# Flux density expressions at solenoid centre

1. Oct 9, 2016

### moenste

1. The problem statement, all variables and given/known data
A long air-cored solenoid has two windings wound on top of each other. Each has N turns per metre and resistance R. Deduce expressions for the flux density at the centre of the selenoid when the windings are connected (a) in series, and (b) in parallel, to a battery of EMF E and of negligible internal resistance. (In each case the magnetic fields produced by the currents in the two windings reinforce.)

2. The attempt at a solution
Maybe the formula for the flux density should be B = (μ0 N I) / (2 r), where N is the number of turns on the coil, I is current and r is the radius of the coil?

Though I don't quite understand what is being given. Something like this that has two metal cables wound around it? How can they be connected in series or in parallel?

2. Oct 9, 2016

### Jonathan Scott

Firstly, in the question as set N is not the number of turns (which is the usual input to the flux formula); it is the number of turns per metre, so it is already in the form of number per unit length. Take care how you use the flux formula; the radius is not given and is not relevant.

To determine the current in each of the two possible configurations (a) and (b), each coil can be considered as a resistor with resistance R as mentioned in the question.

3. Oct 9, 2016

### moenste

So the formula is B = μ0 n I?

We have in series:

And in parallel:

4. Oct 9, 2016

### Jonathan Scott

It's N (upper case) rather than n (lower case) in the question as you gave it, but that would be fine for each of the separate windings.

Now you need to calculate the current in terms of E and R each of the two cases and then work out the answers noting that the flux also involves two windings, not just one.

5. Oct 9, 2016

### moenste

B = (μ0 N1 I1) + (μ0 N2 I2)

I = E / R

B = (μ0 N1 (E1 / R1)) + (μ0 N2 (E2 / R2))

?

6. Oct 9, 2016

### Jonathan Scott

You seem to have the general idea, but you need to think for yourself, so I'm not going to sort out the details for you. Work out the current in each of the two cases (serial and parallel) in terms of the values given in the question then work out the total flux in each case to give the required answers.

7. Oct 9, 2016

### moenste

So for the series part we have: B = μ0 (N1 * ((E - V2) / R1) + N2 * ((E - V1) / R2)).

And for the parallel part: B = μ0 (N1 * (E / RParallel) + N2 * (E / RParallel)), where RParallel is: 1 / RParallel = 1 / R1 + 1 / R2.

8. Oct 9, 2016

### Jonathan Scott

You're probably making it unnecessarily general and confusing for yourself by using all those -1 and -2 subscripts and splitting it all up. If you're going to use symbols which aren't in the question, you should write down exactly what each one means. In this case, I think it's simplest just to write down the answer to part (a) step by step using the original variables, then do the same for part (b). Be careful about the current in each winding in the parallel case, as each winding only gets half of it, which I think you've overlooked.

9. Oct 9, 2016

### moenste

I don't see a way of presenting it otherwise...

1 means winding one and 2 means winding two.

I think this should be it:
(a) B = μ0 (N1 * ((E - V2) / R1) + N2 * ((E - V1) / R2)), where V are the voltages of the respective resistors.

(b) B = μ0 (N1 * ((E / RParallel) - (E / R1)) + N2 * (E / RParallel) - (E / R2)), where RParallel is: 1 / RParallel = 1 / R1 + 1 / R2.

10. Oct 9, 2016

### Jonathan Scott

If the problem is as you've stated it, the windings are identical. For the serial case, the current through both is given by the total EMF over the total resistance. For the parallel case, the current through each one is given by the same total EMF over the resistance of that separate winding; although you could use the parallel resistance formula and then split up the result again, that doesn't seem very useful in this case.

In the problem as stated, the only quantities are N, R and E so the answer needs to be given in terms of those quantities. You may define additional working variables if really necessary, but if so you must explain what they are, and they should not appear in the final answer.

11. Oct 9, 2016

### moenste

So, in series: B = 2 * μ0 * N * (E / R), two because we have two resistors and in parallel: B = μ0 * N * (E / R1 + E / R2).

12. Oct 9, 2016

### Jonathan Scott

You're not being very careful. The total resistance in the series case is not R. And you've got references to R1 and R2 which do not appear in the question.

13. Oct 10, 2016

### moenste

I don't see what's wrong with the series one.

I did a graph in my notebook and since we have two identical resistors then their voltage and resistance is identical or half the given one (E / 2 and R / 2).

And so B = μ0 * N * ((E / 2) / (R / 2)) for one resistor. And so in total we have B = 2 * μ0 * N * (E / R).

And in parallel: B = 2 * μ0 * N * (4E / R). The voltage is the same but the current is half of it and as well as resistance. So we have I / 2 = E / (R / 2) which gives us I = 4E / R.

