If we hold the voltage constant, will (and how) the current through a coil change if we put a paramagnetic material in the coil ?


2 Answers 2


Yes, it will change.

Simply take the equation for the energy stored in a coil, which is equal

$$U_L = \cfrac{1}{2}L I^2 \quad .$$

Assuming that you had first an air coil, the inductance $L$ will increase if you put a paramagnetic material inside the coil. Therefore, more magnetic energy can be stored in the coil and therefore the current will increase until

$U_{L\,\mathrm{new}} = \cfrac{1}{2}L_\mathrm{new} I^2, \qquad \mathrm{whereby} \qquad L_\mathrm{new} > L .$

When the coil is "complity loaded" the voltage across the coil is zero!
The current, $I$, tends to be the same for both cases $t \rightarrow \infty $, assuming you have a non-zero resistance. If you make a circuit analysis don't forget to include a resitor ($R= R_\mathrm{source} + R_\mathrm{coil}$) into your equivalent ciruit diagram, otherwise the current would increase towards infinity.

|   R         L       |
+                     |
-                     |
|                     |
-----------------------      .

You can also make an exact analysis using Maxwell's $3^{rd}$ law:

$$\nabla \times \mathbf E = - \cfrac{\partial \mathbf B}{\partial t}\quad . $$

Writing this equation in an integral form and using Stoke's theorem yields to the emf $\mathcal E$:

$$\mathcal E = \oint \nabla \times \mathbf E \, \mathrm d \mathbf s = - \int_\mathrm{surface} \cfrac{\partial \mathbf B}{\partial t} \; \mathrm d \mathbf S = - \cfrac{\partial \mathbf \psi}{\partial t}\, .$$

Since the definition of inductance is $L = \cfrac{\psi}{I}$, you get

$$\mathcal E = - \cfrac{\partial (L \cdot i)}{\partial t} = - i \cdot\cfrac{\partial L} {\partial t} - L \cdot \frac{\partial i}{\partial t} \, .$$

This result can be used for a correct circuit analysis!

Setting up the equations of this mesh, gives on the left side the total emf, and on the right the resistor's voltage:

$$U_0 - \mathcal E = R \cdot i\, . $$

Making use of the second last equation results in

$$ U_0 = \Big( R + \cfrac{\partial L} {\partial t}\Big) \cdot i + L \cdot \frac{\partial i}{\partial t} \; .$$

Now finding the solution is very straightforward, and the result a little surprising (at least for me).

Discussion of Energy and Power

It is interesting to discuss the power of this equivalent circuit shown above!

By simple multiplying the last equation by $i$, we get power equation,

$$ p(t)_{\text{DC-source}} = U_0 \cdot i = \Big( R + \cfrac{\partial L} {\partial t}\Big) \cdot i^2 + L \cdot i \frac{\partial i}{\partial t} \; \\ = \Big( R + \cfrac{\partial L} {\partial t}\Big) \cdot i^2 + \cfrac{1}{2} L \frac{\partial i^2}{\partial t}\,.$$

This assumes that the material has no initial magnetic field! Otherwise mechanical work has to be done, and has to be involved to this equation.

Knowing that energy $E = \int p \,dt $ yields:

$$E = \int R i^2 \, dt + \int i^2 \, dL + \int \cfrac{1}{2} L \, d(i^2)\, . $$


magnetic dipoles in paramagnetic materials can align under the influence of an outside magnetic field, but this alignment much weaker and never stays permanent in comparison with feromagnetic materials, so placing a paramagnetic object in a coil with current flowing through it would temporarily increase this current depending on the shape of the object and how fast we move it in, but in time the current will come down to the new value which will be lower than the original current


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