Motional EMF

Рет қаралды 38,894

Күн бұрын

Physics Ninja loos at calculating the motional EMF, current (direction and magnitude), Force, Power when a conducting rod is dragged in a region with a magnetic field. At the end of the video i also look at a different approach to the same problem using Faraday's Law.

Пікірлер: 50

@krishnannarayanan8819 5 ай бұрын

The final part where you said to compensate for the change in magnetic flux due to increase in area, a current is induced such that to counteract that magnetic flux made so much sense. Thank you!

@sydneyfarina7879 3 жыл бұрын

Thank you so much!! This helped me a lot for my final tomorrow :)

@jtotherock7244 3 жыл бұрын

Great video! Both ideas connect!

@guillaume6373 2 жыл бұрын

Damm, incredible explanation. Thanks!

@hesarafonseka9756 Жыл бұрын

Thank you so much, he explains better than my professors

@bluefoxf5963 3 жыл бұрын

Thanks for sharing your knowledge

@sophiem7770 Жыл бұрын

THIS is so incredibly helpful thank u so much...i am ur number one fan #NINJANATION

@sridharchitta7321 2 жыл бұрын

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. kzfaq.info/get/bejne/irqkp5Vpx5fIiaM.html and ii. kzfaq.info/get/bejne/bqiBgMKp3Ji2lqM.html 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.

@juniorcyans2988 Жыл бұрын

Perfect!!! Thank you very much!!!

@YadelewAsrie 4 ай бұрын

I love you too because of your best explanation keep it up!!!

@markkennedy9767 9 ай бұрын

Hi, thanks for this. Just about the F=ILB acting on the rod where I is the current in the rod (due to the vertical velocity of the charge carriers in the rod). Can you explain why we do not consider the rod's horizontal velocity also when getting the magnetic force B on the charge carriers in the rod. Why ignore this horizontal component v of the rod (and its charge carriers)? More broadly, is there a F=ILB on the other sections of the loop due to the velocity along the wire of their charge carriers. I know it won't cause displacement of these sections but does such a force exist due to their velocities? This is something I don't understand about the extent of self-induction that you may be able to clarify.

@Yzwskdj1 3 жыл бұрын

Where does the positive charges come from

@ndiogouniang83 Жыл бұрын

You really called the lecture. At some point I tought you'd say "Touchdown Faraday!!"

@MuqtadaMO Жыл бұрын

I watched the video from Iraq and thank you my friend

@heybaboo4020 3 жыл бұрын

Thank you 🥺❤️

@roseyousif125 3 жыл бұрын

bro, like from where did we actually get the original v isn't it from external work? and also yes the current will move to the opposite direction of velocity but this happens in a fraction of a second the force will come back from the other end of the wire and it will be in the same direction of velocity so, in the same way, it decreases it will increase in the fraction of a second. the thing is we should apply force from the beginning to get the charges to move in the first place but there is friction so we should keep adding force and also there is resistance in the wire which will get the charges to slow down so that is why we should add the external force, I really need to understand this but it is just not logical to me

@minhokim8263 Жыл бұрын

Hi Ninja, I got a question on your explanation on 3:15 of the video. Electrons are supplied by the loop, so there isn't a buildup of positive charges! The charge balance would be neutral or zero over the entire loop. Could you please reply to me? Thanks very much, Ninja. I like to watch your vids very much and I am your fan.

@PhysicsNinja Жыл бұрын

In this case there is no buildup because the conductor form a close loop. If it was just a rod moving there would be a voltage between the ends.

@minhokim8263 Жыл бұрын

@@PhysicsNinja Hi Ninja, Thanks for the reply but the problem still remains unsolved. Let me reiterate the problem in this way; Suppose a resistor(R) is placed on the left side of the loop and the resistance of the rest of the loop is negligible. The conductive rod is moving at a constant velocity of v and a current, I is flowing along the loop. If I integrate E-field along the closed loop, the result would be IR + BLv , where IR is the voltage drop across the stationary resistor and BLv is due to Lorenz force along the moving rod. Now I apply Faraday's law, IR + BLv = BLv , where the right hand side is -d(BA)/dt . Something should be wrong but I don't know what it is. Ninja, could you please help me on this?

@user-ny3vn4ii1l 2 жыл бұрын

I am an arabian but i study physic in english language I' ve understand you and thanks for great job

@PhysicsNinja 2 жыл бұрын

Thanks for your support.

@kassyg3897 4 жыл бұрын

It sounds like he's trying to hold in a fart the whole time

@PhysicsNinja 4 жыл бұрын

Best comment this week

@fatemusik 5 ай бұрын

I'm trying to solve a certain acceleration the rod gets from rest when the circuit has a dc source in it while in a uniform magnetic field. I still have no clue how to approach the final value formula. All I know is that it has something to do with Lorentz's force, Newton's II law and Faraday's law. unfortunately this video didn't help me get on the right tracks either. (I think it's a bit more advanced computation?)

@zacharialameck1980 12 күн бұрын

Yes concepts are so great making me more knowledgeable bt the only challenge I face is all about Fb and Fapp directions why are in those sence.

@user-fq3qm5kq4m Жыл бұрын

does the direction of force Fb =ILB should be to the left because the charge is negative ?

@fatemusik 5 ай бұрын

the Lorentz's force is against the motion in the wire. The magnetic field tries to slow down the wire/rod by all means. The faster the wire moves, the more the magnetic field tries to resist the motion.

@raneemalbadi1526 Жыл бұрын

Is induced current by frhr the direction of e flow or conventional current?

@PhysicsNinja Жыл бұрын

Conventional

@HK-cq6yf 3 ай бұрын

What about the forces on the rod due to the parallel part of the circuit? They both have current which create fields on each other, which then would create attractive forces, right? Or is it negligible?

@PhysicsNinja 3 ай бұрын

Negligible if you assume distance is large

@fatenali5128 3 жыл бұрын

NOW velocity in the equation E=vLB is the velocity of charges that are moving inside the wire or it is the velocity of the wire ? Got confused ☹️

@PhysicsNinja 3 жыл бұрын

Wire

@PhysicsNinja 3 жыл бұрын

wire but also the velocity of the free electrons in the wire. There is a magnetic force acting on the electrons which drives a current.

@utkarshprithviraj2238 2 жыл бұрын

If magnetic force is in the direction of displacement of the electrons ....work is done by magnetic field but this isn't possible right....?

@fatemusik 5 ай бұрын

in this case, magnetic force doesn't do work and emf is not generated from nowhere, the mechanical work done from elsewhere converts to electricity

@squidzooka8744 2 ай бұрын

If conductor is moving to the right and magnetic field is into the page then force is upwards

@saharbahloul9212 Жыл бұрын

if the forces are equal it must not move shouldn't the force applied be bigger ?

@AndyMBBallSkills Жыл бұрын

It can still move if the forces are equal, it will move at constant velocity.

@cosmolearner5562 4 жыл бұрын

@12:55 how you calculated terminal velocity value I am getting 25 rather than 2.5 Please guide.!

@fatemusik 5 ай бұрын

emf/Bd = v_max. the velocity given is already known in this exercise.

@michaellamoreaux4402 Жыл бұрын

The emf is motional emf. The change of flux linking the circuit has no effect. There are two independent principles called induction. They are motional emf and transformer emf. Transformer emf is expressed by the Maxwell-Faraday Law, which is one of Maxwell's equations. An intrinsically changing flux is accompanied by an electric field. A change of flux due to a change in exposure of a circuit to a magnetic field is not. Therefore, Faraday's Law is not a law, just math.