6. Motional EMF in a conductor( EMI ) | 12th physics | physics handwritten notes

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[@sridharchitta7321]

Faraday’s law is very general and any change in magnetic flux, whether due to a change in magnetic field or, a change in area, produces an emf equal to the rate of change of flux. In fact, both may be present simultaneously, as when you drag a metal bar along metal rails with one end of the rails bridged by a resistor and there is a uniform magnetic field in the area between the rails and bar and resistor. Assume, the magnitude of the uniform magnetic field is increasing. The flux is BA, where B is the magnetic field and A the area enclosing it, and the rate of change of the flux is due in part to the change in the magnetic field, and in part to the change in the area enclosed by the changing path. The resistor will heat up! The equation of the rate of change of flux....without the details of the derivation are = (dB/dt) A + BLv (Change of B, fixed path) (Change of path) The calculation is a bit mysterious. Physically, where did the emf represented by BLv come from if it is not associated with a changing magnetic field? Befire we provide an answer, it would be instructive to understand Current, the conduction process and Voltage at the fundamental level. Electrostatics and circuits belong to one science not two as in the following two videos: i. kzbin.info/www/bejne/ioXXpWVul5aXj9E and ii. kzbin.info/www/bejne/bnO0fpKurJeFnNE Now, returning to the experiment of the moving bar on rails, we see that we have a conducting metal bar and we are moving it. Note carefully that we are moving the jumpy energetic conduction band electrons in it as well. As a result of the magnetic force (due to the presence of the uniform magnetic field), the free electron sea shifted, and the bar polarized and a Coulomb electric field develops and that is how an induced emf is produced. It is not possible in this post to discuss in more detail the process of polarization and the consequent production of an emf. The last frame References in video #1 lists textbooks 3 and 4 which discuss these topics in more detail using a unified approach and provide an intuitive understanding of the generation of an emf in generators and motors.

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Short & accurate explaination 👍

[@mukulyadav967]

Sir in this video point Q is High Potential (+) and P is low Potential (-) and the direction of current is HP to LP

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[@ErenYeager-yc2do]

Bhai p point hai hp and q lp aur direction High to low hoti hai

[@AnonSukhoi30MKI]

@@ErenYeager-yc2do bhai current high to low potential jaata hai Agar upar Jaa rha hai, ie Q to P, then Q high potential hai na

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[@PHYSICSwithUmeshRajoria]

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[@sarcasticsoul21]

2:14 Sir aapne angle magnetic field aur area vector ke bich m bataya h . Aapne bataya ki loop ke piche vale surface aur magnetic field m 0° ka angle h . But hum loop ke aage vale surface ka angle magnetic field se kyo nhi lete ???

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5:50 direction concept

[@firdousjahan2266]

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Sir vo 'Flemings left hand rule'hota h or last m v sin(theeta ) downward direction m hoga

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[@PHYSICSwithUmeshRajoria]

Welcome 🙂

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[@shreyanshmishra3979]

In last DX/dt will be taken negative so we get emf +ve

[@STUDENTPERSON_127]

Sir Motional emf ke derivation mein kyu Constant velocity chahiye instead of Accelerating velocity?

[@mohitnehra8593]

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[@PHYSICSwithUmeshRajoria]

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Is, it mendatory to use the sign of (-) in each case ?

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[@sri78563]

Sir, I think Vsin(tetha) ka direction downward across the conductor rod hoga

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[@dankkrrish6780]

Sir ye rod ek battery ki tareh act karti rahe uske liye hame ise bahar external force se kheenchte rehna hoga but uske liye to rails bhi lambi honi chhaiye so this is not cost effective so is cheez ko solve karne ke liye kya karna padta h?

[@yogeshvishwakarma3916]

sir at 7:40 if current is flowing from Q to P how can P be at higher potential??????????????????????????

[@PHYSICSwithUmeshRajoria]

Here rod is equivalent to a voltage source

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[@Mamsad1348]

Agar hum kisi circular loop ko uniform magnetic field mn rakenge tho kya koi emf produce hoga???plzz reply

[@arsabhisekff03]

Yes, by changing 'theta'/ by orientation wala case dekho...... NBa(omega)sin(omega×t)

[@arsabhisekff03]

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7:44 sir current toh high potential se low potential ki taraf flow hota hai na

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[@Itsybitsyak]

Sir I did not understand how vcos theta is not responsible for EMF production. If it’s making theta=0 then costheta=1 .

[@PHYSICSwithUmeshRajoria]

Here v cos theta is parallel to the magnetic field and due to this component magnetic field lines will not cut by the road and emf is not induced

[@b2.rai.]

In special case ..(v sin theta) has to be in downward direction

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[@PHYSICSwithUmeshRajoria]

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[@anish39]

Sir jo cos thita liya tha matlb vo koi nhi magnetic field cut nhi karega to isme koi bhi emf nhi produces hoga ,mf lines aur angle ektaraf hoga to yeh to possible hi nhi

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2:32 Hum area vector ka direction Dekhnee ke liye loop ke peeche ki surface hi q dekhnge? Aage wali q nahi dekhii??? I just have this doubt otherwise u explained really very well 😀

[@PHYSICSwithUmeshRajoria]

To get the magnitude of emf positive

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Sir idhr to permenantly +ve or -ve aa rha hai but humhe agar polarity reverse krni ho to hum conductor ki direction opposite kren gy na??

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[@nikhilbarnaval1202]

Rod ko aaga badhayenga toh sir magnetic mai kya koi farkh nhi padhega woh move karega toh

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[@ayushsshorts6183]

Sir magnetic force ki direction kaise pata karenge due to motional Emf

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[@saimawan4817]

sir what would be the induced emf if the rod is rotated parallel the field {such that the axis of rotation in the center along the length of rod }? I have a bit doubt here cuz i think that the angle between velocity of rod and field will continously vary across either ends{ from the axis of rotation} of the rod cancelling each other out making the net emf zero am I right please currect me ,if not Thank you

[@physicslover9597]

Sir ji, mera ek question tha, Agar ek area ka vector direction nikalna ho toh kaise nikale, matlab ek area ka two surface hota hai, toh mera question ye hai ki, konsi surface se area vector ka direction nikal kar ayega ?? Please reply sir ji

[@PHYSICSwithUmeshRajoria]

You can consider any one of them

[@physicslover9597]

@@PHYSICSwithUmeshRajoria thank you sir 😍😍