# Why are there two terms of both mutal and self inductance

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1. Nov 25, 2016

### loesung

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) !

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Last edited: Nov 25, 2016
2. Nov 25, 2016

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.