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enter image description here

From my understanding, the magnetic field inside a solenoid is constant, so at points C and D, the magnetic field strength is the same. Magnetic field strength outside solenoid is minimal and is regarded as zero, so point B display this. But what about point A, it is just outside the solenoid, would the magnetic field strength at that point is zero? If so why? Is the magnetic field strength taken to be zero, even at a point just outside?

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  • $\begingroup$ First of all, please clarify that is that solenoid an infinite solenoid (as we can consider some approximations as well if this is the case) or a normal solenoid. $\endgroup$ Jan 29, 2023 at 15:23
  • $\begingroup$ I'm not sure any answer will be better than reading the wiki en.wikipedia.org/wiki/Solenoid. Check the section "Finite continuous solenoid". and specifically the formula for $B_z$ on axis (the last formula for that section). You might be surprised how non-constant a finite length solenoid's field is. This is why most real solenoids used in experiments have "corrector coils" at the ends to make the field more constnat. Phillip wood's answer is also very useful - if the solenoid is very long, at the end the field is half the field in the center. $\endgroup$
    – AXensen
    2 days ago

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Suppose point A to be on the axis of the solenoid and at its end (rather than just beyond). For a long solenoid (length >> diameter) the field at A is $$B_A=\tfrac 12 B_D$$ in which $B_D$ is the field at the point D, which we'll take as the central point. We can deduce this result from symmetry... The field at D must be due equally to the turns either side of D. At A, though, there are turns only on one side of the point, so there is only half the field that there is at D.

But, you say, there are twice as many turns to the left of A as there are to the left or to the right of D. This, though, is irrelevant if the solenoid is LONG: turns far from D or far from A (that is more than a distance equal to several diameters away) will hardly contribute to the field at these points. Half a long solenoid is still a long solenoid!

It is for this reason, too, that the field is $\mu_0\tfrac nlI$ inside the solenoid not just at the centre, but everywhere except near (within a few diameter-lengths of) the ends.

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To a good approximation, the magnetic field inside a solenoid is constant. The ends of a solenoid, on the other hand, are, like magnetic poles, apparently sources of a diverging B field which decreases in intensity as R**(-2).

An infinite-length solenoid does have uniform internal B field. A solenoid with cylinder radius much smaller than its length, has similar B field character, except near endpoints.

A finite solenoid can be treated by exact calculation (which is messy) or by approximations for which the chief merit is that they are 'often good enough', because the approximation captures some of the truth of the field nature, without being entirely true.

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To add to the (correct) answers above, the general rule in understanding pictures like this one is that magnetic field strength at any given point is proportional to the density of "field line" that they draw. This is not a lucky coincidence, this is how Sir Michael Faraday came up with the idea of magnetic field in the first place.

So for instance the density of field lines near point B is very low (in fact, there is none drawn. Which means that with the level of precision of this sketch the field strength is zero there). The density of field lines is the same at points C and D hence the field strength is the same there. By the time you get to the point A the field lines are not bundled up together anymore as the solenoid is over, and so the spread a bit as the figure suggests, so the field strength will decrease as compared to the value inside solenoid.

If you are interested in how exactly field strength scales with distance $R$ from the solenoid edge, here are some rules-of-thumb. Here we assume the diameter of the solenoid $D$ is much less than its length $L$. For $R<<D$ the field strength is roughly constant and crudely equal to the value inside of the solenoid (in fact one half, see Philip Wood's answer). Once outside, the field lines flow out of the edge of the solenoid like water from an open hose. In other words, the edge of the solenoid seems like a localized source of field, i.e. a monopole, so for $D<<R<<L$ the field strength goes as $|\vec{B}| \propto 1/R^2$. Here the first condition $R>>D$ is to ensure the edge of the solenoid can be treated as a point, and the second condition $R<<L$ is to be sure the field "flowing" from the other solenoid tip does not interfere. Finally, when $R>>L$ the contributions from both solenoid tips/edges are equally important, and the entire solenoid is nothing but a magnetic dipole with the standard field dependence $|\vec{B}| \propto 1/R^3$.

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  • $\begingroup$ Sir Michael Faraday? I believe that Faraday turned down, on religious grounds, an offer of a knighthood. $\endgroup$ Jun 2, 2023 at 17:24
  • $\begingroup$ @Philip Wood You must be right. He is not a "sir" indeed. My bad. $\endgroup$
    – John
    Jun 2, 2023 at 17:57

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