6
$\begingroup$

If you were to have a solenoid (0 current) floating still in space, and shot a magnet through it, would the solenoid move, or would it only create a DC current (what if it has a closed/open circuit)? Does the magnet slow down (if so by how much)? What if you increased the speed/power of the magnet drastically, would anything change?

(Feel free to link anything that could help explain solenoids/E&M fields better.)

$\endgroup$
1

3 Answers 3

1
$\begingroup$

No current, no back emf, no reaction force if the solenoid circuit is open. But if you close the circuit by shorting the solenoid input leads you will see current, and a reaction force. You need a closed loop to have current flow

$\endgroup$
0
$\begingroup$

For current to flow you need a closed path and here as there is no complete circuit the current won't flow. If the circuit is closed the current would definitely be produced which is directly proportional to the speed with which you move the magnet through the solenoid. The more the speed the more is the current


remember it would only be momentarily for example when you start the motion of magnet there is a change in flux associated so current is induced but once the magnet moves with same velocity no current is induced, again when you stop the magent a current is induced due to the same reason

$\endgroup$
1
  • $\begingroup$ The point of the question was the motion of the solenoid. You don't mention this in your answer. $\endgroup$
    – nasu
    Dec 20, 2020 at 17:49
0
$\begingroup$

You have two answers to this old question saying that there will be a momentum exchange if the solenoid is part of a closed circuit, as in the famous magnet-in-a pipe demonstration, where a hollow copper cylinder functions as a series of low-inductance, low-resistance, self-connected solenoids. Clearly if one were to send a magnet through a copper pipe in free space, the energy lost to eddy current heating would have to come from the kinetic energy of the combined system of the magnet, the pipe, and the electromagnetic field, and you could use the ordinary rules of partially- or completely-inelastic collisions to figure out the final momentum of the pieces.

Your existing answers also say that, if the solenoid is not part of a complete circuit, such as having its ends disconnected, no current will flow and there won't be any momentum exchange between the magnet and the solenoid.

However, that's only approximately correct.

Let's pretend your solenoid is very long, you can describe it as several smaller solenoids connected in series, each of which is still much longer than your magnet.

--( ( ( ( ( ( (  5 L ( ( ( ( ( ( (---  one very long solenoid

--((L((--((L((--((L((--((L((--((L((--  several not-so-long solenoids

(I've never tried to make an ASCII inductor 
symbol before, but you get the idea.)

The "lumped circuit element" approach to the solenoid says that, while the magnet is moving within a solenoid (say, the second one, $L_2$), the magnetic flux in that solenoid is constant, so there is no e.m.f. generated and no current flows. However, as the magnet leaves $L_2$ and enters $L_3$, the decreasing flux in $L_2$ generates an e.m.f. which produces a potential difference between $L_1$ and $L_3$. Likewise the increasing flux in $L_3$ prompts that sub-solenoid to generate an e.m.f. with the opposite sign, so that there's also now a potential difference between $L_2$ and $L_4$. These two e.m.f.s are equal and opposite: a volt-meter from the far end of $L_1$ to the far end of $L_5$ wouldn't measure any signs of these internal, cancelling e.m.f.s.

But in a resistive solenoid, these internal voltage changes are maintained by having the charges move around. Moving charges in a resistive medium generate heat, so the magnet's trip through the long solenoid is subject to the same sort of eddy-current friction as the magnet's trip through the copper pipe, and you would find the final motion of the long solenoid using the tools for partially-inelastic collisions.

In the limit where the solenoid is superconducting, the eddy-current effect doesn't steal any energy, but the magnet is prevented from ever entering or leaving the solenoid: the total magnetic flux through a superconducting material is a constant, which has interesting effects. Whether a "collision" between a moving magnet and the empty bore of a superconducting solenoid would be perfectly elastic or not depends on a question about electromagnetic radiation which is beyond the scope of this answer. But it would totally look like an elastic collision: the magnet would bounce off of the empty core of the superconductor.

To compute how much momentum the magnet would transfer to the solenoid, you would have to make some assumptions about the magnet's speed and field strength and the solenoid's inductance and internal resistance per unit length. It would be an interesting senior-physics-major kind of problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.