# Homework Help: Question about strength of magnetic field down coil (solenoid)

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1. Jul 20, 2016

### RoboNerd

1. The problem statement, all variables and given/known data

Hi everyone.

I was reading through a book and came across the following question and explanation:
"
You are given two 200 meter strands of identical copper wire. With one strand you create a coil whose radius is 2 cm. With the second strand you create a 4 cm coil. Assuming the current is the same in both, which coil will have the greater B field down its axis?

Solution: It turns out that the magnetic field expression for a coil is µoin, where µo is a constant and n is the number of turns per unit length in the coil. You might expect that B would have something to do with the coil's radius. After all, a bigger radius would mean more distance between the wire in the coil and the axis. But a bigger radius also means more length of wire per loop for the current to pass through, and the two parameters counteract one another. In any case, the magnetic fields should be the same.
"
I have tried to figure out what they were saying in this solution, but I was not able to figure i out.
I will explain what I tried to do in my attempt

2. Relevant equations
Since a coil of wire is basically a solenoid, what we have is ampere's law and its usage in deriving an equation for a solenoid.

The strength of a magnetic field down a solenoid is:

B = mu0 * (Number of loops total / total solenoid LENGTH) * length of solenoid section * I

3. The attempt at a solution

Simplifying the equation, I examine the entire length of the solenoid, so I have

B = mu0 * Number of total loops * length of ENTIRE solenoid * current

Now, since I have one coil of 2 cm and another coil of 4cm, I will naturally have a greater number of loops "N" for the coil of 2 cm radius, so the B must be larger for that, is it not?

Then could someone please explain how I am mistaken and how I might recognize the mistake.

Thanks in advance for your input.

2. Jul 20, 2016

### Charles Link

The equation for the magnetic field in a solenoid is $B=\mu_o n I$ where $n=N/L$. $N$ is the number of loops and $L$ is the solenoid length. (A couple of your equations above where it talks about the length of a solenoid section are incorrect.) The magnetic field $B$ points along the axis of the solenoid everywhere inside the solenoid and is quite uniform except near the ends where the value decreases somewhat. The question really needs to specify the relative lengths of the 2cm and 4cm solenoids. Are they the same length, so that N/L for one is 2x the other, or is one twice the length of the other so that the N/L are the same? The problem doesn't seem to be clear on this. (I'm of course assuming N for the 2cm solenoid is 2x that of the 4cm solenoid.)

Last edited: Jul 20, 2016
3. Jul 20, 2016

### Staff: Mentor

Correct.
Not so. The information provided is that the solenoids are wound using identical wire, meaning turns/metre will be the same for each (assuming close-wound with wire of fixed gauge). As for whether one solenoid is longer than the other, it doesn't matter because length does not appear in the formula for B. ( I believe we can infer that the fatter coil will be shorter by half, however, this is just not relevant.)

4. Jul 20, 2016

### Staff: Mentor

Correct.
length of solenoid section does not belong in the equation for B.

5. Jul 21, 2016

### Charles Link

@NascentOxygen Please take another look at this. This problem is elementary. The relevant parameter is $n=N/L$ and this can be adjusted by how the solenoid is wound. If you keep them both the same length $L$, all you need to do is make it a double layer and you will have twice the magnetic field. The wire needs to be insulated in any case. A double layer if the wire is thin really will not change the r=2cm appreciably.

6. Jul 21, 2016

### Staff: Mentor

In the problem statement, there is nothing to suggest other than closewound, single-layer coils, so N/L is fixed here.

7. Jul 21, 2016

### Charles Link

Many times solenoids have multiple layers. There is also nothing to suggest that they wound it on two spools/non-magnetic canisters of different lengths. I guess we are going to have to agree to disagree.

8. Jul 22, 2016

### RoboNerd

Why do you say N/L is fixed? How can you prove it is a fixed ratio, provided that I can change the radiuses of the loops and with that the total length of the solenoid?

9. Jul 22, 2016

### Charles Link

The claim by NascentOxygen is that the solenoid is made with a single layer of wire that is wound as closely spaced as possible. Thereby the N/L is constant and determined by the width of the wire. This is certainly one possibility, If this is the case, I think it needs to be spelled out in the statement of the problem (For this case the magnetic fields are of course the same for both solenoids). I don't think in this problem there is any disagreement with the physics that will result. If a double layer is used for the 2cm solenoid with both solenoids the same length, I think the agreement is unanimous that the magnetic field B for the 2cm solenoid will be 2x that of the 4cm solenoid. ..editing...Meanwhile, I think it even makes it a better homework problem if both of the above cases are considered.

Last edited: Jul 22, 2016
10. Jul 24, 2016

### RoboNerd

But why would you say this is so for sure? Are there any formal arguments that you can make to support this claim?

11. Jul 24, 2016

### Charles Link

In that sentence, I am only repeating what NascentOxygen is claiming. If you read my postings carefully, you will see I very much disagree with NascentOxygen's interpretation of the question. Please read the complete listing of postings thoroughly. I believe any questions that you may have, have been addressed.

Last edited: Jul 24, 2016
12. Aug 2, 2016

### RoboNerd

Thank you everyone for your inputs regarding this problem!

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