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(English is not my native language; kindly ignore certain grammatical errors.)

Reference: Problems in General Physics - IE Irodov Q 3.241 A very long straight solenoid carries a current I . The cross-sectional area of the solenoid is equal to S, the number of turns per unit length is equal to n.

Find the flux of the vector B through the end plane of the solenoid.

Doubt:

As far as I can remember, when we make simplifying assumptions about a solenoid we say that it should have tight winding, no resistance, and be very long (relative to the radius). In such a case the field lines are almost parallel, and uniform. There is no magnetic field leakage outside the solenoid, and hence the field strength is 0 outside and $N(\mu_0)I$ inside.

If that is the case, then why exactly does the field strength become half at the edge?

If I take a surface deep within the solenoid or at the surface, it should not make much of a difference, since there is no leakage of the lines.

There is one reasoning I have heard: "You add two semi-infinite solenoids to make an infinite one, hence the field strength of each should be half."

But my problem is that the "loss" of half of the strength at the edges makes the assumptions dubious, as the results we get do not match. Those field lines must escape through the solenoid itself. Therefore, the above assumptions become pointless.

Kindly correct me if I am wrong. Thank You.

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  • $\begingroup$ Yes, there has to be a magnetic return path because the field lines have to be closed. The external field is weak but finite. The longer the solenoid becomes relative to its diameter the weaker the external field becomes because the field lines are extending further and further into the space outside of the solenoid. Do we need to make any of these assumptions? No. We can calculate the actual field using Biot-Savart's law, it's just a lot more work than this simple argument. $\endgroup$
    – FlatterMann
    Commented Jul 30 at 19:11
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    $\begingroup$ You can avoid invalidating any assumptions by just describing the field as the sum of terms coming from each turn of the coil. If you then leave out one half of the coil you will miss exactly half of the field in the plane where the split is made. No assumptions needed, so we don't have to worry about their validity! (And we do not actually have to compute anything with Biot-Savart. Symmetry assures us that we will miss half the field, whatever it is...) $\endgroup$
    – Jos Bergervoet
    Commented Jul 30 at 19:13
  • $\begingroup$ What is the end plane of the solenoid? $\endgroup$
    – Claudio Saspinski
    Commented Jul 30 at 23:05
  • $\begingroup$ @ClaudioSaspinski The plane of the last turn of the winding. Think of the end of the equivalent cylinder. The question asks by how much the field weakens at the end of a long solenoid. Not a bad question, really, and the OP has a point. One can not model the ends of even the longest solenoid as ideal. It just happens that the region in which the field is not almost perfectly parallel is always on the order of the diameter of the solenoid for any lenght l. So if we make the ratio d/l small, then we can neglect the ends because they only give O(d/l) error terms. $\endgroup$
    – FlatterMann
    Commented Jul 31 at 0:04

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Its a good idea to be very careful how you describe position on the solenoid. For example "edge" what do you mean? do you mean a point on the axis through the solenoid at one or other end of the solenoid or do you mean any point on the cylindrical surface of the solenoid (at radius r from the axis)?

To find the magnitude of the vector field B anywhere in the geometry of the solenoid you have to use Ampere's equation and integrate for contribution from turns dn along the length of the solenoid dy.

Magnetic field magnitude does not drop to zero anywhere inside the solenoid or anywhere outside the solenoid (unless you get far away from the solenoid).

In the case of an infinitely long solenoid the magnitude of B (let's say on the axis inside the solenoid) there are an infinite number of turns so the magnitude of B is infinite. This is because B is proportional to n and to the current I. Current stays the same but if n is infinite then so is B.

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