# Calculating flux with a magnet and a coil

## Homework Statement

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Hey, I'm having some difficulties with my physics project and I hoped someone could help me out. We have to calculate the magnetic field, but first we need to know the flux.

This is the setup we got:
We have to let a magnet fall through a PVC tube. The tube has a coil around it (it covers just a small part of the tube) which is connected to a voltage probe. The voltage probe is connected to a laptop with LoggerPro.

Then we have to calculate the flux ( Φ ) with the results we got in LoggerPro.

## Homework Equations

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I already know that I have to use the formula U = -N • d Φ / dt
(d, not Δ!)

## The Attempt at a Solution

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I do know how many windings the coil has (N) and LoggerPro shows the time (t) and the induction voltage (U). The flux has to be calculated with a diagram.
But I don't really know how to do this due to the 'd' in the formula. I've been told that I have to do something with the derivative and the surface in the diagram, but I'm not sure how to do this.

Does anyone have an idea how to do this?
Thanks in Advance!

Anne

## Answers and Replies

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kuruman
Science Advisor
Homework Helper
Gold Member
I assume that LoggerPro has given you U(t) as a function of time. Note that you can write dΦ/dt=(U(t)/N). (The minus sign is irrelevant here.) Then the total flux is Φ = ∫(dΦ/dt)dt = ∫(U(t)/N)dt. Got it?

Welcome to PF, AnneClara.

I assume that LoggerPro has given you U(t) as a function of time. Note that you can write dΦ/dt=(U(t)/N). (The minus sign is irrelevant here.) Then the total flux is Φ = ∫(dΦ/dt)dt = ∫(U(t)/N)dt. Got it?
Thank you!!
LoggerPro gives the U as a function of the t indeed. But I don't really know how to put this in a graph to get the flux.

kuruman
Science Advisor
Homework Helper
Gold Member
You don't really need a graph to get the flux. Look at what ∫(U(t)/N)dt, which is equal to the flux, is saying you should do.
∫ means "add", add what? Answer: the ratio U(t)/N at a given time t (provided by LoggerPro in a table) multiplied by dt. OK, so what's dt? Answer: the constant time interval between measurements (also provided by LoggerPro). In other words, you want the area under the curve, not the curve itself. You approximate this area by the sum of areas of rectangles of height U(t)/N and width dt. If you don't know how to (or can't) do this directly in LoggerPro, I suggest that you export the file to spreadsheet format which you can then process as you please.

You don't really need a graph to get the flux. Look at what ∫(U(t)/N)dt, which is equal to the flux, is saying you should do.
∫ means "add", add what? Answer: the ratio U(t)/N at a given time t (provided by LoggerPro in a table) multiplied by dt. OK, so what's dt? Answer: the constant time interval between measurements (also provided by LoggerPro). In other words, you want the area under the curve, not the curve itself. You approximate this area by the sum of areas of rectangles of height U(t)/N and width dt. If you don't know how to (or can't) do this directly in LoggerPro, I suggest that you export the file to spreadsheet format which you can then process as you please.
Thank you so much!!! :)