# Can non-conservative fields store potential energy?

I was taught that a time-varying magnetic field generates an electric field which is non-conservative in nature, and my teacher also told me that when a conducting coil is placed in a region with a time-varying magnetic field, with appropriate orientation, some EMF is induced in the coil and this EMF is due to the presence of that induced electric field.

Now, this is what is given in my physics NCERT book (I have highlighted the text): The statement "work against the back emf" is the same as "work against the induced electric field", right? If that's true , then how can that energy get stored?

Negative work done by conservative fields is stored in the fields as potential energy. I don't think the same is true for non-conservative fields, otherwise, there is no difference.

So how can the induced electric field store energy ?

• Definition of a conservative field is that it is the field for witch work done is independent of the path taken from, e.g., point A in the field to point B in the field, which in other words mean, that in circular path, there is no net change in energy. Now, for non-conservative fields, this is not the case, work done depends on the path, but I dont think that means you can not store any energy in the field. Nov 17, 2021 at 16:15
• Using text instead of images wherever possible is usually recommended, see meta.stackexchange.com/q/320052.
– gmz
Nov 25, 2021 at 17:55
• Hard to say, what do you mean by "store"? By construction you cannot define the energy function of a non-conservative force. Nov 25, 2021 at 19:28
• electronics.stackexchange.com/a/551548/236654 Nov 26, 2021 at 5:51
• electronics.stackexchange.com/questions/551244/… Nov 26, 2021 at 5:51

The statement "Work against the back emf" is the same as "work against the induced electric field" (Right ?). And if that's true , then how can that energy get stored ?

Yes. This work against the forces of induced electric field is stored as magnetic part of EM energy in the region. As induced electric field is present and is doing negative work (other forces such as electrostatic forces due to battery act against the induced forces and can do the positive work), greater magnetic field is created and this is associated with greater magnetic energy. It does not matter whether the EM field is conservative or not, if it is not vanishing everywhere, energy can be stored as EM energy.

• what did you mean by this "if it is not vanishing everywhere, energy can be stored as EM energy." ? Also a non conservative field can store energy as long as it doesn't "vanish" ??? Nov 18, 2021 at 15:12
• Well, when EM field vanishes in some region, this means it is zero at every point of that region. And then this region does not contain Poynting EM energy. Yes, also non-conservative EM field can store energy, the formula for EM energy is the same, it does not matter whether the field is conservative or not. Nov 18, 2021 at 16:39

A conservative field $$\vec F$$ is one which satisfies $$\oint_{\Gamma}\vec F\cdot \vec{dl}=0$$ for any path $$\Gamma$$. This just means that work done from moving a particle interacting with that field is independent of the path taken i.e $$\int_{\Gamma_1}\vec F\cdot \vec{dl}=\int_{\Gamma_2}\vec F\cdot \vec{dl}$$ if $$\Gamma_1$$ and $$\Gamma_2$$ have the same endpoints and orientation. Due to this path independence one can write down the work done as a function of just the two endpoints without any mention of the path taken. This function is the potential energy and can be interpreted as the energy stored in the field. This does not mean that only if one can define a potential energy function then energy can be stored in the field. A non-conservative field can store energy too.

• didn't you mess up with the definition of non conservative force ? It's work depends on the "path" it took Nov 25, 2021 at 16:24
• Yes I did, it was a typo Nov 26, 2021 at 1:00

No. By the nature of nonconserative fields, potential energy is not stored.

From the equation $$\frac{dW}{dt}=|\varepsilon|I=(L\frac{dI}{dt})I\;\;\;\;\;$$(1)

that can be rewritten $$\;\;dW=LIdI$$

a simple integration of the equation gives $$W=\frac{1}{2}LI^{2}$$

In a solenoid, the following results are valid :

$$B=\frac{\mu NI}{l}\;$$, magnetic field, where $$\mu$$ is the relative permeability, $$N$$ the number of turns of the coil, and $$I$$ the system current.

$$L=\frac{\mu N^{2}A}{L}\;$$,where $$l$$ is the length of the solenoid and $$A$$ its cross section.

We can replace the stored energy terms in the following expression, where the volume of the system is given by $$Al$$ :

$$u=\frac{energy}{volume}=\frac{\frac{1}{2}LI^{2}}{Al}$$

hence, the magnetic energy density (energy per unit volume) in a region of permeability space $$\mu _{0}$$, containing a magnetic field $$\vec {B}$$ is written :

$$u=\frac{1}{2}\frac{B^{2}}{\mu_{0}}\;\;\;(2)$$