]

5:25 method 1 18:44 method 2 25:29 Question 1 29:00 generalised case 47:00 next 49:00 next 01:01:01 next 01:11:11 01:12:00

[@shivamsingh-hn1lp]

Thanks for the timestamp

[@Prachita516]

Here method 1 and method 2 are for derivations right ?? Cuz i'm just skipping the derivations part

[@ayushbangar1955]

@@Prachita516 yes

[@tanyadagar3190]

@@Prachita516 isn't this topic deleted for boards?

[@Prachita516]

@@tanyadagar3190 even if its deleted its important for next lectures so i have done it. nd just left the derivations part

[@Amirhussain_17]

To the ones interested, the formula E = (v×B).l is nothing but Scalar triple product ( to be studied in Vectors in maths) ... All the properties of Scalar triple product are valid here... So if you know what scalar triple product is, then you can ace this topic easily...

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Sir in last qs last part 1:23:00 , we know that pd for (3) was 2Bvr , and as in (4) the path taken by current is half as that of (3) we can simply say that potential drop is half of 2Bvr = Bvr

[@abhaykumar-rt4uo]

yaar mera bi 4 mein ans -Bvr aaraha hai sir ne l=root2 r kyu nahi liya plzz clarify anyone

[@yurathore9007]

Sir ne jaldi jaldi mai kr diya ans - bvl aayega. L=✓2r lege na ki 2r

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@@abhaykumar-rt4uo he forgot most probably

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[@solitary6002]

56:38 this can also be solved using vectors.... Magnetic field is going into the plane of the field. Therefore, B vector = -Bḱ v vector can be resolved as v.cos(37°)î + v.sin(37°)j Or 4î/5 + 3j/5 Length vector can be taken +lî or -lî as you choose. Now solving (v×B).l gives 3/5vBl

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[@harshitgarg9703]

sir at 1:00:00 isnt the angle between v cross b and length vector 180 degrees, then E = -BVL and answer will be E = -BV2R

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[@Sasivardhan-dz4vs]

1:22:48 sir here l=root2*r so the final ans will be -bvr

[@parthdudhe5236]

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[@Bella_ciao117]

​@@parthdudhe5236bhai √2r = l hai Toh sir put nhi kiye hai

[@Bella_ciao117]

bhai √2r = l hai Toh sir put nhi kiye hai

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Yeah

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1:18:37 vq-vs 2VBR hoga

[@RaKtaN_]

18:00 motional emf proof for ncert ❤️

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1:12:02- FULL CONCEPT QUESTION

[@vinodsachan9136]

Please help me how to find direction of length vector ????

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​@@vinodsachan9136shortest distance between final and initial path

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[@alexmercerind]

In the last question, why haven't you taken the l as l√2 ... As the diagonal displacement was l√2

[@sanamansari1197]

Here displacement length is √2 r so why sir haven't putted it in equation. Plz tell anyone , physics wallas.....

[@harkiratsinghchana9466]

arre bhai question mein thoda radius r diya hai v toh sir r laekr hame smjhaya k displacement length aise laeni hai na k PQ ki arc se. or vo sir do side r laeke angle nikala hai cos 45.question mein given hi nai tha toh sir ne nai daala sir ne bilkul sai kiya hai

[@yashubhasin8644]

@@harkiratsinghchana9466 1:12:25 It's given that radius of coil is 'r'

[@rahulshuvin5332]

Sir could've taken l=r√2 but he didn't, i guess he forgot. you are free to put the value, the answer will be e=BVr which is correct :)

[@aryananhsirk7604]

where is lecture no. 2 ??? can anyone please reply

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[@imyasharya]

1:14:20 1. Net emf = 0 2. Vp - Vr = 0 3. Vq - Vs = 2Bvr 4. Vp - Vq = Bvr

[@imyasharya]

Well, the last one had a sign problem

[@ShivamKumar-hx2tc]

When the velocity vector and length vector are not perpendicular then we also break the components of velocity vector along and perpendicular to length vector. we already know that the components of velocity vector along length vector provided 0 emf. Then only solve the perpendicular component. e=Vperpendicular*B*L

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[@balwankundu2460]

At 1:15:00 , as there is no change in flux in coil , there is no change in the flux in rod too , so the emf of rod must be zero tooo

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Did you find lecture no 2?

[@motivation27293]

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THANK YOU ALAKH SIR THANKS A LOT

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[@rajraj-pu5uo]

Sir we can also take component of length along v and perpendicular to v . e along v will be zero and we can directly find Vp-Vq

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Students can directly apply right hand palm rule to find dir of v×B

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You are doing revision or completing chepter

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@@sonuraja5072 completing

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Right ....

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Very good classes..keep going. Those who are looking for math classes in simple english please watch math kafe...

[@decorousdivina3295]

u tube sahab ka bahut dhanyawad....physics wallah sir ji ko hamare pass pahunchane ke liye!

[@_iamld]

🙏🙏🙏

[@tabaahi_07]

Neither there is need of coaching institute 😉😉😉nor any online subscription🧐🧐 If u wanna fall in love with phy ,just approach physics wallah😍😍😍🤘🤘

[@me0wAnna]

There is no words to thank you sir You are best.

