Induced current in a wire loop embedded in a changing B-field

[jjson775]

Homework Statement

Find the current through section PQ of length a = 65 cm as shown in the attached picture. The circuit is located in a magnetic field whose magnitude varies with time according to the expression B = (1.00 x 10 ^-3 T/s)t. Assume that the resistance per length of wire is 0.100 ohms/m.

Relevant Equations

E = d/dt BA cos theta

 

[Delta2]

I think there is an arithmetical mistake in the 5th line (I simply don't get $\frac{1}{5.13}+\frac{1}{1.54}$ if I put $a=0.65m$), but there is something more important:

Why do you calculate the total R in that way and not as follows
$$R=R'_{eq}+0.1(3a)$$
$$\frac{1}{R'_{eq}}=\frac{1}{0.1(5a)}+\frac{1}{0.1a}$$

You didn't give explicitly an equivalent lumped circuit model, but I think that model you have in your mind and in which you base the calculation of total R in that way, is wrong. And the above way is also wrong I believe.

I have a different lumped model in my mind where we have two EMF sources, one in the left loop (of total surface area $2a^2$) and of value $E_1=-10^{-3}2a^2$ and one in the right loop (of total surface area $a^2$ ) and of value $E_2=-10^{-3}a^2$ . What do you think?

 

[haruspex]

The difficulty I see in approaching the problem is that opposing emfs in PQ due to the circuits either side effectively reduce the resistance of PQ.
So I represented PQ as two resistors, one in each loop. Their resistances are so as to make the potential difference across PQ the same in each circuit, and consistent with the net current through PQ.
Necessarily, this will make one of these virtual resistances negative.
Writing V for $\dot B.a^2$, this gave me a current from Q to P of $I=\frac V{23\Omega}$. Correspondingly, the current through the right hand three resistors is 8I and that through the left hand three is 9I.
Edit: I may have all current directions reversed, but the magnitudes should be right.

Edit 2: I forgot the resistances are only $0.1\Omega$, so I should have written of $I=\frac V{2.3\Omega}$.

 

[Delta2]

No @haruspex the polarity of the two EMF sources is such that if we omit the segment PQ , then in the total loop the two EMFs will add up to give a total EMF of $E=10^{-3}3a^2$. That is the two EMFs are in series in the large loop.

P.S OK, sorry it seems i blundered, the two EMFs create opposite currents in PQ.

 

[jjson775]

I think there is an arithmetical mistake in the 5th line (I simply don't get $\frac{1}{5.13}+\frac{1}{1.54}$ if I put $a=0.65m$), but there is something more important:

Why do you calculate the total R in that way and not as follows
$$R=R'_{eq}+0.1(3a)$$
$$\frac{1}{R'_{eq}}=\frac{1}{0.1(5a)}+\frac{1}{0.1a}$$

You didn't give explicitly an equivalent lumped circuit model, but I think that model you have in your mind and in which you base the calculation of total R in that way, is wrong. And the above way is also wrong I believe.

I have a different lumped model in my mind where we have two EMF sources, one in the left loop (of total surface area $2a^2$) and of value $E_1=-10^{-3}2a^2$ and one in the right loop (of total surface area $a^2$ ) and of value $E_2=-10^{-3}a^2$ . What do you think?

As you saw, I treated the model as a “lumped” circuit. The square on the right consists of 2 resistors in parallel, one of .1(.65) and one of .1(3*.65) ohms. The remainder of the circuit on the left consists of 5 resistors in series, 5*.1*.65. The current in the circuit is continuous at a particular time. I did look at the alternate of two separate magnetic fields and EMF’s with Ipq being due to the difference between the oppositely directed EMF’s. I didn’t get the right answer (although I may have made a mistake) and can’t reconcile the idea of separate currents.

 

[haruspex]

As you saw, I treated the model as a “lumped” circuit. The square on the right consists of 2 resistors in parallel, one of .1(.65) and one of .1(3*.65) ohms. The remainder of the circuit on the left consists of 5 resistors in series, 5*.1*.65. The current in the circuit is continuous at a particular time. I did look at the alternate of two separate magnetic fields and EMF’s with Ipq being due to the difference between the oppositely directed EMF’s. I didn’t get the right answer (although I may have made a mistake)

But your method in post #1 is clearly incorrect since it ignores the field in the square on the right.
Do you know what the answer is? If so, does it match my resultin post #3?

can’t reconcile the idea of separate currents.

I take that to refer to post #3.
Ignore resistance for the moment. We can consider the effective PD around each loop - rectangle on left, square on right - separately. This considers two copies of PQ, one in each of two unconnected circuits. We would then be able to merge the two copies of PQ, summing the two currents and two PDs along it. In each case they would partially cancel in the same way, leading to a consistent result.

When we try to reintroduce resistance, the same method would overtax the flows in PQ. It would impede currents that aren't there because they cancelled.
To resolve that, I allowed the two copies to have lower resistances. In particular, because the actual flow in PQ is opposite to what results from only considering the square, the resistance of the right hand copy would be negative.

Once you accept that model, it is a matter of writing down the usual equations and solving.

HOWEVER, I can see another way that gives a different result.
Considering each loop separately, there is a voltage 2V around the left circuit, spread over six sections length, so V/3 across PQ. On the right, V/4 on each section.
Since these oppose on PQ, the net PD along PQ is V/12.

Edit: I very much doubt the validity of that second approach since it seems to give a result that violates conservation of current if we change the resistance in one of the sections length a.

 

[jjson775]

But your method in post #1 is clearly incorrect since it ignores the field in the square on the right.
I believe I did include the field in the square on the right because I used 3a^2 for A in the formula.

