Given that your solenoid has inductance L and resistance R with initial current I such that there is uniform B inside the coil, I would like to figure out how much time will pass before B goes to zero.

Using the equations shown here, I think I have some idea of how to solve for \$\Delta t\$.

First, we know that the energy stored in the inductor is \$ E = \frac{1}{2}LI^2\$. Then, power, the rate of energy change, is \$ P = L I \dot{I} \$. Also, we know that \$ P = I^2 R \$.

From the link above, given the energy density inside the coil spring (solenoid), the energy stored can also be written as \$ E =\frac{B^2 V}{\mu} \$, where \$V \$ is the volume.

Take its time derivative, \$P = \frac{BV}{\mu} \frac{dB}{dt} = I^2 R \$.

Since \$ I = \frac{Bl}{\mu N} \$, \$P = \frac{BV}{\mu} \frac{dB}{dt} = (\frac{Bl}{\mu N})^2 R \$.

Can I now just solve for \$ dt \$ to figure out the time lapse for the magnetic field to go to zero?

  • \$\begingroup\$ Er... Coil spring?. What do springs have to do with this? \$\endgroup\$ – Edgar Brown Jan 14 at 23:32
  • \$\begingroup\$ @EdgarBrown: I suspect he means a typical 1-layer solenoid coil. OP: please edit your question. \$\endgroup\$ – TimWescott Jan 14 at 23:33
  • \$\begingroup\$ "with current I" If the current is present, the magnetic field will never go to zero. Again: edit your question. \$\endgroup\$ – Oldfart Jan 14 at 23:38
  • \$\begingroup\$ Vdrop across R rises with I, and V drop on L controls LdI/dt \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 15 at 1:52

It is almost evil to answer your question as stated, but I'm going to do it anyway.

If you are talking about a coil with internal resistance that is shorted, or if you are talking about an explicit LR circuit, then it probably matches the schematic below. The equation that controls the circuit is way simpler than you make out: $$ R i_L = -L\frac{di_L}{dt} $$

This quickly simplifies to \$\frac{d i_L}{dt} = -\frac{R}{L} i_L\$

This is a simple 1st-order linear ordinary differential equation with solution $$ i_L(t) = A e^{-t/\tau} $$ where \$\tau = \frac{L}{R}\$ and \$A\$ is found from some boundary condition on the differential equation. For simplicity's sake, just say that we don't know what all else happened before \$t = 0\$, but at \$ t = 0\$ \$ i_L(0) = i_0\$. Then $$ i_L(t) = i_0 e^{-t/\tau} $$

Now note that \$ e^x \$ is strictly positive for any finite value of \$x\$, and the answer to your question is -- never. That's it!


simulate this circuit – Schematic created using CircuitLab


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