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Let's say we have a DC source connected to an inductor. The orientation of the magnetic field through the inductor depends on which direction the coil was wrapped (right hand rule). But why do we not care about it's polarity when we use inductor in the circuit? I guess we do care about polarity in terms of mutual inductance, but what about when only one inductor is used in the circuit?

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    \$\begingroup\$ Simple inductors are just a wire coiled up. They are symetrical. \$\endgroup\$ – Fredled Sep 3 '20 at 14:54
  • \$\begingroup\$ @Fredled This isn't true. Sure, you can always re-orient two unpowered coils wound in the opposite direction to be in the same orientation since there is nothing to differentiate the ends, but things are different when you send current through it since the act of sending current through it a front and back by virtue of polarity which produces a defined CW or CCW direction. \$\endgroup\$ – DKNguyen Sep 3 '20 at 21:10
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    \$\begingroup\$ @DKNguyen Yes but if the polarity depends on the curent but the part itself is materially symetrical, then we are talking about the polarity of a circuit, not of a part. I don't really understand the question. \$\endgroup\$ – Fredled Sep 3 '20 at 21:23
  • \$\begingroup\$ @Fredled Materially maybe. We use them magnetically and when magnetized they are definitely not symmetrical. OP is basically asking why we don't care which end is north and which end is south when sticking an inductor into the circuit. I am not even sure what the plane of symmetry is for a helix. Can't seem to pin it down. It must fall under a more complete math definition of symmetry. \$\endgroup\$ – DKNguyen Sep 3 '20 at 21:35
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    \$\begingroup\$ @DKNguyen You're introducing complexity where none need exist. OP asked about an inductor not a transformer, and made no mention of centre taps etc. For a simple inductor, the only time a polarity might be detectable is when it is when it is in a magnetic field, including (to a very small extent) the Earth's. \$\endgroup\$ – Mark Morgan Lloyd Sep 4 '20 at 6:59
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We only care about the polarity of an inductor when the polarity of the magnetic field is important.

Such as with electromagnets or common mode chokes and measurement transformers.

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  • \$\begingroup\$ Okay, that's what I thought. Thanks!! \$\endgroup\$ – user207787 Sep 3 '20 at 12:41
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    \$\begingroup\$ @user207787 When we do care, that's when the inductor has the dot on one end of it to indicate polarity relative to another coil. \$\endgroup\$ – DKNguyen Sep 3 '20 at 13:14
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You might care. For example, if you have a filter choke that has a permanent magnet biasing it.

enter image description here

Or if there is another inductor magnetically coupled to it, then the relative polarity matters.

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  • \$\begingroup\$ I love how a pirate is related to a Joule thief. :D \$\endgroup\$ – mberna Sep 5 '20 at 17:37
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Suppose you have two identical inductors, except one has a right-handed winding, and the other has a left-handed winding. Ramp up the current through the right-handed inductor. This produces a changing magnet field which induces a voltage in the inductor. That's self-inductance. Now do the same with the left-handed inductor. The handedness reversal reverses the relationship between the current and the magnetic field, but it also reverses the relationship between the magnetic field and the induced voltage. With the double reversal, you get an identical induced voltage, so the self-inductance is the same.

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enter image description here

Actually we do care about Inductors, especially with DC circuits. The polarity depends on the direction of the current flowing through it. If a current is moving from point A to point B (say), the Inductor stores the current in the coil inturn producing magentic field. Now if the device is turned off, the inductor that had the current stored in it must dissipate. But because of the reverse EMF (according to your right hand thumb rule) the current travels form point B to A. This current can spike up and destroy your circuitry. That's why you have bypass diodes connected across node A and B such that it creates a loop. Eventually the current dissipates with all its might saving the circuitry behind it. Hope this helped.

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    \$\begingroup\$ But inserting the inductor with the connections reversed doesn't change anything in this situation. You may have missed the point of the question. \$\endgroup\$ – Transistor Sep 3 '20 at 13:48
  • \$\begingroup\$ I wanted to convey that inductance polarity matters and there a certain significance taken into consideration. At least that was my interpretation of the Question. But Apologies for the inconvenience. \$\endgroup\$ – N0m4d Sep 3 '20 at 14:16
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    \$\begingroup\$ Not my interpretation of the question but the comments are very useful and on topic. So I voted it up. \$\endgroup\$ – Fredled Sep 3 '20 at 14:51
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    \$\begingroup\$ The polarity depends on the direction of the current flowing through it. This is not polarity of the component. And it doesn't answer the question. \$\endgroup\$ – Graham Sep 3 '20 at 23:16

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