Integral of voltage time means?

[almaand]

Homework Statement

Hi! In my physics class we've been doing this classical experiment where we measured the induced voltage in a coil when letting a magnet fall through it using a PASCO system. Now I understand perfectly fine most of the theory behind this, and understood how to calculate the magnetic flux (the integral of the area under the voltage-time graph). But what I'd just can't get my head around is exactly what that number tells me? From time t to time t+h the flux was about 0.586? Is this the average flux during this time or the total flux? Thank you for helping me with my stupid question!

 

[Andrew Mason]

Homework Statement

Hi! In my physics class we've been doing this classical experiment where we measured the induced voltage in a coil when letting a magnet fall through it using a PASCO system. Now I understand perfectly fine most of the theory behind this, and understood how to calculate the magnetic flux (the integral of the area under the voltage-time graph). But what I'd just can't get my head around is exactly what that number tells me? From time t to time t+h the flux was about 0.586? Is this the average flux during this time or the total flux? Thank you for helping me with my stupid question!

Welcome to PF!

I am sure it is not a stupid question, but it is not a very clear one. You have to follow the rules for posting your question and follow the template.

I expect that the graph of induced voltage vs. time tells you something about the rate of change of flux. This derives from Faraday's law:

EMF_{induced} = \oint \vec{E}\cdot d\vec{s} = \frac{d\phi}{dt}

so:

\int (\oint \vec{E}\cdot d\vec{s}) dt = \int d\phi = \phiAM

 

[almaand]

Welcome to PF!

I am sure it is not a stupid question, but it is not a very clear one. You have to follow the rules for posting your question and follow the template.

I expect that the graph of induced voltage vs. time tells you something about the rate of change of flux. This derives from Faraday's law:

EMF_{induced} = \oint \vec{E}\cdot d\vec{s} = \frac{d\phi}{dt}

so:

\int (\oint \vec{E}\cdot d\vec{s}) dt = \int d\phi = \phi


AM

Oh sorry, I'm going to read them through again and remember to be more clear next time! And thank you for your answer!

As for the question; what I've done is that I've calculated the area below the voltage time graph and also divided the value by the amounts of turns and thus I should have the magnetic flux. Since:

V = -N*dø/dt --> -V/N = dø/dt

But this value of magnetic flux, what is it. Like if it had been a velocity-time graph the area would have represented how far I would had come. But in this case what does the value of magnetic flux represent?

 

[Andrew Mason]

Oh sorry, I'm going to read them through again and remember to be more clear next time! And thank you for your answer!

As for the question; what I've done is that I've calculated the area below the voltage time graph and also divided the value by the amounts of turns and thus I should have the magnetic flux. Since:

V = -N*dø/dt --> -V/N = dø/dt

But this value of magnetic flux, what is it. Like if it had been a velocity-time graph the area would have represented how far I would had come. But in this case what does the value of magnetic flux represent?

Flux per unit area is a measure of the strength of the magnetic field. So flux can be thought of as the integral of the magnetic field over an area.

\phi = \int \vec{B}\cdot d\vec{A}


If you think of the magnetic field strength as being represented by "lines of force", the flux is the total number of lines of force. Lines of force don't really exist physically but it can be a helpful model to use.

AM