tun1ju2k1ki

2022-10-05

Assume you have an iron core in the interior of the solenoid. It is well known that the strength of the field should increase by a factor of several hundred inside the solenoid as a result of the iron core.
However, at the boundary between the iron core and the surrounding air, what happens to the magnetic field strength?

Mckenna Friedman

At the boundary of air and iron core, the following relations hold:
$\stackrel{\to }{n}\cdot \left(\stackrel{\to }{{B}_{1}}-\stackrel{\to }{{B}_{2}}\right)=0,\phantom{\rule{1em}{0ex}}\left[\stackrel{\to }{n},\stackrel{\to }{{H}_{1}}-\stackrel{\to }{{H}_{2}}\right]=0$
$\stackrel{\to }{n}$ is normal to the surface of the core. The surface-normal induction component is continuous. The tangential component of the magnetic field is also continuous. In practice, in the case of a cylindrical core and with sufficiently tight winding, the magnetic field induction (measured in Tesla) at the ends will be as in a core. And on the lateral surface of the core (adjacent to the winding), the tangential component $\stackrel{\to }{B}$ changes abruptly.

dripcima24

Solenoid has a core in the shape of a cylinder. When magnetized, such cylinder produces its own magnetic field $B$, orders of magnitude stronger than the external field due to electric current, but this is apparent mostly near its poles - ends of the cylinder and it is also true inside the core.
This magnetic field of the core is continuous when crossing the pole face disks, but not so when crossing the cylindrical surface of the core such as near the core center. The field just outside the core and above the core center is quite weak, as opposed to the field inside just below the surface, which is strong. So there is a jump in $B$ when crossing the cylindrical surface.

Do you have a similar question?