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Average Current Through A Coil Over An Interval

  • Thread starter rzermatt
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8
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[SOLVED] Average Current Through A Coil Over An Interval

1. Homework Statement
The problem comes from a homework assignment I have. It has to deal with average current flowing through a wire over an interval. It is stated as follows:

A five-turn circular coil of radius 15cm is oriented with its plane perpendicular to a uniform magnetic field of 0.15T. During a 3s time interval, this field increases to 0.2T. If the resistance of the coil is 8[tex]\Omega[/tex], find the average current that flows in the coil during this interval.

2. Homework Equations
I know that to find EMF in a setting like this:
E = NAB [The rest doesn't matter since it is perpendicular]
V = iR [Ohm's Law]

3. The Attempt at a Solution
Let E1 be before the change, and E2 at the end.
E = NAB
E1 = (5 Turns)(pi*(0.15m)^2)(0.15T) = .0530V
E2 = (5 Turns)(pi*(0.15m)^2)(0.2T) = .0707V

Using these numbers, V = iR and the known resistance, the currents are as follows:

I1 = (.0530V)/(8Ohm) = 0.006625A
I2 = (.0707V)/(8Ohm) = 0.008837A

The average between them over 3 secs, is I1+I2/3s or:

(0.006625A+0.008837A)/3s = 0.005154A

Is my methodology correct? For some reason I feel I may be doing this incorrectly. Any help would be much appreciated.
 

Answers and Replies

cepheid
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The computation of the average makes no sense. You have current/time in A/s, and yet somehow you call this quantity a current. It's not dimensionally consistent.
 
cepheid
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The average current would be equal to the total charge that flowed across the coil in the 3s time interval, divided by the time interval. In other words, it would be equal to the integral of the instantaneous current i(t) with respect to time (from t = 0 to t = 3 s) divided by the time interval.

Now, the question is, what is i(t)? Well, IF the EMF increases linearly, then since Ohm's law applies in this situation, then we can say that the current increases linearly, in which case the integral is fairly straightforward. Whether the EMF increases linearly or not depends upon whether the magnetic field increases linearly.

But actually, that's not true. I think your formula for the induced EMF is wrong (E=NAB). The EMF is not proportional to the magnetic field. It is proportional to the rate of change of the magnetic field (E = NAdB/dt, perhaps). Only a magnetic field that is changing with time produces an EMF.

If that is the case, then provided that the magnetic field increases linearly, the voltage (induced EMF) will be *constant* over the 3 seconds, and so will the current. That simplifies the problem.
 
8
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Alright... that took a few minutes to make sense of.

Allow me to explain my recalculations:

[tex]\Phi[/tex] = BACos([tex]\Theta[/tex])
Emf = -N([tex]\Delta[/tex][tex]\Phi[/tex]/[tex]\Delta[/tex]t)

Thus:
[tex]\Delta[/tex][tex]\Phi[/tex] = (0.15T*(pi*(.15)^2)) - (0.2T*(pi*(.15)^2)) = -0.003534
EMF = -(5 Turns)*(-0.003534)/3s = .00589V

The average current would be:
i = V/R
i = (.00589V)/8[tex]\Omega[/tex] = 7.363x10^-4 A

I hope this is correct.
 
cepheid
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The method looks alright to me. I didn't really double check the calculations.
 
8
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No problem. I just wanted to really make sure that my concepts were correct. I can always recheck the number as I write it down in my actual homework.

Thanks for all the help! Greatly appreciated.
 

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