# Electromagnetic Induction

1. If switch S' is closed and kept closed, the end of the core that is market W becomes:

I answered "a North pole temporarily." I used the right hand to rule to get North-pole, which is correct in the answers, but got the second part wrong. I stated temporarily, while the answer states permanently. My explanation is that it's temporary because there is not a change in magnetic field once the switch is closed, that would induce a current.

The textbook's explanation is as follows: "The coil on the left becomes an electromagnet. Conventional current goes from plus to minus outside the battery. Grasp the coil with the right hand so that the fingers point in the direction of conventional current; the outstretched thumb points in the direction of the North pole: in this case, W."

Which is correct?

1. The instant after switch S' is closed, the conventional current in wire YZ will be from...

I realised that the magnetic field at around YZ is facing in the direction roughly of the vector YZ (as field lines go from North to South). However, I was not able to find a direction of current that would oppose this change in magnetic field.

Could an explanation be offered for both questions?

• For the second question, The textbook's explanation is as follows: A changing current in coil X induces a current in the other coil. Current in coil X builds up slowly to its maximum value because of the large self-induced emf in coil X. Until this maximum value is reached, current is induced in the second coil in a direction opposing the current buildup in coil X. For this to occur, the left end of the second coil should be a South pole, which will tend to weaken the left magnet. Using the right hand rule, conventional current will flow from Y to Z. – StopReadingThisUsername Sep 19 '16 at 14:40

1 As long as current is flowing in the left hand coil end $W$ will be a north pole.
I think that you are confusing induced currents which are cause by a changing magnetic flux and continuous currents produced by a battery?

2 when the switch is closed the magnetic field in the left land soft iron core increases with a north pole on the left and a south pole on the right.

This is just like moving the south pole of a magnet closer to the left hand side of the right hand coil.

The induced current in the right hand coil will oppose that by forming a south pole at the left hand end of the right hand coil.

Looking at the polarities of the two coils means that the currents in those two coils are in opposite directions.

So the induced current will flow from $Y$ to $Z$.

Your textbook is right in both cases. As the first circuit has a battery. The current will flow permanently. Following Biot-Savart's law you get a permanent electromagnet. Note, a theoretical battery can provide current until the end of time. Your direction with the right hand rule is correct, though.

As the field was zero before the switch was closed, you will get a change of flux in the second coil. What happens there is described by Lenz' law. A current will flow, hence, to counteract the field of the first coil, i.e. south pole to the left. This is virtually done by a battery where $+$ and $-$ are opposite the the battery shown left, i.e. current will flow from $Y$ to $Z$. So in principle you are right again. The wire $Y-Z$ if in the plane of field lines, so no induction there. The important point is, that it is part of the circuit including the coil. And your right hand rule of that coil producing a south pole on the left results in the current as described by the textbook.