Why is induced emf maximum when magnetic flux is minimum? (2 Viewers)

[ragingcurry]

Can someone please explain to me why induced emf is maximum when magnetic flux threading a coil is minimum and vice versa. I have seen this in a few exam papers (e.g. Multiple choice question 10 in 2002 HSC physics paper)

Any help would be very much appreciated

 

[leehuan]

Terribly explained:

Induced EMF is maximised when the CHANGE in magnetic flux is maximised. This simultaneously occurs when the flux THROUGH the coil is minimised.

In a hypothetical motor, magnetic flux through the coil is maximised when the coil cuts the magnetic field, as we know. However, at this point, the change in magnetic flux to reach this point is actually 0. Whereas when you allow the maximum amount of flux to pass through the plane of the coil (breaking that awkward wording down in another way - when the most amount of magnetic field lines are able to pass through that circular shape the coil makes), the change in magnetic flux to arrive at that point is maximised.

(Mathematically, this has to do with the fact that d/dx cos(x) = -sin(x))

 

[Nailgun]

you know I was just thinking about something
If you have an electromagnet, and somehow you could turn it off instaneously (idk black hole whatever)
the slope of the flux graph would be a vertical line, and so m=infinity
lelelele

infinte emf induced lol

 

[Ambility]

Can someone please explain to me why induced emf is maximum when magnetic flux threading a coil is minimum and vice versa. I have seen this in a few exam papers (e.g. Multiple choice question 10 in 2002 HSC physics paper)

Any help would be very much appreciated

As leehuan pointed out, the induced emf is proportional the the rate of change of flux. At the instant the plane of the coil is parallel to the magnetic field, the flux threading the coils is changing at a high rate. This is because the "catchment area" of the loop for magnetic flux lines increases/decreases quickest at this point, resulting in a high emf. When the plane of the coil is perpendicular to the magnetic field, the "catchment area" changes fairly slowly, resulting in a low emf.

 

[ragingcurry]

Thanks guys

 

[leehuan]

you know I was just thinking about something
If you have an electromagnet, and somehow you could turn it off instaneously (idk black hole whatever)
the slope of the flux graph would be a vertical line, and so m=infinity
lelelele

infinte emf induced lol

I gave a brief thought about it.

If hypothetically, the flux itself could be instantaneously reestablished from any value (say y) to 0. The graph of the EMF curve will therefore show a discontinuity at that point when the flux changed.

At a discontinuity, the function Φ(t) is not differentiable as it is not continuous there. Thus, φ'(t)=ε(t) is not strictly defined, and hence the behaviour of the induced EMF is unknown.

The fact that EMF is voltage means that it is essentially impossible for EMF to be infinite. This occurs because V=W/q, and because the charge of an electron is fixed, we are implying that we are putting an infinite amount of energy in.

 

[InteGrand]

I gave a brief thought about it.

If hypothetically, the flux itself could be instantaneously reestablished from any value (say y) to 0. The graph of the EMF curve will therefore show a discontinuity at that point when the flux changed.

At a discontinuity, the function Φ(t) is not differentiable as it is not continuous there. Thus, φ'(t)=ε(t) is not strictly defined, and hence the behaviour of the induced EMF is unknown.

The fact that EMF is voltage means that it is essentially impossible for EMF to be infinite. This occurs because V=W/q, and because the charge of an electron is fixed, we are implying that we are putting an infinite amount of energy in.

It'd still be possible to have a function's graph have infinite slope somewhere yet be continuous there, e.g. The graph of the cube root function at the origin. A place where the slope is infinite makes the function non-differentiable there (even if it is continuous there) because the derivative does not exist there.

But yeah, generally infinities don't exist in nature, and if infinity appears as your answer to a physics problem, you've probably made a mistake somewhere in your calculation.

 

[leehuan]

It'd still be possible to have a function's graph have infinite slope somewhere yet be continuous there, e.g. The graph of the cube root function at the origin. A place where the slope is infinite makes the function non-differentiable there (even if it is continuous there) because the derivative does not exist there.

But yeah, generally infinities don't exist in nature, and if infinity appears as your answer to a physics problem, you've probably made a mistake somewhere in your calculation.

Oh true, but Nailgun's example technically has a jump discontinuity though so I was just thinking along the lines of that

 

[Nailgun]

It'd still be possible to have a function's graph have infinite slope somewhere yet be continuous there, e.g. The graph of the cube root function at the origin. A place where the slope is infinite makes the function non-differentiable there (even if it is continuous there) because the derivative does not exist there.

But yeah, generally infinities don't exist in nature, and if infinity appears as your answer to a physics problem, you've probably made a mistake somewhere in your calculation.

doesn't a black hole (or any singularity) have infinite density?

 

[InteGrand]

doesn't a black hole (or any singularity) have infinite density?

Yes. I believe the laws of physics as we know them break down there.

 

[leehuan]

doesn't a black hole (or any singularity) have infinite density?

That's why they say that the laws of physics as we know them break down for a black hole.

Because we can never get out of one once we reach the event horizon, there's virtually no means through which we can find out exactly what goes on as the information cannot be transmitted back out.