Loops in loops

Wil will use upper case symbols for the amplitudes and lower case symbols for the time dependent values as in $i_1=I_1 \cos \omega t$ , where $\omega = 2 \pi f$ is the angular frequency.

Let us first look at a more general problem, depicted on the Figure. The mutual inductance connects the emf of one coil to the derivative of the current in the other coil. For example,

$u_3= - L_{23} \frac{d i_2}{d t}$ .

In the current situation the time derivative of the current generates a voltage $u_2$ in the second coil. In response to that there will be a current in the second coil, $i_2=u_2/R$ . This current flows through coil #2' and its time derivative will generate a voltage $U_3$ in the third coil. Therefore

$u_3=\frac{d^2i_1}{dt^2} \frac{L_{12}L_{23}}{R}$ .

Our job is to determine $L_{12}$ and $L_{23}$ . Let us start with the easy one, the interaction of coil #2 and #3 on the original Figure in the problem.

The magnetic field in a long solenoid is $B=\mu_0 N i/L$ , where $\mu_0= 4 \pi 10^{-7} T/Am$ . The flux in that solenoid is $\Phi = AB=r^2 \pi \mu_0 i N/L$ , where $A$ is the cross sectional area. Therefore

$u_3= - N' \frac{d \Phi}{dt} = - r^2 \pi \mu_0 \frac{N N'}{L} \frac{d i_2}{dt}$ .

We can identify the mutual inductance as $L_{23}=r^2 \pi \mu_0 \frac{N N'}{L}$ . This is independent of the dimensions of the third coil, only the number of turns matters.

Next we need to determine $L_{12}$ . Let us go back to coil #3 and put the current into that coil. Can we determine the voltage in coil #2? It helps to know that there is a symmetry between interchanging the current and voltage terminals: the mutual inductance is the same, $L_{23}=L_{32}$ , where $L_{32}$ describes how much voltage is generated in coil #2 if we drive a current into the terminals of coil #3. Note however, that coil #1 has the same properties as coil #3 and therefore $L_{12}=L_{32}=L_{23}$ . Finally we get

$u_3= - \omega^2 I_1 \frac{L_{12}L_{23}}{R} \cos \omega t= - [ I_1 (\omega r^2 \pi \mu_0 \frac{N N'}{L})^2/R]\cos \omega t$ .

With $u_3= U_3 \cos( \omega t - 180^{\circ})$ we get the amplitude $U_3= I_1 (\omega r^2 \pi \mu_0 \frac{N N'}{L})^2/R =9.95mV \approx 10mV$ .