Confused about how inductors work and produce a back EMF?

[lillybeans]

Homework Statement

A coil with zero resistance has its ends labeled a and b. The potential at a is higher than at b. Explain why the two following statements are consistent/correct with the situation.

a) The current is increasing and is directed from a to b.
b) The current is decreasing and is directed from b to a.

The Attempt at a Solution

The reason why I am confused is because I thought:

1. If the current is increasing from a to b, then the EMF induced should want to produce a current from b to a (opposite to the original direction of the current). In order to produce a current from b to a, you need the voltage at b to be higher potential than a, not a at a higher potential than b. So I don't see how the first statement is consistent with the situation.

2. A similar problem. If the current is decreasing and is directed from b to a, then the EMF induced would try to INCREASE/reinforce the current from b to a. That would still require b at a higher potential than a,not a at a higher potential than b as the problem suggested.

I'm quite sure I am misunderstanding something here, but I don't understand why my reasoning is wrong and why the textbook's reasoning is right. Please help!

Thanks.

 

[gneill]

Imagine the two cases cast in terms of typical circuits. For case (a), take a series connection of battery V, resistor R, and inductor L. At t=0 the battery is connected, perhaps by the closure of a switch. What's the initial current at t=0⁺) in this scenario? What then is the initial potential across the inductor?

For case (b), assume that an initial current is flowing from b to a in the inductor at time t=0⁺. The circuit contains a resistance R that will dissipate power, so the current will be decreasing as the energy stored in the inductor wanes. What will be the potential across the resistor and hence the inductor?

 

[lillybeans]

Imagine the two cases cast in terms of typical circuits. For case (a), take a series connection of battery V, resistor R, and inductor L. At t=0 the battery is connected, perhaps by the closure of a switch. What's the initial current at t=0⁺) in this scenario? What then is the initial potential across the inductor?

For case (b), assume that an initial current is flowing from b to a in the inductor at time t=0⁺. The circuit contains a resistance R that will dissipate power, so the current will be decreasing as the energy stored in the inductor wanes. What will be the potential across the resistor and hence the inductor?

Hi gneill,

Please check my reasoning. I even drew a diagram this time.

1. If the current were flowing from A to B and increasing, the inductor would try to reduce this current by producing a current in the opposite direction, I_L.
2. In order to produce this induced current from B to A, B needs to be at a higher potential than A (or else the current will flow from A to B).
3. This is not in agreement with the situation that the textbook suggests, which says Va>Vb.
4. The textbook also says that when the current is INCREASING, the EMF induced MUST always be opposite as the direction of the EMF of the battery. But in this case, if Va>Vb, then the two EMFs are in the SAME direction (as suggested by the polarities that I drew. The EMF's direction is that Va>Vb. and so is the induced EMF, how does that make sense?)

Thanks,

Lilly

 

[gneill]

Hi gneill,

Please check my reasoning. I even drew a diagram this time.

1. If the current were flowing from A to B and increasing, the inductor would try to reduce this current by producing a current in the opposite direction, I_L.
2. In order to produce this induced current from B to A, B needs to be at a higher potential than A (or else the current will flow from A to B).
3. This is not in agreement with the situation that the textbook suggests, which says Va>Vb.

Alternatively, the inductor produces an EMF that opposes the change, V = L*di/dt, which means it makes Va larger than Vb so that it tries to force current "back" through the circuit. That is, it raises its potential difference so that the external potential difference trying to increase current through the inductor is "bucked" by the higher potential across the inductor.

4. The textbook also says that when the current is INCREASING, the EMF induced MUST always be opposite as the direction of the EMF of the battery. But in this case, if Va>Vb, then the two EMFs are in the SAME direction (as suggested by the polarities that I drew. The EMF's direction is that Va>Vb. and so is the induced EMF, how does that make sense?)

That would be opposite the direction of the EMF of the battery if you do KVL around the loop. No fair drawing the inductor opposite the battery and declaring the polarity the same!

 

[lillybeans]

Alternatively, the inductor produces an EMF that opposes the change, V = L*di/dt, which means it makes Va larger than Vb so that it tries to force current "back" through the circuit. That is, it raises its potential difference so that the external potential difference trying to increase current through the inductor is "bucked" by the higher potential across the inductor.


That would be opposite the direction of the EMF of the battery if you do KVL around the loop. No fair drawing the inductor opposite the battery and declaring the polarity the same!

Aha, got it. Thank you so much!