Sliding Metal Bar in Magnetic Field

[justadream]

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Can someone explain why decreasing resistance doesn't do anything to the voltage?

 

[Cawolf]

The voltage is being generated by the changing flux of the magnetic field.

This voltage v = -dFlux/dt and does not depend on the resistance of the circuit.

The voltage will be proportional to the changing flux and the resistance is constant so what will change is the current!

 

[justadream]

@Cawolf

"The voltage will be proportional to the changing flux and the resistance is constant so what will change is the current!"

Wait but how is the resistance constant when you are explicitly quartering the resistance?

 

[Cawolf]

In each scenario you design, the resistance is constant.

So if you use resistance = R, then you will generate some current I but V will be proportional to flux.

If you set resistance = .25R, then the current will be I = V/.25R = 4x larger than before. But the voltage is still not changed.

 

[justadream]

@Cawolf
Okay, I get that.

But where in the equation does magnetic field strength come in?

 

[Cawolf]

The magnetic field is constant - but the area is changing.

So V = - flux/dt = - dB/(dA)(dt) (note constant B field)

V = - B/(dt)(dA)

 

[justadream]

@Cawolf

so delta(B)/delta(A) = flux?

 

[Cawolf]

I should stop answering questions today - time for bed....... the dA should be a dot product not divided.

It is technically Flux = surface integral (B . dA)

So with constant B, dFlux/dt = (d/dt) B surface integral (dA) = BdA/dt

Therefore, V = - BdA/dt which is proportional to the change in area over time.

Sorry for the confusion.

 

[justadream]

@Cawolf

So can I just get rid of the "d's" in that formula and know:

V = -B * delta(A) / delta(T)
or
V = -B * A / T

 

[Cawolf]

Sure, you will never need to calculate it though as far as I know. I would mainly focus on the concept.

It isn't technically correct but I wouldn't focus on it....it is a dot product though, so a cosine would be involved.