Do you know what the answer is? If so, does it match my resultin post #3?
The answer is 283 x 10^-6 A. Ba^2 = 1 x 10^-3x(.65^2) = 4.225 x 10^-4= your V so your answer would be V/23 = 18x 10^-6.

Thanks for your patience, I still don’t understand why my attempt didn’t work.

.

 

[haruspex]

I did include the field in the square on the right because I used 3a^2 for A in the formula.

Yes, I should have written that your work in post #1 ignores the fact that PQ is to the left of the square. It treats PQ as another path around immediately adjacent to the three other members of the square.

I have made a correction to post #3 leading to V/2.3, so now it is the right order of magnitude, but still below the given answer.

 

[haruspex]

I'm trying another tack.

In principle, the changing field exerts an emf along each section of wire independently. I have always considered magnetic field lines to be neither created nor destroyed. If the field through a loop increases then field lines migrate into the loop from an infinite reserve outside. The emf results from these lines crossing the wire.
But there is a puzzle here. How do we know the rate at which field lines cross any given bit of wire? We only know the rate at which the number inside the loop increases.
I believe this doesn't matter. We can take any point in the plane as a fixed field line and have all others migrate towards it. The difference between choosing one fixed point and another is equivalent to shifting a field of constant intensity relative to the loop. The emfs generated cancel, giving rise to no current.

With that in mind, we can pick a convenient fixed point and for each section of wire consider:
- the emf generated by the field
- the current in the section
- the 'back emf' representing the resistance to that current
- 'forces' (emfs) from adjoining sections
The net force at any junction should be zero, likewise the net current.
The remaining equation is power in = power dissipated.

In the hope of gaining some reassurance regarding algebraic correctness from the symmetry of the equations, I addressed a general two node three path circuit.
Three paths run from A to B. The current, applied emf and resistance of path i are $I_i, E_i, R_i$.
I got the following equations:
$\Sigma I_i=0$
$\Sigma E_i=\Sigma R_iI_i$
$\Sigma I_i^2R_i=\Sigma E_iI_i$.

Unfortunately, solving that system even for three paths becomes extremely messy. (For more paths we would need more equations. Maybe it becomes indeterminate.)

As regards a convenient fixed point, P or Q look good. If I can solve the above equations I'll try two fixed points to prove the choice does not matter.

 

[jjson775]

Your analysis is over my head. The problem comes from Serway Beichner Physics for Scientists and Engineers. The book categorizes problems as straightforward, intermediate or challenging. Believe it or not, this one is categorized as straightforward!
If you remove wire PQ, the EMF induced is -1.268x10^-3V. If you replace it, all you do is add resistance to the circuit, which reduces the current. Is that right?
I have been playing with the numbers trying to back into the answer and will keep trying. I am afraid I am missing something very straightforward and basic. Thanks again for your support.

 

[Steve4Physics]

Using @Delta’s 'lumped model' approach (Post #2) we can treat the system as having 2 sources of emf – one generated by the loop left of PQ and the other generated by the loop right of PQ.

|Emf induced in left loop| = A|$\frac {dB}{dt}$| = 2*0.65² * 0.001 = 0.000845 V
|Emf induced in right loop| = A|$\frac {dB}{dt}$| = 0.65² * 0.001 = 0.0004225 V

Applying Lenz’s law + right hand rule, the emfs’ polarities are as shown below.

Let R be the resistance of a 65cm length of wire (R = 0.065 Ω). We can represent the circuit as follows:

Kirchhoff’s 2nd law, left loop: 0.000845 + 0.065I₂ + 5*0.065I₁ = 0
Kirchhoff’s 2nd law, right loop: 0.0004225 + 3*0.065(I₁ – I₂) - 0.065I₂ = 0

Doing a bit of algebra and using an online ‘solver’ for simultaneous equations gives
I₂ = -0.000283A (so I guessed the direction of I₂ wrongly).

 

[haruspex]

Using @Delta’s 'lumped model' approach (Post #2) we can treat the system as having 2 sources of emf – one generated by the loop left of PQ and the other generated by the loop right of PQ.

|Emf induced in left loop| = A|$\frac {dB}{dt}$| = 2*0.65² * 0.001 = 0.000845 V
|Emf induced in right loop| = A|$\frac {dB}{dt}$| = 0.65² * 0.001 = 0.0004225 V

Applying Lenz’s law + right hand rule, the emfs’ polarities are as shown below.

Let R be the resistance of a 65cm length of wire (R = 0.065 Ω). We can represent the circuit as follows:
View attachment 288204
Kirchhoff’s 2nd law, left loop: 0.000845 + 0.065I₂ + 5*0.065I₁ = 0
Kirchhoff’s 2nd law, right loop: 0.0004225 + 3*0.065(I₁ – I₂) - 0.065I₂ = 0

Doing a bit of algebra and using an online ‘solver’ for simultaneous equations gives
I₂ = -0.000283A (so I guessed the direction of I₂ wrongly).

I believe my analysis in post #9, choosing P or Q as the fixed field point, is completely equivalent to that. Where I struggled was finding enough equations. There must be a better choice than the quadratic power equation.

 

[Steve4Physics]

I believe my analysis in post #9, choosing P or Q as the fixed field point, is completely equivalent to that. Where I struggled was finding enough equations. There must be a better choice than the quadratic power equation.

Our analyses are certainly equivalent in part, because they both directly use Kirchhoff's laws.

But, like you, I don't see how it is possible to "know the rate at which field lines cross any given bit of wire"; so I did not determine the emfs that way. Also, I did not need to use power. The problem was easily reduced to solving 2 simultaneous linear equations.