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Assume we have this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

At t=0s we decide to close the switch, then current will start flowing through the circuit and around the inductor a magnetic field will be induced.

My questions: According to Faraday's law, the electric voltage generated on a coil will depend on the change of the magnetic flux through the coil. Since the magnetic field grows at a constant rate shouldn't the back emf of the inductor remain the same? Why does it have the form it does?

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  • \$\begingroup\$ It may be interesting to note that this switch can ONLY be switched on one time. It can not be switched off due to a contradiction concerning the steadyness of the current. See my answer here:electronics.stackexchange.com/questions/458814/… \$\endgroup\$
    – xeeka
    Mar 28, 2021 at 23:07
  • \$\begingroup\$ MissMulan, do you have any questions or are you done with this subject now? Are you able to state what Why does it have the form it does means so I can provide more detail? \$\endgroup\$
    – Andy aka
    Mar 31, 2021 at 12:17

3 Answers 3

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Once the switch is closed the voltage across the inductor is constant and equal to the voltage of the source. Based on the schematic you have drawn, and assuming ideal elements, there is no other possibility.

The rate of change of the current through the inductor, \$\frac{di}{dt}\$ will therefore also be constant.

So, the current will increase linearly.

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  • \$\begingroup\$ Please add to your answer that if the inductor has a magnetic core, at some point it will saturate, the inductance will drop, and the current will increase at a faster rate. Also at a certain point, the resistance of the wires will cause a voltage drop equal to the applied voltage and the current will no longer increase at that point. \$\endgroup\$ Mar 28, 2021 at 21:53
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    \$\begingroup\$ @MathKeepsMeBusy I am only talking about circuits using ideal components...no resistance in the wires. The OP seems to be working through some circuit analysis homework. \$\endgroup\$ Mar 28, 2021 at 22:03
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    \$\begingroup\$ Yup. This is a basic circuits question, so assume you're in Plato's ideal universe, where every inductor is actually just the idea of an inductor, and infinite currents can exist. \$\endgroup\$
    – TimWescott
    Mar 28, 2021 at 23:37
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Since the magnetic field grows at a constant rate shouldn't the back emf of the inductor remain the same?

For an ideal inductor (no saturation and no losses) the current grows at a constant rate as governed by this equation (basically a spin-off of Faraday's induction): -

$$\dfrac{di}{dt} = \dfrac{V}{L}$$

And, of course, flux grows at a constant rate. This means that the back emf is also constant and equal to the applied voltage.

Why does it have the form it does?

If you can explain what that means I might be able to answer it.

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  • \$\begingroup\$ @MissMulan, do you have any questions or are you done with this subject now? Are you able to state what Why does it have the form it does means so I can provide more detail? \$\endgroup\$
    – Andy aka
    Mar 31, 2021 at 9:31
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Since the magnetic field grows at a costant rate shouldn't the back emf of the inductor remain the same?

No, magnetic field grows as an (Constant + exponential function) if there is resistance and inductor (in series) e.g

$$\phi= \phi_o[1-e^-(tR/L)]$$

and grows linearly if only inductor is there ! Assuming inductor in linear region.

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  • \$\begingroup\$ Ah really? thanks! \$\endgroup\$
    – Miss Mulan
    Mar 28, 2021 at 20:03
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    \$\begingroup\$ But how does the magnetic field grow exponentially?How is that even possible? \$\endgroup\$
    – Miss Mulan
    Mar 28, 2021 at 20:12
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    \$\begingroup\$ A more intuitive way to word it might be that it exponentially approaches a fixed value. \$\endgroup\$
    – Hearth
    Mar 28, 2021 at 21:03
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    \$\begingroup\$ Also please add to you answer the fact that if the inductor has a magnetic core, as the current increases, the core will get closer to saturation, and the inductance will decrease. (I.e. the "linear" or "constant" growth part, is only linear or constant up to a point. \$\endgroup\$ Mar 28, 2021 at 21:56
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    \$\begingroup\$ Unless you're in a basic circuits class, in which case it's an ideal inductor -- in that case, with a constant voltage the current is a constant ramp, forever. The exponential part comes about if you're talking about a real inductor. But you need to get your head wrapped around the ideal inductor case first -- then you won't be confused by a more-real inductor model. \$\endgroup\$
    – TimWescott
    Mar 28, 2021 at 23:39

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