Understanding Inductance and Induced EMF in Simple Circuits

[Meow12]

Homework Statement

The inductor in the figure has inductance 0.260 H and carries a current in the direction shown that is decreasing at a uniform rate, $\displaystyle\frac{di}{dt}$ = -0.0180 A/s.
(a) Find the self-induced emf.
(b) Which end of the inductor, a or b, is at a higher potential?

Relevant Equations

$\displaystyle\mathcal E=-L\frac{di}{dt}$

(a) $\displaystyle\mathcal E=-L\frac{di}{dt}$ = -0.260 x -0.0180 = 0.00468 V = 4.68 mV

(b) Current always flows from higher potential to lower potential. So, b is at a higher potential.

My textbook says that my answer for part (a) is correct but part (b) is wrong. Please help!

 

[kuruman]

Think of the inductor as being ideal which means zero resistance. An inductor that has a steady current through it has zero potential difference across its ends. What will happen when you try to change this current from its steady value to a lower value? What does the negative sign in $\displaystyle\mathcal E=-L\frac{di}{dt}$ signify?

 

[Meow12]

What does the negative sign in $\displaystyle\mathcal E=-L\frac{di}{dt}$ signify?

It signifies that:
If $i$ is increasing, $\mathcal E$ is in the direction opposite to $i$.
If $i$ is decreasing, $\mathcal E$ is in the same direction as $i$.

 

[Delta2]

I have an explanation on why (a) should be at higher potential but I don't think it is intended for college or first year/second year university students. And yes it might be wrong lol, its gonna sound like "@rudeman thinking "in his insights. Let's wait to hear from @kuruman, maybe he puts us on the right track.

 

[DaveE]

Current always flows from higher potential to lower potential.

Nope, not always.

This is roughly analogous to a car moving from left to right with it's speed decreasing at .01 m/sec² because some one is pushing on it. Which side of the car is the guy pushing on?

Hint: the answer isn't that cars only move in the direction of the applied force.

 

[Delta2]

I think the key here is that the problem asks for (scalar) potential while the EMF isn't always the same as scalar potential difference.

 

[Delta2]

I think I now get the college level explanation on why (a) is at higher potential, but still I think it tell us half the truth and half a lie.
@Meow12
at the OP: Imagine that the inductor acts as a voltage source. Which side (a) or (b) should be positive (and the other negative) in order to tend to produce a current in the circuit that augments the decreasing current?

 

[Meow12]

@Meow12
at the OP: Imagine that the inductor acts as a voltage source. Which side (a) or (b) should be positive (and the other negative) in order to tend to produce a current in the circuit that augments the decreasing current?

Are you saying that the inductor acts like a battery? Inside a battery, current flows from lower potential to higher potential. (Is this correct?) So, a is at a higher potential.

 

[Delta2]

Are you saying that the inductor acts like a battery? Inside a battery, current flows from lower potential to higher potential. (Is this correct?) So, a is at a higher potential.

Yes exactly that. However I believe this is not exactly half a lie but for sure it is half the truth. I don't know if the moderators will allow me to present what I believe is the total complete truth.

 

[DaveE]

OK, let's just take a break with the analogies to cars and batteries and such. Analogies are always wrong, at least by just a little bit. Inductors aren't that complex, but they might be a new concept to you. We don't need a new theory that Orsted, Faraday, or Maxwell didn't figure out 2 centuries ago. I think you just need to study a bit more about this new, to you, concept. Khan academy is a good place to start, but the web is full of good choices for learning this if you prefer other versions.

In particular, I think your question is best answered by Lenz's law. This basically states that the polarity of the induced voltage will oppose any change in the magnetic field (or flux, or current, etc.).

 

[Delta2]

This basically states that the polarity of the induced voltage will oppose any change in the magnetic field (or flux, or current, etc.).

According to this and since the current is decreasing in order to try to increase the current the polarity of the induced voltage will be (b)+ (a)- because then the electric field resulting from this polarity points from b to a and augments the decreasing current. You have to view the inductor as battery and say that inside a battery current flows from lower potential to higher potential, there is no other way suitable for college level explanation.

