Calculating Voltage Drop with Non-Constant Speed

[Son-of-Shem]

so I'm working on a physics lab report, and I've derived the following formula

V=-N∫dB/dt(dot)dA

then I used the chain rule to turn it into

V=-N∫dx/dt * dB/dx(dot)dA

my question is, can I pull the dx/dt out of integral? I was doing the experiment, and measured the voltage drop, and varied the speed. the speed isn't constant, as it's the speed of a (frictionless) cart going down a ramp

'cause this formula ( V=-Nv∫dB/dx(dot)dA ) would be much better for my data anaylsis

 

[tiny-tim]

Welcome to PF!

Hi Son-of-Shem! Welcome to PF!

What is your A ?

You can pull the dx/dt outside the ∫ if it doesn't depend on your variable of integration (A)

 

[Son-of-Shem]

thanks,

yeah... I should have mentiond that, shouldn't I... oops.

A is the area. I just wasn't sure since v depended on x, and there was a dx in there, if it was ok to pull that out...

 

[tiny-tim]

I'm completely confused …

I guessed A would be area, but area of what?

 

[Son-of-Shem]

sorry, I'll start over from the beginning.

I send a cart with a magnet down a ramp, and through a coil of wire with 200 turns, sitting perpendicular to the tabletop, and measure the maximum voltage drop. The formula I found (using magnetic flux) is V=-N∫dx/dt * dB/dx(dot)dA

so A is the area of the coil.

if I can pull out dx/dt, then ∫dB/dx(dot)dA=-V/Nv, which is a measurable quantity :)

my question is: can I do that?

 

[tiny-tim]

Hi Son-of-Shem!

(just got up :zzz: …)

isn't A a constant?

why are you integrating?

 

[Son-of-Shem]

yeah, it is constant...

I was trying to find voltage drop, which is -dΦ/dt

and dΦ=B(dot)dA

so Φ=∫B(dot)dA

so d/dt(∫B(dot)dA)=dΦ/dt

then V=-d/dt(∫B(dot)dA) (for a 1 turn coil)

or V=-d/dt(N∫B(dot)dA) (for coil w/ N turns)

and since area doesn't change with time...

then V=-N∫dB/dt(dot)dA

and by chain rule

then V=-N∫dx/dt * dB/dx(dot)dA

so V=-N∫v * dB/dx(dot)dA

my TA seemed to give the impression that the dB/dx term would interfere with my ability to remove v from the integral... that coupled with the fact that the velocity is actually changing (sqrt(2gh)) with x (height depends on x)

I believe that the magnetic field felt at the coil depends on the area of it, so the db/dx term can't be taken out...