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- Thread starter Sanosuke Sagara
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Please , I really need someone help me figure out where I have done wrong.

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learningphysics

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Sanosuke Sagara said:Please , I really need someone help me figure out where I have done wrong.

I think you did everything correct. The flux linkage is: [tex]N\Phi[/tex] not [tex]\Phi[/tex] (this is only for 1 turn, not the entire coil). So for coil 1 you get:

[tex]N\Phi=100*(1.5*10^{-6})=150\mu WB[/tex]

and coil 2:

[tex]N\Phi=200*(9*10^{-8})=18 \mu WB[/tex]

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Andrew Mason

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I take it that you are having problems with the flux linkage questions because you seem to have figured out the induced voltages. You may want to ask some engineers about flux linkage. It is a concept used more in engineering, but I will give it a shot.Sanosuke Sagara said:I have my question,solution and the problem I faced in the attachment that followed.Thanks for anybody that spend some time on this question.

Flux linkage is a measure of how much magnetic field produced by the coil is enclosed by the coil windings. It seems to be a relationship between flux enclosed by the coil and the actual current - I think. Inductance expressed in terms of the actual current I is

(1) [tex]L = nB\cdot A/I[/tex]

where n is the number of windings, B the magnetic field in the coil and A the cross-sectional area.

The term [itex]\lambda = nB\cdot A[/itex] is called the flux linkage.

But since you are given L and I, you can work out flux linkage from (1).

Does that help?

AM

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- #6

learningphysics

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Sanosuke Sagara said:

I think an easier way to answer part a) is just to use the definition of inductance.

[tex]L_1=N_1\Phi_1/I_1[/tex]

[tex]25*10^{-3}=100(\Phi_1)/6*10^{-3}[/tex]

[tex]\Phi_1=1.5*10^{-6}[/tex]

[tex]N\Phi_1=100(1.5*10^{-6})=150 \mu WB[/tex]

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