# Question regarding mutual inductance of two coil

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.

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## Answers and Replies

Please , I really need someone help me figure out where I have done wrong.

learningphysics
Homework Helper
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: $$N\Phi$$ not $$\Phi$$ (this is only for 1 turn, not the entire coil). So for coil 1 you get:

$$N\Phi=100*(1.5*10^{-6})=150\mu WB$$

and coil 2:

$$N\Phi=200*(9*10^{-8})=18 \mu WB$$

Last edited:
Andrew Mason
Science Advisor
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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.
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.

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) $$L = nB\cdot A/I$$

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

The term $\lambda = nB\cdot A$ 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

Thanks for your help ,Andrew Mason and learningphysics.I now can understand with the question already.

learningphysics
Homework Helper
Sanosuke Sagara said:
Thanks for your help ,Andrew Mason and learningphysics.I now can understand with the question already.

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

$$L_1=N_1\Phi_1/I_1$$
$$25*10^{-3}=100(\Phi_1)/6*10^{-3}$$
$$\Phi_1=1.5*10^{-6}$$
$$N\Phi_1=100(1.5*10^{-6})=150 \mu WB$$