# Magnetic field outside the solenoid

I found solution to problem 257 from "300 Creative Physics problems" hard to understand. In that problem we have very long solenoid with coil wounded in one layer. Data such as: density of turns, magnetic field at distance of 5 cm form the axis of solenoid and diameter (1 cm) was given. We hevo to calculate field inside the solenoid. Solution states, that "Because of the symmetry of arrangement, the magnetic field produced by the coil outside the solenoid is similar to that of a straight, current-carrying wire, as it is shown in the figure on page 474 [picture below]."

Then, author used that fact and applied Maxwell second law to circle of radius 5cm. IMO, quoted text is meaningless. Magnetic field lines, which are eascaping interior of solenoid must "return" and re-enter that (Gauss law for magnetism). Also, they will never be perpendicular to the lines inside the solenoid, so they cannot be treated as circular field (produced by straight conducting wire). Please help!

• That's a poorly written book, but to be honest, there is a lot of such nonsense about the solenoid out there. For one thing most people forget that the current has to flow in a loop, so there is always a wire going back from one end of the solenoid to the other. That wire together with the effective current flow mentioned by this author forms another field component... which is almost never mentioned. – CuriousOne Dec 26 '15 at 0:10
• Thank you so much for your analysis. But it entails fact that the wire connecting two ends of solenoid must lie in the axis of symmetry of the solenoid (to form that circular field component), which is, hm, peculiar configuration. Is my reasoning correct? – Z. Naglik Dec 26 '15 at 20:03
• The wire connecting the two ends usually lies outside of the solenoid and gives rise to an asymmetric field component. Making nice, symmetric magnetic fields is not easy! – CuriousOne Dec 26 '15 at 20:39