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enter image description here

We did this derivation in class. We took a cross section of an ideal solenoid and took a rectangular loop to apply ampere's circuital law. It is pretty much the same as in the diagram but the sides 'ad' and 'bc' were extended such that the loop covered the complete solenoid including the current coming 'out of page'. The current going in from the top cancels the current moving out from the bottom so the summation of B.dl from the four sides of loop will be equal to zero. The sides bc and ad are perpendicular to B so B.dl for them is zero.

Now we were told that as the sum of B.dl for sides cd and ba is zero, B outside is zero (keep in mind this is not the diagram we were working on, the side cd has engulfed the solenoid, so both segments, ab and cd, are on the outside of solenoid).

But it could also be the case that B.dl for cd is negative of B.dl for ab because after all element dl is in opposite direction for both while B must be in the same direction (I mean B outside the solenoid). So how can you justify that B outside= 0?

Hope you got my question.

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The magnetic field inside, and near the center, of a long narrow current carrying solenoid is very close to being uniform. Out side, it is close to zero. Toward the ends it diverges more strongly than indicated on your sketch. Ampere's law is nearly accurate near the center (inside or out), but cannot be simplified near the ends.

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The magnetic field outside the selenoïd is not null. If it was it would mean that $\nabla \cdot {B} \neq 0$ or equivalently $\int_\Sigma B \cdot \textrm{d}\sigma \neq 0$ where $\Sigma$ is the surface of a cylinder enclosing the selenoïd. This violates the fact there is no magnetic monopoles.

I'm not sure about this derivation, but one should be careful when using amper's law. The current takken into account are just the ones passing any surface with border accounting for the magnetic field (abcd in this case). So the elements of current of the top ("current out of page") are not accounted.

$$ \oint_C B\cdot \mathrm{d} l = \iint_\Sigma \mu_0 J\cdot \mathrm{d} \sigma $$

where sigma is any surface with border $C$.

Best

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The way you drew your diagram the B field is parallel and lies in the same direction as the integration contour hence $\mathbf{B}\cdot \rm{d}\ell$ is positive over the contour. This is so because the B-field lines of a finite solenoid must form closed lops and consequently the "number" of lines inside the solenoid and outside solenoid are the same. Hence outside where they spread out the line density representing the strength of the field must be weaker than it is inside.

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  • $\begingroup$ oh dear God.... $\endgroup$ Mar 20 at 9:52

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