# Relationship between flux, current and voltage

1. Aug 3, 2011

### rohitchaky

Consider the following system - A coil is wrapped around a soft iron core as shown below. A permanent magnet is placed at one end of the coil.
Code (Text):

[FONT="Courier New"]
+--+-----+--+
|   |        |   |
Coil +ve Side -->   |   |        |   | --> Coil -ve Side
|   |        |   |
+--+-----+--+

+-----+
|        | --> Permanent Magnet
+-----+
[/FONT]
Consider the magnet being moved from right of the screen (initial position) to the left of the screen. As the magnet moves towards the perfect alignment with the core (lets call it center), the flux linkage of the coil increases.
-- Since there is a rate of change of flux linkage, I have voltage induced in the coil
-- Assuming the circuit to be complete, I get a current flowing through the coil whose value is current = flux linkage / coil inductance

As the magnet moves away from the center, the flux linkage decreases. However the magnitude of flux linkage is positive
-- From the previous equation, current is still positive since flux linkage is positive
-- Since the rate of change of flux linkage is negative, voltage polarity is reversed

In the former case if I had assumed the polarity to be +ve, I have a situation where I get -ve voltage and + current

Is this the reason why we say in an inductor current lags the voltage?

If to close the circuit I had a simple resistor then how do I explain the presence of -ve power in this case? I do not know how to account for the negative power. Can someone please explain this to me.

Last edited by a moderator: Aug 3, 2011
2. Aug 3, 2011

### MisterX

Your equation for current is not correct in this case. For the total flux linkage to be proportional to the coil current, there would have to be no other source of flux besides the coil current. In this case, the flux through the coil would be the sum of the flux from the coil current and the flux from the permanent magnet.

Current lags (sinusoidal) voltage for an uncoupled inductor because the flux linkage is proportional to the inductor current, the voltage is proportional to the time derivative of flux linkage (thus proprotional to the time derivative of inductor current), and the time derivative of a sine wave is that sine wave shifted in the -t direction by pi/2 radians.

Last edited: Aug 3, 2011
3. Aug 3, 2011

### sophiecentaur

Could you be confusing the effect of self inductance for a coil and the voltage induced in an inductor due to a changing external magnetic flux? The current through your load resistor (R) would just be Vi/R. (where Vi is the induced voltage in the coil) The power would be Vi2/R, which would, of course, always be positive.
The value of Vi would depend on dφ/dt, which will be positive - then negative, as you say.

4. Aug 3, 2011

### rohitchaky

Yes I was confusing myself with the two. But even in this case, until the direction of current changes, isn't there a -ve power?

5. Aug 4, 2011

### rohitchaky

So since this is an uncoupled inductor, current lags correct?

Thanks for the info regarding relationship between flux and current.

6. Aug 4, 2011

### sophiecentaur

To start, there's +V and +I then there's -V and -I. In both cases the product will be positive. ( in the same way that vsquared is always positive.

The current through the resistor has to be in phase with the volts across it. How. Would it 'know' where the volts were coming from at any time? This can be confusing, I know.

7. Aug 4, 2011

### MisterX

Not neccessarily, because of the changing component of flux linkage from another source. I could have made that more clear. For the hypothetical situation you described, the total flux linkage is not proportional to the current, so the reasoning I supplied for why current lags voltage for an uncoupled inductor is not valid.

And, as sophiecentaur has posted, current and voltage will always be in phase for a resistor in circuit theory.

8. Aug 4, 2011

### rohitchaky

thanks MisterX and sophiecentaur