 

[DaveE]

You have to view the inductor as battery

No, I don't, and you can't make me.

 

[Delta2]

No, I don't, and you can't make me.

Ok so when you say "The polarity of the induced voltage" what exactly do you mean by induced voltage? How this induced voltage augments a current if it doesnt act as some sort of battery?

 

[DaveE]

some sort of battery

Sorry, I don't know what you mean by this.

There is energy stored in the magnetic field which may come and go. But there is no chemistry involved; there are no electric dipoles involved. I completely fail to see the value of this analogy. Frankly, I think inductors are much simpler devices than batteries. Why not just study what inductors are/do. Inductors are like inductors.

 

[DaveE]

exactly do you mean by induced voltage?

https://phys.libretexts.org/Bookshe...:_Electromagnetic_Induction/13.03:_Lenz's_Law

 

[Delta2]

Ok can you explain with the most detail possible why a polarity (+) at (a) and (-) at (b) opposes the decrease in current (thats what Lenz law says right, we got to oppose the decrease in current).

 

[Charles Link]

I agree with @Delta2. This same thing came up previously a couple times. See https://www.physicsforums.com/threa...-voltage-across-inductor.880100/#post-5529838 post 25.

 

[Charles Link]

See also https://www.physicsforums.com/threa...duction-lecture-16.948122/page-6#post-6857043 post 189.

 

[PhDeezNutz]

(a) $\displaystyle\mathcal E=-L\frac{di}{dt}$ = -0.260 x -0.0180 = 0.00468 V = 4.68 mV

(b) Current always flows from higher potential to lower potential. So, b is at a higher potential.

My textbook says that my answer for part (a) is correct but part (b) is wrong. Please help!

$\frac{di}{dt} = -0.0180$ implies $i(t) = -0.0180t$ so current is always negative. If current is always negative then the current is actually flowing opposite to the way the arrow is indicated in the figure. So current is actually traveling from A to B. So A is at a higher potential.

@vela I see you are dismayed at my post. Sorry if I am incorrect. Can you point me in the right direction? Apologies if I mislead OP.

 

[Delta2]

The current is given that it flows in that direction, we don't solve a circuit here using Kirchoff's laws and assuming some arbitrary direction of currents.

 

[Charles Link]

Normally we're not supposed to supply answers in the homework forum, but since even the homework helpers aren't in complete agreement, let me propose one other explanation:

We can consider the inductor to be a voltage source with the inductor on the left in the circuit, kind of like a battery, with other resistors on the right with current running clockwise.
We then write $\mathcal{E}=-L \frac{di}{dt}=iR$.

We see with $\frac{di}{dt}$ negative that the voltage will be positive that gets delivered to the resistors.

That tells us the upper end of the inductor is positive.

We can alternatively treat the inductor as a component like a resistor or capacitor, and have a different voltage source $V$ in the circuit, and then we write $V=+L \frac{di}{dt}+iR$, and we see there will be a voltage drop across the inductor when $\frac{di}{dt}$ is positive, so that when $\frac{di}{dt}$ is negative, the more clockwise end of the inductor will be at the higher voltage.

 

[PhDeezNutz]

The current is given that it flows in that direction, we don't solve a circuit here using Kirchoff's laws and assuming some arbitrary direction of currents.

I see that now, also just because the derivative is negative doesn’t mean that the current is negative. Just that it’s decreasing.

Yikes I dun goofed.

 

[Delta2]

I see that now, also just because the derivative is negative doesn’t mean that the current is negative. Just that it’s decreasing.

Yikes I dun goofed.

Yes the accurate solution to that ODE is that $I=C-at$ and since it is given that the current flows in that direction and that it is decreasing we have to assume that the constant C is positive.

 

[kuruman]

It signifies that:
If $i$ is increasing, $\mathcal E$ is in the direction opposite to $i$.
If $i$ is decreasing, $\mathcal E$ is in the same direction as $i$.

