Mutual Inductance of a conducting loop and a solenoid

  • Thread starter kiltfish
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  • #1
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Homework Statement



A 0.100m long solenoid has a radius of 0.05m and 15000 turns. The current in the solenoid changes at a rate of 6.0 A/s. A conducting loop of radius 0.0200m is placed at the center of the solenoid with its axis the same as that of the solenoid. Determine the mutual inductance of this combination.

Homework Equations



Es=-M([tex]\Delta[/tex]Ip/[tex]\Delta[/tex]t)

The Attempt at a Solution



The answer has to be the induced emf divided by the rate of current change(6 A/s), but I just don't understand where and how we can glean the emf from the information given. I'm having a hard time on this emf stuff. I thought I understood very well how and when it is induced, but this is the second instance that I've had to solve for two variables and I just don't know how my teacher expects us to do it.
 

Answers and Replies

  • #2
gneill
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The emf induced in a loop due to changing flux inside the loop is

[tex] \left| E \right| = \left| \frac{d \Phi}{dt} \right| [/tex]

The flux in side the loop depends upon the magnetic field strength inside the loop and the area of the loop.

So, I suggest you start with the equation for the field strength inside the solenoid. Differentiate to find rate of change.
 
  • #3
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Ok, I see that I need the change in magnetic flux to calculate the Emf of the inner conducting loop, but I still can't calculate that without a single hint about magnetic field strength. I thought I could figure out the change in field strength by doing two calculations of
B=uo(N/L)I.
I made I=6 in one equation and I=12 in another, to assume a one second time variable in every other equation to come. So I got Bo=1.131T and B=2.262T. I multiplies each with the cross sectional area of the primary solenoid ([tex]\pi[/tex].052) and I took the difference to get .00888 for the change in magnetic flux over 1 second inside the solenoid. From there, I thought I could set that number equal to the negative mutual inductance times the rate of current change. But that would mean that .00888/-6 would equal my answer, and I know the answer isnt -0.00148 H. I'm expecting a positive number to the 10-4 power
 
  • #4
gneill
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You have

B=uo(N/L)I

and the flux in an area A is just

Φ = B*A

So what's dΦ/dt for the area of the loop? (Hint: the only variable that changes in B is I).
 
  • #5
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ok, ok. so I didn't need to do two magnetic field strength equations? If I let I=6, I get B=1.131. The flux inside the conducting loop then is 0.00142, and Es would be -.00142 for one second. Then I can finally solve for M to get 2.369*10-4. I hope I've done it right this time. It makes a little more sense now. I don't know why I thought the magnetic field strength would change throughout the system
 
  • #6
gneill
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20,813
2,786
ok, ok. so I didn't need to do two magnetic field strength equations? If I let I=6, I get B=1.131. The flux inside the conducting loop then is 0.00142, and Es would be -.00142 for one second. Then I can finally solve for M to get 2.369*10-4. I hope I've done it right this time. It makes a little more sense now. I don't know why I thought the magnetic field strength would change throughout the system
You got to a correct answer, but you should be able to get there without plugging in arbitrary values for the current and assuming a particular time interval (sort of a "poor man's derivative").

I think that the intention of the problem writer was to get you to recognize that the given rate of change of current is in fact dI/dt, which you could use directly to find dΦ/dt for the loop, hence the induced emf, E. Then E/(dI/dt) yields the mutual inductance.
 

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