Flux through a coil due to Self-Inductance

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In summary, the maximum average flux through each turn of a coil with 460 turns and self-inductance of 7.50 mH, with a current varying over time according to i=(680mA)cos[πt/(0.0250s)], is 1.11*10^-3 Wb.
  • #1
betamu
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Homework Statement


A coil has 460 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i=(680mA)cos[πt/(0.0250s)].

What is the maximum average flux through each turn of the coil?

Homework Equations


##\varepsilon_{ind} = N \frac{d\Phi}{dt} , \varepsilon_{ind} = L \frac{di}{dt} ##

The Attempt at a Solution


I set the above equations equal to find ##N \frac{d\Phi}{dt} = L \frac{di}{dt} ## and integrated both sides wrt t to find ##N\phi = Li##, input the given equation for i with the entire cos() portion = 1, as it is asking for the maximum average flux, set ##N = 460## and solved for ##\Phi = 1.11*10^{-5}## but this doesn't seem to be the correct answer.
Is it the "average" flux bit that's throwing me off?
 
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  • #2
The average flux is zero, the maximum flux is as you say. But we need to understand what exactly we mean by the term "maximum average" flux.

Can you provide the exact statement of the problem (perhaps via a screenshot from the book , or from the page of the e-document)?.
 
  • #3
Attached is a screenshot of the problem. It asks for the "maximum average flux" through each turn of the coil. I was thinking the average would be zero, but MP didn't accept that as an answer, either.

I'm confused as to what the problem means by "maximum average." Is it asking for the average of all of the "positive" values of flux?

edit: Sorry, I'm not sure why I can't upload it in higher resolution. I saved it much higher resolution than it appears on the forum - does PhysicsForums downsize?
 

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  • #4
The screenshot is OK.

I am also confused about what it means by maximum average. Maybe it asks for ##\frac{1}{T}\int_0^T|\Phi(t)|dt## where T is the period (inverse of frequency) of the flux (or of the current).
 
  • #5
Hm, okay. It wouldn't be solvable with the given information in that case, right?

I'll email my professor about it and hope I get something back.
 
  • #6
It is solvable, the period is given, you can find it from the given term inside cos() that is the ##\frac{\pi t}{0.0250}##
 
  • #7
Thanks for the help the other day. Sorry to get back to this thread so late.

My initial answer was correct - I wasn't paying attention to the units used for the answer which were already given by the problem. I had the correct answer, but to the wrong power of 10.
 
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1. What is self-inductance?

Self-inductance is a property of a coil or circuit in which a changing current through the coil induces an opposing electromotive force (EMF) in the same coil. This opposing EMF is known as self-induced EMF.

2. What is the formula for calculating flux through a coil due to self-inductance?

The formula for calculating flux through a coil due to self-inductance is: Φ = L * I, where Φ is the flux, L is the inductance of the coil, and I is the current flowing through the coil.

3. How does the direction of current affect the flux through a coil due to self-inductance?

The direction of current flowing through a coil affects the direction of the induced magnetic field, which in turn affects the direction of the induced EMF. This means that the direction of current can affect the direction and magnitude of the flux through the coil due to self-inductance.

4. What are some applications of self-inductance?

Self-inductance is used in a variety of applications, including inductors for electronic circuits, transformers for power transmission, and in electromechanical devices such as motors and generators. It is also a key concept in understanding and designing circuits for electromagnetic compatibility.

5. How can self-inductance be reduced in a circuit?

Self-inductance can be reduced in a circuit by using materials with lower permeability, increasing the distance between the coils, or adding a magnetic shielding material around the coil. Additionally, using capacitors in parallel with the coil can help to balance out the inductance and reduce its effects on the circuit.

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