No. Compare with the expression $F=-kx$ expressing the force exerted by a spring that is displaced from its stretched position by an amount $x$. The negative sign means that when the displacement is positive (to the right) the force is negative (to the left) and when displacement is negative (to the left) the force is positive (to the right). One concludes that what is on the left-hand side (force) always opposes what is on the right-hand side (displacement).

Here, we have $\frac{di}{dt}$ on the right hand side not $i$. One concludes that the induced emf $\mathcal{E}$ always opposes the change in current $\frac{di}{dt}$. How does it do that? By having a polarity such that it would generate an induced current $i_{ind}$ that would oppose the change. That means and induced current that would flow from b to a.

The question now is which point must be at higher potential so that we get that induced current. Look at the picture below, left. It shows a closed circuit with current flowing from b to a. What generates this current is hidden. However, it is clear that point a is at higher potential than b otherwise the current in the resistor would not be flowing left to right. If you are allowed to see what is hidden, it might be the inductor that we have here (middle picture) or a battery (right picture.)

 

[jbriggs444]

(b) Current always flows from higher potential to lower potential. So, b is at a higher potential.

Current across an ideal resistor always flows from higher potential to lower potential.

Ohm's formula for current across an ideal resistor tells us this. The equation for an ideal inductor does not say the same thing.

 

[Steve4Physics]

Maybe this pulls together various points already made.

We have an ideal inductor - there is no resistance. For example, imagine it’s a superconducting coil.

There are no ohmic losses in an ideal inductor - whatever current flows. So with a steady current, the p.d. between $a$ and $b$ is zero .

If $i$ is steadily reduced, there is an induced emf generated by the changing magnetic field; this is the only source of p.d. between points $a$ and $b$.

Applying Lenz’s law, the induced emf is in the direction which tends to increase $i$. Hence $a$ is at a higher potential than $b$ as nicely explained/shown in @kuruman’s Post #24.

 

[hutchphd]

this is the only source of p.d. between points a and b.

I would implore everyone to be more careful about nomenclature here because the term "potential difference" seems to inevitably lead to its misappropriation. One can define the emf, and a path integral that has units of volts, but it that does not mean one should talk about it as ordinary "voltage", and indeed attempts to do so lead to comprehensive misunderstandings because the value is dependent upon the path as skillfully reiterated by MIT prof. Walter Lewin. Attempts to twist the pedagogy to admit to such a definition, while perhaps well intentioned, are both counterproductive and foolish IMHO. There are no batteries in the coil etc etc etc etc etc. So include the emf in your Kirchhoff sum and never forget that the schematic drawing in your lab notebook is a crude representation of a much more complicated system of fields and imprecise terminology makes us lazy to our detriment.

 

[DaveE]

Y'all need to build a flyback or boost SMPS, or maybe fiddle with the ignition coil on a 1965 Mustang. This polarity question will then become quite obvious. The EEs that made the power supply for whatever device you using right now don't argue so much about arcane physical descriptions. This may be why we had to take a lab course in college.

 

[Steve4Physics]

this is the only source of p.d. between points a and b.

I would implore everyone to be more careful about nomenclature here because the term "potential difference" seems to inevitably lead to its misappropriation. One can define the emf, and a path integral that has units of volts, but it that does not mean one should talk about it as ordinary "voltage", and indeed attempts to do so lead to comprehensive misunderstandings because the value is dependent upon the path as skillfully reiterated by MIT prof. Walter Lewin. Attempts to twist the pedagogy to admit to such a definition, while perhaps well intentioned, are both counterproductive and foolish IMHO. There are no batteries in the coil etc etc etc etc etc. So include the emf in your Kirchhoff sum and never forget that the schematic drawing in your lab notebook is a crude representation of a much more complicated system of fields and imprecise terminology makes us lazy to our detriment.

Not quite sure if you’re telling me off or not!

I think I used ‘p.d.’ appropriately because the original question is specifically about the relative potentials of points a and b. I avoided use of 'voltage'.

