Solving Faraday's Law Problem: Finding EMF Needed to Stop Current at Middle R

[snir]

Homework Statement

If I have the circuit below, and the white circle is Infinite Coil with current of I=xt, Counterclockwise, with a radius a. from Faraday's law we know that will be Electromotive force .

2. Homework Equations
what the EMF in needed to stop the current at the middle R ?

The Attempt at a Solution

(the question ask what the x is needed but that easy if I know the EMF )
I tried to create more battery but where I can connect her? there is different if it will be at the middle wire or another wire .

 

[berkeman]

Homework Statement

If I have the circle below, and the white circle is Infinite Coil with current of I=xt, Counterclockwise, with a radius a. from Faraday's law we know that will be Electromotive force .
View attachment 112882
2. Homework Equations
what the EFS in needed to stop the current at the middle R ?

The Attempt at a Solution

(the question ask what the x is needed but that easy if I know the EFS)
I tried to create more battery but where I can connect her? there is different if it will be at the middle wire or another wire .
View attachment 112882

Welcome to the PF.

What do you mean by "infinite" coil? And what is EFS?

 

[snir]

Welcome to the PF.

What do you mean by "infinite" coil? And what is EFS?

that mean we can use that the magnet field is just inside the coil and the magnet field is: µ0 *n*(xt) - n is the density . (I forget that we know what the n).
the EMF * I mean for Electromotive force form Faraday's law.

 

[berkeman]

that mean we can use that the magnet field is just inside the coil and the magnet field is: µ0 *n*(xt) - n is the density . (I forget that we know what the n).
the EMF * I mean for Electromotive force form Faraday's law.

Ah, thanks for the correction.

But the circuit with the battery and resistors is a DC circuit, no? What do you have to do to the current in a coil in that position to make it generate a DC counter-current that opposes the DC current from the battery?

 

[snir]

Ah, thanks for the correction.

But the circuit with the battery and resistors is a DC circuit, no? What do you have to do to the current in a coil in that position to make it generate a DC counter-current that opposes the DC current from the battery?

yes.

 

[berkeman]

yes.

Then how do you generate and sustain a DC voltage with flux linking through that circuit loop of the battery and resistors?

 

[snir]

Then how do you generate and sustain a DC voltage with flux linking through that circuit loop of the battery and resistors?

I have DC voltage, and the current of coil is x*t (t is time), the flux is a^2*B (B I wrote), now I can derivative it and get the EMF. but what the EMF need to be? -v or -v/3. that depend where I put the EMF on the circuit.

 

[snir]

I have DC voltage, and the current of coil is x*t (t is time), the flux is a^2*B (B I wrote), now I can derivative it and get the EMF. but what the EMF need to be? -v or -v/3. that depend where I put the EMF on the circuit.

the voltage on the middle is 1/3V so I confused

 

[berkeman]

the current of coil is x*t

Ah, I see that now -- your current is increasing forever to make the DC counter voltage.

but what the EMF need to be? -v or -v/3. that depend where I put the EMF on the circuit.

Since it's an infinite solenoid (coming up out of the page), it has negligible flux elsewhere, right? So it only couples into the right-hand side loop of the voltage source and the two resistors. I think you only have to have enough counter-EMF to cancel out the battery, since that is the whole voltage drop around the loop with no coil current.

The other resistor to the left in an interesting twist, but I still think that once your counter-EMF is equal to the battery voltage, that the whole voltage will go down to zero.

 

[snir]

Ah, I see that now -- your current is increasing forever to make the DC counter voltage.

Since it's an infinite solenoid (coming up out of the page), it has negligible flux elsewhere, right? So it only couples into the right-hand side loop of the voltage source and the two resistors. I think you only have to have enough counter-EMF to cancel out the battery, since that is the whole voltage drop around the loop with no coil current.

The other resistor to the left in an interesting twist, but I still think that once your counter-EMF is equal to the battery voltage, that the whole voltage will go down to zero.

ok thanks but why we can't say that the counter-EMF is not on the middle wire? like a another battery at the middle - than when we do so the answer is changes.

 

[berkeman]

ok thanks but why we can't say that the counter-EMF is not on the middle wire? like a another battery at the middle - than when we do so the answer is changes.

By Faraday's law of induction, the Counter-EMF is generated around the whole right-hand loop...

 

[snir]

By Faraday's law of induction, the Counter-EMF is generated around the whole right-hand loop...

TY!
but I can also said that if I made it on middle that will generated around the whole right-hand loop also. no?

and also - now .. they move the coil to the side left, between the R. now we can get a EMF cancel the current at the middle? may that will EMF = v ?

 

[snir]

By Faraday's law of induction, the Counter-EMF is generated around the whole right-hand loop...

sory If not understand the meaning of EMF. the EMF not mean that all the voltage of all R in the right-hand loop will down the same ? .. if we make the EMF like battery the ratio of the generated change between the right R.

 

[Diegor]

sory If not understand the meaning of EMF. the EMF not mean that all the voltage of all R in the right-hand loop will down the same ? .. if we make the EMF like battery the ratio of the generated change between the right R.

Remember the circuit schema is a model the resistance may be a discrete component or a distributed resistance the same hapen with the DC power source.

The magnetic induction is generating an electric field that is distributed over the wires of the circuit and can move charges doing work that is acting like a distributed dc power source. Think of it as a lot of tiny power sources in series over the circuit.

 

[snir]

Remember the circuit schema is a model the resistance may be a discrete component or a distributed resistance the same hapen with the DC power source.

The magnetic induction is generating an electric field that is distributed over the wires of the circuit and can move charges doing work that is acting like a distributed dc power source. Think of it as a lot of tiny power sources in series over the circuit.

Thanks
for I can understand where I put the EMF I founded, if the coil was at the left circuit? the EMC need to be V if I want no current at middle?

and if I have a circuit like I made (new image) . I can know where to put the DC for replace the coil?

 

[Diegor]

Thanks
for I can understand where I put the EMF I founded, if the coil was at the left circuit? the EMC need to be V if I want no current at middle?

and if I have a circuit like I made (new image) . I can know where to put the DC for replace the coil?

Yes to the first question.

And for the second i think the original problem is intended for the student to learn about faraday law, flux, induction etc. But no to go further to complicated configurations. I mean in more complicated circuits you would need to analize the electric field in every point.

I general when the problem involves field theory one has to forget about circuit models and analize the problem using maxwell equations in space for that more information is needed like charge density. Charge distribution over space. And how the conductor or diferent materias of the circuit are distributed in space.

 

[Diegor]

and if I have a circuit like I made (new image) . I can know where to put the DC for replace the coil?[/QUOTE]

Besides what I posted before I see your second circuit is interesting. I would think of it like being two loops with triangular shape one inside the other and the emf induced would be like adding a dc source to every branch of the circuit. Then maybe could be analized using kirchhoff law. (still as i posted before would be something aproximated)