My knowledge of electromagnetism is quite elementary but I think I understand WPE and the Law of Conservation of Energy well enough.

Here's what I understand of electromagnetic induction in a coil in the form of a solenoid:

enter image description here

When a bar magnet with a specific pole facing the solenoid is moved rapidly, it produces an emf in the coil causing an electric current to move in a direction that would produce a magnetic field of the solenoid that opposes the motion of the bar magnet.

In this case, the bar magnet is moved rapidly towards the solenoid with its north pole facing the solenoid. The solenoid's magnetic field thus developed repels the bar magnet. This means that north pole is developed on the side of the solenoid which faces the bar magnet.

Now I understand that in moving the bar magnet towards the solenoid, some kinetic energy is being given to it.. say by my hand. When the solenoid acts to repel the bar magnet, it does negative work on the bar magnet, reducing the bar magnet's kinetic energy in the process. Therefore, the energy goes from my hand to the solenoid in the form of electrical energy (and heat produced in the wires)

But the transformation of energy in the case where the bar magnet is kept stationary, and the solenoid is moved towards it doesn't make sense to me.

enter image description here

This is the case when the bar magnet is held stationary and the solenoid is moved towards it. In moving the solenoid towards the bar magnet, kinetic energy is given to it.. say by my hand. The bar magnet repels the coil, thus doing negative work on the coil. Shouldn't this cause energy to go from the solenoid to the bar magnet? How is the current driven in the solenoid in terms of energy conservation then?

Is my way of viewing the transformations in energy correct? Should I be considering the field lines and potential energy between the magnet and the coil somehow?

  • $\begingroup$ Referring to the second example, with steady magnet: no displacement, no work. $\endgroup$
    – basics
    Commented Dec 2, 2022 at 23:03
  • $\begingroup$ @basics so no work is done on the magnet but negative work is done by the magnet on the solenoid, right? (as the solenoid is not stationary) what is the effect of this work in terms of the transfer of energy $\endgroup$
    – Kayen Jain
    Commented Dec 3, 2022 at 8:30

2 Answers 2


Suppose you push an elephant. The elephant does not move. Where does your energy go? The energy of the system ( you + elephant) does not remain constant because some energy is lost due to heat produced with friction (External force).

Similarly, the energy should be transferred to the magnet. The magnet should be repelled and thus move having some Kinetic energy. However, if the magnet remains stationary as in your case, it is understood that there is an agency (external force) on the magnet.


Few modification to the setup. Let's replace the generator with a resistor in the electric circuit to avoid strange behaviors.

Governing equations. Let's write down some equations, before doing an energy balance,

  • dynamical equation of the moving magnet:

    $m \ddot{x} = F^{m, em} + F^{ext}$,

    being $F^{m, em}$ the force acting on the magnet due to the induced field produced by the electromagnet, and $F^{ext}$ the force you provide to the magnet with your hand;

  • Faraday-Lenz law,

    $v = \dot{\Phi}_{S}(\mathbf{b}) = \dot{\Phi}_{S}(\mathbf{b}^{mag}) + \dot{\Phi}_{S}(\mathbf{b}^{em}) = \dot{\Phi}_S(\mathbf{b}^{mag}) + \mu \dfrac{N}{\ell} \dfrac{d i}{dt} = \dot{\Phi}_S(\mathbf{b}^{mag}) + L \dfrac{d i}{dt}$,

    being $\Phi_{S}(\mathbf{b}^{m})$ the magnet flux in the coil due to the magnet, and $\Phi_{S}(\mathbf{b}^{em}) = L i$, the flux of the magnetic field produced by the induced current in the coil;

  • circuit equation

    $R i = v$

Energy balance. Now let's multiply the mechanical equation by $\dot{x}$ and the circuit equation by $i$, we get

$\dfrac{d K}{dt} = \dfrac{d}{dt}\left(\dfrac{1}{2} m \dot{x}^2 \right) = m\dot{x}\ddot{x} = \dot{x} F^{m, em} + \dot{x} F^{ext} = \dot{x} F^{m,em} + P^{ext}$
$Ri^2 = i v = i \dot{\Phi}_S(\mathbf{b}^{mag}) + i L \dfrac{d i}{dt} = i \dot{\Phi}_S(\mathbf{b}^{mag}) + \dfrac{d}{dt} \left(\dfrac{1}{2} L i^2 \right) $.

Now we can add these two equations, to get

$\dfrac{d}{dt}\left(\dfrac{1}{2} m \dot{x}^2 + \dfrac{1}{2} L i^2 \right) - \dot{x} F^{em,m} + i \dot{\Phi}_S(\mathbf{b}^{mag}) = -Ri^2 + P^{ext}$

Energy balance, conservation and force on the magnet. If we assume that electromagnetic and mechanical processes involved in the process are conservative, beside the dissipation $-R i^2$ in the resistor, we can conclude that

$\dot{x} F^{em,m} = i \dot{\Phi}_S(\mathbf{b}^{mag}) = i \dot{x} \dfrac{d }{dx}\Phi_S(\mathbf{b}^{mag})$

providing a relationship between the force acting on the magnet due to the interaction with the electromagnet, namely

$F^{em,m} = i \dfrac{d \Phi_S(\mathbf{b}^{mag})}{dx}$.

Relativity. Changing the reference frame:

  • the electromagnetic contributions are the same;
  • the power of the forces are the same, since both the forces and displacement of the moving parts change signs;
  • the kinetic energy is zero if we assume that we can neglect the mass of the electromagnet. So what? Does the total energy have not the same in these two cases? Remember that we don't care about the absolute value of energy, in classical mechanics at least, but only about the difference of energy values. To be more explicit, try to have a look in these reference, for the evaluation of kinetic energy w.r.t different reference frames:
  • Work required to change an object's speed in different reference frames, https://physics.stackexchange.com/q/734777,
  • Velocity in power calculations in different inertial frames, https://physics.stackexchange.com/q/736748
  • Conservation of kinetic energy and external forces, https://physics.stackexchange.com/q/735204

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