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Why are there two terms of both mutal and self inductance

  1. Nov 25, 2016 #1
    1. The problem statement, all variables and given/known data

    Given two inductors, connected in parallel connected to a battery, why do the following emf relations hold?

    $$\mathcal{E}_1 = - N_2 A \frac{d B}{d t} = -M\frac{d I_1}{d t } \\
    \mathcal{E}_2 = -L_2 \frac{dI_2}{dt}- M \frac{dI_1}{dt}$$

    See attachment !

    2. Relevant equations

    I know for example that the mutual inductance on coil 2 caused by the change in current I_1 is
    $$\mathcal{E}_2 = - N_2 A \frac{d B}{d t} = -M\frac{d I_1}{d t },$$

    3. The attempt at a solution

    It's not clear to me how there are two terms of the rhs of each of the equations in part (1) !
     

    Attached Files:

    Last edited: Nov 25, 2016
  2. jcsd
  3. Nov 25, 2016 #2

    Charles Link

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    Homework Helper

    Mutual inductance ## M ## is defined as ## M=\Phi_{1,2}/I_2=\Phi_{2,1}/I_1 ## (the inductance is mutual, i.e. ## M_{1,2}=M_{2,1}=M ##), where ## \Phi_{x,y} ## is the magnetic flux in x due to y. You can then write ## \varepsilon_{1m}=-d \Phi_{1,2}/dt=-M d I_2/dt ## for the effect of change in current in inductor 2 on inductor 1, but you also need to include the self inductance: ## L_1=\Phi_{1,1}/I_1 ## to get another source of EMF in inductor 1: ## \varepsilon_{1s}=-d\Phi_{1,1}/dt=-L_1 dI_1/dt ## . ## \varepsilon_1=\varepsilon_{1m}+\varepsilon_{1s} ##. ... Similarly for inductor 2.
     
    Last edited: Nov 25, 2016
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