# Inducing a current

1. Apr 9, 2012

### jsmith613

Is an induced current (conventional current) in the same direction or in the opposite direction to the induced emf.
I ask this in relation to electromagnetic induction. We can predict the direction of the current using the right hand rule but how do I know the direction of emf?

Another question, related to this, is how could an EMF be induced (and hence a current be induced) in a strip of wire, such as an aerial, if it is not in a complete circuit? Surely the whole point of EM induction is to oppose the change in magnetic flux. If a current cannot be induced, how is this done? For example when a car moves forward, an EMF is induced across the ends of the aerial BUT there is no complete circuit

Thanks alot guys!!

2. Apr 9, 2012

### DragonPetter

Hello,

To your first question: First, when you say direction of induced emf, do you mean polarity of the emf voltage?

Think of it this way, the emf reduces the amount of current flowing, or in the perspective of current, the decreasing change in current creates an increase in the amount of emf.

The equation for induction is:

$EMF = -\frac{d\Phi}{dt}$

The minus sign shows that EMF opposes the change in flux.

In circuits with inductors, this looks like a voltage drop as the current passes through the inductor, so the current flows into the EMF's positive terminal, and out its negative terminal.

I hope that helps and makes sense.

3. Apr 9, 2012

### jsmith613

Well I presume I mean polarity of EMF voltage although I am not quite sure of the difference :S

for the second paragraph, surely the induced EMF induces a current. if I take a strip of wire no current initially flows, so if I move it in a field and an EMF is induced a current HAS to flow (it cannot be reduced)

4. Apr 9, 2012

### sophiecentaur

That's right. The Right hand rule tells you the direction of the EMF, which works even with an open circuit. The current direction follows from the EMF and the value depends on the load it drives..

5. Apr 9, 2012

### DragonPetter

What happens when current flows in an open wire? Do you think you can compare this to an antenna?

6. Apr 9, 2012

### sophiecentaur

My point is that the EMF is what is induced. The Current is a secondary issue and depends upon the Impedance of the load. No current will flow in a short length of unconnected wire but there will be an EMF.
The theory of how an antenna couples to an EM wave is more involved because there is a varying E field as well as an H field. In the case of simple Magnetic induction, any E field is ignored because we aren't dealing with free space conditions.

7. Apr 9, 2012

### jsmith613

well my first question really was to they flow in the same direction.
maybe I have misunderstood emf and that it doesn't flow?

8. Apr 9, 2012

### sophiecentaur

It's a good thing to get the terminology right in things like this.
An EMF doesn't "flow"; it is a Potential Difference and exists 'across' to parts of a circuit. A current will flow as a result of an applied EMF.
PD is like a difference in heights. Things can fall from one height to another - like a current can flow.

9. Apr 9, 2012

### DragonPetter

Another thing to consider in the equation I first wrote out is that there is no current term. Magnetic flux can be dependent on a current, but it is not always the case. EMF is directly dependent on magnetic flux.

10. Apr 9, 2012

### jsmith613

well surely one end of the emf is +/- and the other end 0
so lets say + is on the left and 0 is on the right. the emf will go from + to 0 (left to right).
if this is a closed wire, in which direction wil the current flow?

11. Apr 9, 2012

### DragonPetter

The same way current flows with any other + to - voltage source.

12. Apr 9, 2012

### jsmith613

so from - to +

therefore, the direction of the induced emf can be worked out using the LEFT hand rule, right?

13. Apr 9, 2012

### sophiecentaur

How do you come to that conclusion?
EMF=−dΦdt and
I=EMF/Resistance
so where does your other sign change come from?
If you connect a battery, which way will the current flow? What's different for an induced emf?

14. Apr 9, 2012

### FOIWATER

induced voltages result in current flow that opposes the current flow that created the field that induced the voltage.

This is the reason inductors exhibit impedance. Apply a voltage across an inductor, current flows through it, which creates a magnetic field around it. Assuming AC voltage and that field varies with time depending on frequency, the magnetic field expands and collapses over time. Each time current passing through it increases, the field increases, each time the current collapses, the magnetic field collapses. the collapsing of the field has the opposite effect that the initial current-produced magnetic field did. It induces a voltage back into your inductor, this causes current to flow which bucks the source current. This is the nature of impedance in an inductor, and why it does not account for actual power loss as well, it just stores energy in a magnetic field.

This phenomena also accounts for CEMF inside an electric motor. The reason why motors draw more current bases on load, is because of the fact EMF is induced opposing source voltage. At slower speeds, less EMF is induced that bucks the source voltage, hence why a motor draws more current when loaded.

Research lenz' law

Right hand rule gives direction of current flow based on field direction and relative motion direction.

Left hand motor (lorentz' force) rule, right hand (faraday's induction) generator rule. Notice only your two middle fingers will oppose one another, which accounts for generator action in all motors, and all loading phenomena

15. Apr 9, 2012

### FOIWATER

basically, it depends how the magnetic field interacts with the conductor that is having a voltage induced in it.

If the field is expanding from the source, and collapsing on the conductor, the voltage is the opposite direction.

For a static field and relative motion of the conductor, use the right hand rule.

16. Apr 9, 2012

### sophiecentaur

Absolutely BUT the 'back emf' is never greater than the initial induced EMF. Add the original EMF to the EMF generated by the induced current and it will always produce a current that is less it would have been but still in the 'right' direction . If you look upon it as being analogous to the force of reaction against you when you are pushing a massive object along, that reaction force is never great enough to move the object towards you.

17. Apr 9, 2012

### BruceW

$$\oint \vec{E} \cdot d \vec{l} = - \int \frac{d \vec{B}}{dt} \cdot d \vec{A}$$
This is the equation for induction, just written out in more detail. On the right-hand side of the equation is an area integral, where the direction of the area vector is the normal to the surface. And on the left-hand side is a line integral over a closed path bounding the same area.

The convention in physics is that the line integral goes anti-clockwise around the area which it bounds. So from this, you can work out the direction of the electric field.

18. Apr 9, 2012

### jsmith613

well then as there is no change in sign I would presume therefore that EMF and current are in the same direction

19. Apr 9, 2012

### sophiecentaur

Right. (Thank god - or nothing would work. :surprised)

20. Apr 9, 2012

### FOIWATER

yeah i totally agree, there has to be power lost during the transfer of energy, as well all the flux that the current created can not be linked, much is leaked.

I know you know this. :D