[@AbhishekPandey-sg1eb]

4) ans:-BVr (l=√2r)

[@jyotishmankalita.1754]

Thank you sir.. Imagine the whole generation thinking sir ji

[@arshiyakhan1170]

Best teacher......u are........

[@shatanikmahanty]

Thanks sir. You are awesome. Was scratching my head for a long time trying to understand this concept. You solved my problem.👍👍

[@krishnakumarimishra4285]

Sir in 55:03, the 2nd question where V is with angle of 37 degree with L vector, we can also solve by resolving the velocity into two components i.e. Vsin37 and Vcos37. Vcos37 is parallel to l vector so e=0. Therefore e will only be due to Vsin37 component i.e. e=BlVsin37=(3BVl)/5. This way is also correct right sir??

[@rashmijurel7083]

How can angle between V cos37 and L vector is zero . I think it should be 180. Please explain

[@IIT24Aspirant]

100%correct

[@StudyRate]

I know that 99% will ignore this but for the 1% who is reading this are a very sweet person and have a great life and good luck in the future ??

[@niharikasrivastava1039]

THANKS......HEHE

[@VipinKumar-cm6ve]

Why question mark

[@sarbaridas653]

❤

[@ShivamSingh-hy1tf]

Nice

[@pradeeshadithya3641]

Physics with Humour is a Beautiful combo

[@niladrimondal1078]

Once I used to watch these lectures Now in a medical college Alakh sir is my God🎉

[@AnuragMishra-ob1er]

attendance please how many of u think sir should awarded by Padmashree award.💕💕😎😎

[@khushisingh..4965]

Inspite of being PCB student, because of u physics is at the top of my favourite subject list,❤️👍

[@foodhacksforteenagers8226]

This topic is not in cbse board syllabus I guess We only have to study self and mutual induction Is this correct

[@priyanath793]

@@foodhacksforteenagers8226 yess this topic is deleted !!

[@dhrubajyotidutta3297]

Tu Kisi rail si guzarti hai one of the most beautiful songs♥️

[@Khatukaladla12]

You are great sir.. ,sir apne to physic ko very easy bana diya h... Thnx sir.

[@lynoragaming8784]

Dil garden garden ho gaya , after completing lecture ❤😊

[@Jannahgoalss]

Pehli baar KZbin pe.. 2 views pe 91 like.😂. sir kya jaadu hai aapke.. 💝 Lecture choro bhaiya pehle like maaro..😂🤗😋😋

[@tigerking4960]

🤣🤣🤣🤣🤣

[@sauravsurbhit5349]

Hiiiii

[@Karanssharma03]

Or kya

[@PawanSingh-sc5qy]

you think that we are fool?

[@HipHopTalks]

Kisi ko pata electromagnetic induction 02 Kaha hai

[@kuldeepsinghrajput2511]

28:46 finally sir roasted tik tok

[@aryansatyam2597]

kzbin.info/door/qSkq8KYV9BQ_h95-hi1zkA plz watch my channel

[@godson200]

Sir was also on tiktok

[@adityadua6867]

@@godson200 was he ?

[@namratasingh8312]

Hlw

[@riyakandpal9305]

kzbin.info/www/bejne/bXPWk4epa9aEoas

[@iqbal__dar]

May almighty have emmense blessings on u 🤲🤲🤲....ameeen❤️❤️❤️❤️

[@Qubit313]

Thankyou Sir. Love and respect from Pakistan.🥰🥰

[@rasmita1885]

Sir ki padhaii acha isiliye lagta hei kyu ki sir game Apne lagte hei ..... 🥰 Thank you sir ... 🙂 Lagta hei jese app nei ham sab ke liye hi Janam liye thee... 🥺 .... Again THANK YOU 🥰 sir .... ... We wish God always bless you sir ....🥺 App hamesha ase hin mushkurate rahe .... 🥰. ... 🥰 ...THANK YOU SIR ..🥰

[@Aaradhyuh]

1:24:16 par answer vbr hoga guys sir l ko substitute krna bhl gye

[@vampire9136]

Hmm mra aa gya

[@satyambarot1748]

my concept is very clear after seeing this ! TY ALAKH SIR❣️

[@shivani794]

Sir I'm starting enjoying physics ☺thanku sir, i like the way that you teach 😊😊all concepts are clear, 😇😇😇you are the bestest bestest physics teacher 😍😍

[@rf7301]

friends i want help from u i am dropper and preparing for jee 2021 my parents support me but other members of family whenever call they just tm nhi kr paoige and laughs on me .... i am thinking of that shit please tell me how to deal

[@lakshyapratapsingh2341]

@@rf7301 bro wo kahenge hi Wo chate ho tum succes full na ho pao Jinki auakt nahi hoti ya unse hi nahi ho pata hai vahi kahte hia Ingnore them 🍁

[@maviyaali1108]

sir you explain better than all teacher because you know what difficulties do students faces...very nice sir

[@abhinavkumarram1136]

Sir hme bhaut darr Lagta tha physics se magar aap ke lecture ki wajah se physics me Maja aane laga hai . I'm very thankful to you sir🙏🙏🙏🙏🙏