14. Oct 10, 2016

### Jonathan Scott

The problem statement says that each winding has resistance R, not that the total resistance is R.

You still have two factor-of-two mistakes in the parallel case. I don't fully understand what you're trying to do, but you seem to be assuming all of the current will flow through both windings, but as the windings are parallel, some of the current flows through each, and as I already mentioned it is easiest to calculate the current for each one separately then add up the flux for the two windings afterwards.

In general, you also need to write out the working in smaller steps, so that each step is clear without having to guess intermediate steps, and so that if you make a mistake it will be easy to spot it. You also need to be careful to make sure your assumptions are the same as those in the original problem statement. I think you should at least state the current in each winding as an intermediate step.

I've always found it useful for simple questions involving voltage, current, resistance and so on to use a mental model based on water flow. Voltage (potential) is equivalent to pressure, which can be represented by the relative height within a water system operated by gravity, or by a pump with a particular pressure for a battery. Current is the amount of water flowing. Resistance is like constrictions in pipes, so parallel constricted pipes have less resistance.

(For dynamic circuits, a capacitor is like a rubber membrane across a pipe, allowing alternating flow but resisting direct flow, and an inductor is like a heavy paddle-wheel or turbine in the flow, which tends to resist changes in current. However, those are only loose analogies and it's probably better to try to get used to the mathematics for describing the electronic properties of the relevant components).

15. Oct 10, 2016

### moenste

In series:
Current is the same in this circuit. Each resistor has R resistance. Each resistor has E / 2 voltage.

General formula for the field is: B = μ0 n I. We know that n = N per resistor. We don't know I.

I = V / R, where V = E / 2 and R = R. So I = (E / 2) / R = E / 2 R.

So B = μ0 * N * (E / 2 R) + μ0 * N * (E / 2 R).

In parallel:
In this circuit the voltage is the same but the current is half of that of the circuit. Therefore I / 2 = E / R → I = 2 E / R.

So B = μ0 * N * (2 E / R) + μ0 * N * (2 E / R).

16. Oct 10, 2016

### Jonathan Scott

It's not actually a resistor; it's a coil winding which happens to have a specific resistance for purposes of calculating the current.

In the standard formula, n is usually the number of turns and there's usually a separate parameter for the length, so for this version of the general formula it might be helpful to add "where n is the number of turns per unit length".

For the final answer to part (a), you should simplify that expression. Also, when writing out your answers you should refer to (a) and (b) to identify the parts as in the original question.

No, this is still confused.

The easiest way to get the current is to consider the windings separately. Each has voltage E across it and resistance R.

A more complicated way is to calculate the total current for the pair of windings, treated as parallel resistors, then to assign half of the current to each one, which cancels out the factors of 2 to get the same as the first method. You've somehow gone the wrong way with one of your factors of two.

And again, for the final answer to part (b), you should simplify the final expression.

17. Oct 10, 2016

### moenste

So (a) B = 2 * μ0 * N * (E / 2 R).

I'm sorry, I don't quite understand what's wrong.

I = 30 A
I1 = 15 A
I2 = 15 A

E = 300 V

RTotal = 10 Ohm
R1 = 20 Ohm
R2 = 20 Ohm

My formula: I / 2 = E / R1 → 30 / 2 = 300 / 20 and also I = 2 E / R1 → 30 = 2 * 300 / 20.

B = 2 * μ0 * N * (2 E / R)

Last edited: Oct 10, 2016
18. Oct 10, 2016

### Jonathan Scott

You should still be able to simplify that some more!

In the parallel case, each separate winding has EMF E across it and resistance R, so what's the current through it? It couldn't be simpler.

If you want to calculate the total current, that's twice the current through each winding, not half of it.

19. Oct 10, 2016

### moenste

Maybe be like: (a) B = 2 * μ0 * N * (E / 2 R) = (2 * μ0 * N * E) / (2 R) = (μ0 * N * E) / R.

I = E / R?

(b) B = (2 * μ0 * N * E) / R?

20. Oct 10, 2016

### Jonathan Scott

I think your answers look right now, although I don't see a need for the "multiply" symbols or the brackets.

You need to be more methodical and careful with your working. Break it down into small steps where you can clearly explain it, as if you were explaining it to convince someone else like yourself. That will help you get it right as well as showing that you understand the subject. It may also help to imagine the questions that the other person might have, to which you need good answers. When you get to the final answer, you should be confident that you're right. If it's a person marking it (and not just multiple choice) then you also need to show your working so that if you make a mistake you can still get partial credit, and the person can also spot exactly where you went wrong.

And always check that you have read and answered the actual question correctly and completely, using the same terms and values as those given in the question.