Faraday's Law and Induced Electric Fields

Explains how Faraday's law can be used to find an induced electric field that is due to a changing magnetic flux.
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[@dr_ruvi]

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Yes! he blew my mind when he did the getting rid of loop thing.

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How did it change it?

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[@ifoxrider]

4:23 that is clear and straight forward in defining the relations

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[@leanlearning2184]

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[@HitAndMissLab]

What I would like to know is, can I measure EMF voltage with voltmeter? If one creates a changing magnetic field inside a loop of wire, and than he connects voltmeter across the loop of wire, will he read the value voltage, even as AC?

[@AshokKumar-rr1zw]

HitAndMissLab Nope, He is only able to read DC, because for AC you need sinusoidal function remember? So magnetic field must be sinusoidal to induce AC. Plus with a voltmeter you will not be able to read correct value of emf cause the internal resistance of voltmeter will screw this all up..You will need a Potentiometer...Hope it helps..

[@wonsilla]

great explanation and examples of a somewhat difficult concept. thanks.

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[@jrsleao]

Great video...very clear on the concepts and details. GREAT material. Thanks!

[@RahulKumar-hc9ze]

nice explanation

[@sunidhidhawan3143]

soo brilliant. thanks!!

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[@gauravo1414]

Does this electric feild exert torque if charge is on ring

[@naveensharma9400]

Yes definitely the ring will rotate

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[@isentient666]

Very clear, except when you assumed E is parallel to dl. Why is E parallel to dl? What if I move the Faradian loop over 1 mm to the right (but within the magnetic flux area)?

[@jinSakai2000]

Both vectors are tangential to the loop at every point on the periphery of the loop. Thus, at every point on the loop, E and dl are parallel as only a single tangent can be drawn to a circle at every point.

[@utf8redactedcharacters]

How do you know that B is equal to that? and thanks for the video

[@dakotaotero4245]

ugh, when taking the derivative of flux, for 2T/s^2, wouldnt we apply implicit differentiation to implicitly put I into terms of R or vise versa?

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Life saver! thank you so much! I hope you are having a happy and beautiful life :) !!!!

[@TizzerFPS]

is the comment section still active?

[@lasseviren1]

It is active. Yes.

[@TizzerFPS]

Nice.....

[@MrAhmed231]

how can i know the direction of the induced electric field?

[@AshokKumar-rr1zw]

MrAhmed231 use the right hand thumb rule..put the thumb in direction on magnetic field and the fingers will tell you the induced emf... So the emf is induced because there is charge difference and electric field wants to go from +ve to -ve charge so the direction of induced emf is the direction of electric field..

[@anushkachauhan110]

brilliant

[@victorguadaluperamoslagune9550]

Isnt the derivative s^3? Because your subtract one to it?

[@chaosinorderrr]

where did the dt go?

[@hoytvolker3]

He is amazing

[@newcomer201]

Great Job!

[@Pepsimaniaco]

Actually it the formula for magnetic flux is closed integral of B dot dA. Since B and dA are parallel, B dot dA becomes B*dA. B gets pull out of the closed integral and Closed integral of dA is just A, thats why the flux is just B times A.

[@jeffersonjackson2783]

at the end should the electric field be pointing the opposite direction because magnetic field is pointing into the paper fake current should be clockwise

[@AshokKumar-rr1zw]

Jefferson Jackson No dude the current is clockwise means the electrons should be moving counter-clockwise so it means the electric field is in opposite direction to movement of electron which means same direction to the induced current...Hope that helps..

[@satadhi]

oh god thank you

[@xHav3n]

thanks for the video!

[@isentient666]

dt in the derivative is just an expression in taking the derivative of the B-flux. It means taking the time derivative of the magnetic flux. Once he took that derivative wrt time, the dt automatically goes away. It's a mathematical thing...

[@kenleulhmmm5808]

Thaaank youu

[@alex1993seattle]

Why you said T/s^2, why is the s^2 come from, it would be wb/s^2 or T only, I think it is a typo

[@tigerresearch2665]

From the magnetic field you would expect the unit T. So, if we have a formula which has quadratic dependence on time, then we have something with seconds squared in there. Therefore, we say that we get "that many Teslas per square(!)-second". So the "per square second" and the square second you get by filling in your time gets canceled out and you end up with just a magnetic field in Teslas. You can always apply this kind of checks to see, if your units in a physical formula make sense. With the same reasoning you'd find out, that you'd run into trouble expressing that the acceleration of your car is " 3 meters per second". That's wrong. With meters per second we are talking about speeds. Therefore you should have something like "3 meters per square second" in your formula. That's an acceleration.

[@prdrer2]

Amazing!!

[@dylanreen6149]

well said! thanks

[@jamesmalone6170]

Why is the electric field pointing in that direction? Isnt is usually perpendicular

[@evildeevid]

no, when you have a current, which is induced as explained by faraday's law, the electric field must also point in the same direction as the current. another way to look at it is when you want a current to be pointed in a particular direction, you must choose an electric field which pushes the electrons in the opposite direction. this is the case for electroDYNAMICS. when you have a situation of electroSTATICS, the electric field is, indeed most of the time, perpendicular to the area of your charged object.

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Wouter David thanks

[@aminemo93]

thank you!!

[@BillyBobby123]

is this patrick lol

[@shashwataditya6685]

That's an assumed value. It significantly tells us that the magnetic field is time varying.

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PW.

[@eaterofzawarudo6764]

2:23 i thought dV=-Edl

[@AshokKumar-rr1zw]

Chicken Nope dude it depends on how you wanna move....Consider this----I wanna take a box from down to some height "H". So I apply a force opposing gravity that is 'mg' cause earth is applying '-mg' and my potential energy increase is 'mg*H'. So if I wanna drag a charge from infinity to 'r' I apply force -E because the field is applying +E that's why there is minus sign...Hope it helps

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