While doing a problem I didn't understand why is the the magnetic flux and the end of a long solenoid is:

$$ \phi = \frac{\mu inS}{2}$$

Where $\mu$ is the permeability of free space, $n$ is the number of turns, $i$ is the current, and $S$ is the area of cross section.

In the solution , it is given that the That the field at the end is:

$$B= \mu ni/2$$

Hence the flux is $BS$.

But isn't the field non uniform at the end and isn't $\mu nI/2$ the field along the axis only ?

Also , how come the flux at the end not equal to the flux somewhere deep in the solenoid ?

Which should be $\phi = \mu niS$

Since the same field lines are passing through the middle and the end ?


1 Answer 1


Hint: If you consider a really long solenoid and cut it in half, the field at the center (in the plane where it'd been cut) comes from equal contributions from both the halves.

The field indeed is non uniform at the end, but for a decent approximation, you can consider your solenoid to be like one of these halves. What field and flux does this consideration give you?

  • $\begingroup$ Ok that gives the correct answer, but that doesn't explain why the flux inside the solenoid is different from the end , since the same no. Of field lines should pass through tt both cross sections $\endgroup$ Jun 27, 2017 at 8:23
  • $\begingroup$ Inside a solenoid, the field lines are contained in it. Here, a lot of the field lines "escape" from the sides. (diverge outward ) This escape effect is countered by two half solenoids. In this case, there is nothing to counteract this. Hope I'm being clear? $\endgroup$ Jun 27, 2017 at 8:25
  • $\begingroup$ Ok , so do you mean filed lines escape from the curved surface of the solenoid ? Because that would explain the decrease in flux $\endgroup$ Jun 27, 2017 at 8:29
  • $\begingroup$ Yes. Also, remember that field lines are a very qualitative thing and mostly won't get you anywhere with calculations. $\endgroup$ Jun 27, 2017 at 8:31

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