I’m not really sure how else I could have phrased it (with appropriate brevity/clarity).

 

[hutchphd]

No I was trying to not turn the spotlight on you. I explicitly said "everyone" because of several previous threads (some referenced here in this one) have been remarkable for the attempts to define "split fields" and a host of other devices to enable the use of term "voltage" without falling into the rathole. I was hoping to simply keep this discussion firmly within the direct view of Maxwell's equations. The use of the term "potential" is fraught. Consider it a pre-emptive strike. My apologies for any collateral damage.

 

[TSny]

[The following is for fun and to check my understanding. If inappropriate in any way, please delete.]

Joline and Tim are considering a circuit that consists of an inductor of inductance $L$ and a load resistor of resistance $R$. Unlike an ideal inductor, the coils of the inductor in this circuit have a non-negligible total resistance $R_C$. At time $t = 0$, there is a current $I_0$ in the direction shown and the current is decreasing. They want to work out the current as a function of time.

The two are arguing about which point $a$ or $b$ is “at the higher potential”. Joline claims $a$ must be at the higher potential to drive current through the load resistor $R$ in the given direction. Tim disagrees, saying that $b$ must be at the higher potential to push current through the resistance $R_c$ of the inductor in the given direction.

They decide to visit their professor to find out who is right. Professor L. smiles and says, “You two have a conundrum because you both believe that ’potential at a point of this circuit' has meaning. It doesn’t. You don’t need to consider potential at all. Electric field is what pushes the charge carriers. Just use Faraday’s law!” $$\oint \mathbf{E}\cdot \mathbf{ds} = -\frac{d \Phi}{dt}.$$
Joline and Tim agree to try this. For the closed path of the integral of the electric field, they choose a path that goes once around the circuit in the direction of the current and the path stays within the conducting material of the circuit. (So, the path through the inductor follows the corkscrew windings of the inductor.) They see that the rate of change of magnetic flux through this path of integration can be expressed as $\large \frac{d \Phi}{dt} = L\frac{dI}{dt}$. Assuming that the hookup wire connecting the inductor to the load resistor has negligible resistance, they write Faraday’s law as $$\oint \mathbf{E} \cdot \mathbf{dl} = \left(\int \mathbf{E}\cdot \mathbf{ds}\right)_{\rm through \, load}+\left(\int \mathbf{E}\cdot \mathbf{ds}\right)_{\rm through \, inductor} = -L\frac{dI}{dt}.$$ From Ohm’s law, the integrals through the load and the inductor can be written as $IR$ and $IR_C$, respectively. Making these substitutions and rearranging, Joline and Tim agree that the differential equation that determines the current as a function of time is $$\frac {dI}{dt} = -\frac{R + R_C}{L} I.$$ From this they derive easily the current as a function of time.

 

[hutchphd]

Looks good to me. No muss, no fuss, no auxiliary fields or magic wires : only Maxwell (Faraday closed loop integral version) required. A required bedtime story for all budding physicists....and EE's

 

[Delta2]

So no one wants to hear from me the full treatment of this I have in mind using Maxwell's equation's the differential version of Ohm's law and concepts like scalar potential, vector potential, conservative and non conservative fields?

 

[Charles Link]

I think @TSny 's is a good question, but it is a trick question. You need to ask, how did it achieve these initial conditions? It appears it would have been necessary to run a (e.g. a DC) voltage source from "b" to "a" without the resistor $R$ in the circuit, in order to establish the current in the inductor. Then at time t=0, the voltage source is removed and the resistor attached. Upon doing so, an EMF occurs in the inductor from "b" to "a" to try to maintain the current, and so from then on "a" is at the higher voltage.

 

[hutchphd]

So no one wants to hear from me the full treatment of this I have in mind using Maxwell's equation's the differential version of Ohm's law and concepts like scalar potential, vector potential, conservative and non conservative fields?

Finite size wires? Non-ideal inductor? Non-uniform currents? Relativistic delays? There are no easy closed form exact solutions. You need to be judiciously approximate.....which means quantifying each assumption.