0
\$\begingroup\$

I'm working through an example with motional EMF and I'm having trouble understanding the directions of vectors so that I can apply induction law.

The magnetic circuit seems complex because the circuit is used to analyze other situations but the air gap 3, the coil 3 and the single open loop coil are the ones relevant to this. We neglect dispersion and the magnetic reluctance of iron. Section S of the circuit is constant and it's a square of side a. The air gaps have thickness $\delta$. All of the coils have the same number of turns. The open coil has a current $i_0$ that is zero.

enter image description here

Now we have i2=1A and i3=i0=0 which originates a flux $\phi=-1mH$ and $B_3=-0.625 T$.

Now my question is about the next paragraph:

"The motional induction electric field only exists when the coil 0 passes the air gap 3 (I understand that, because only there B is not zero). On that conditions, the elecric field as an orthogonal direction to the figure plan, the same as the current i0 (I think I can also see that...). At the bottom side of the coil we will have $u_0=Bva$ (now that is what I don't understand, what is the direction of B? Is it the same as v? Why? I can't see the direction of the vectors!)."

Basically my question is about the direction of the vectors while applying the induction law. Can someone help me clarify it? I only need a small draw or some brief explanation.

\$\endgroup\$
0
\$\begingroup\$
  1. B is roughly along the direction of the core, except in the air gaps.
  2. B is irrelevant to this problem (!!!), or at least mostly so.
  3. A better wording for "when the coil passes the air gap" is "as the coil passes through the air gap". It is the motion of the coil in the air gap that generates a voltage.
  4. The reason that the motion of the coil in the air gap generates a voltage is because as the coil moves through the air gap it cuts across the flux $\phi_3$, getting ever less flux as it moves. This is why the B field is made irrelevant -- all you need to know is that the coil is seeing all of $\phi_3$ when it's inside the core, and none when it's outside, and the change must generate some voltage.

You figure out the direction of B along the core using the right-hand rule. Curl the fingers of your right hand in the direction of current flow in the coil, and your thumb points in the direction of B.

\$\endgroup\$
  • \$\begingroup\$ Hi! Thank you for your answer, I'm still very confused but let's try to work on the excellent info you gave me: I understand that what matters is where the B that is felt by the coil varies. That happens when the coil travels through the air gap, feeling the B field there. The last part is what is confusing me. The right hand rule gives the normal vector of coil. However B must be in the opposite direction to cancel out the minus sign in the induction law (am I being clear?). \$\endgroup\$ – Granger Obliviate Nov 2 '18 at 11:47

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.