0
\$\begingroup\$

From my understanding is the process of induction described by a magnetic field line which approaches a wire perpendicular and creating an local electric field in it which can be detected which measuring electrical voltage at the ends of the wire.

What bothers me is this process in a secondary transfer coil. The magnetic field lines are inside the core and do not cut the coil wire at the surface of the core.

Does one need to differentiate between induction by magentic field lines cutting a wire perpendicular and the sum of magentic field lines inside a loop of wire? How exactly does induction happen in the last case compared to the first case?

\$\endgroup\$
1
\$\begingroup\$

From my understanding is the process of induction described by a magnetic field line which approaches a wire perpendicular

That is the wrong idea. This is the right idea: -

enter image description here

Does one need to differentiate between induction by magentic field lines cutting a wire perpendicular and the sum of magentic field lines inside a loop of wire?

Induction is not due to lines of flux cutting the wire perpendicularly. If you are talking about this scenario: -

enter image description here

Then it is incidental that some magnetic flux lines are passing through the conductor.

\$\endgroup\$
  • \$\begingroup\$ So you mean that even in the last picture I have the so called necessary (one) loop of wire? So it is always about change of lines of flux over an area A in a closed loop of wire. \$\endgroup\$ – hendrik2k1 Jun 14 '18 at 10:10
  • \$\begingroup\$ @hendrik2k1 It does boil down to that. However, if the north and south poles were vertically very high and fully enclosed the loop, there would be no measured induced voltage with motion because there is no net change of flux with motion. \$\endgroup\$ – Andy aka Jun 14 '18 at 10:27
  • \$\begingroup\$ you mean when the loop radius is big and the area of my permanent magnet small in comaprison. When moving this inside the loop the net sum of lines of flux would not change and so I would register no voltage? \$\endgroup\$ – hendrik2k1 Jun 14 '18 at 12:32
  • \$\begingroup\$ No, that isn't what I said. \$\endgroup\$ – Andy aka Jun 14 '18 at 12:35
  • 1
    \$\begingroup\$ The magnet poles are much bigger in area than the coil loop area. \$\endgroup\$ – Andy aka Jun 19 '18 at 8:37
0
\$\begingroup\$

The fundamental thing is how much flux is contained within a closed loop of wire. When that flux changes, a voltage is induced in the closed loop. It doesn't matter what material is within the loop, whether air, ferrite, iron or whatever.

There are two ways to change the flux within the loop. One is to change the strength of the flux, as in a transformer. Second, the loop can move with respect to the field, resulting in the wire of the loop 'cutting' lines of field, as in a motor or generator.

\$\endgroup\$
  • \$\begingroup\$ But is a closed loop necessary? When I only have a straight wire of say 1 meter and pulling the magnet over it in the middle I would not expect this to be a closed loop of wire. \$\endgroup\$ – hendrik2k1 Jun 14 '18 at 9:17
  • \$\begingroup\$ there is always a loop - the connections eventually must loop back somewhere \$\endgroup\$ – Henry Crun Jun 14 '18 at 11:34
-1
\$\begingroup\$

In my opinion, the math obscures the behavior. As you've noted, the induced voltages are orthogonal to the "magnetic flux", so how does this happen?

Relativistically, [ahh the spell checker does not like that word], when the Lorentzian compression maths are developed for masses and accelerations and momentums, some orthogonal forces show up.

\$\endgroup\$
  • \$\begingroup\$ please explain your vote \$\endgroup\$ – analogsystemsrf Mar 15 at 2:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.