# Magnetic Pressure in Solenoid

1. Jun 26, 2012

### Opus_723

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

First, I will describe the solenoid referenced, as it is from another problem.

The solenoid is made by winding a single layer of No. 14 copper wire on a cylindrical form 6cm in diameter. There are 5 turns per centimeter and the length of the solenoid is 30cm. Consulting wire tables, we find that No. 14 copper wire has a diameter of 0.163cm and a resistance of 0.010 ohms per meter at 75°C.

Now here is the problem I'm working on.

Suppose the solenoid described is used to make a very strong field for a very short time by connecting it briefly to a high-voltage source, to force an enormous current through it. Putting all other limitations aside, consider the limit imposed by the tensile strength of the wire. The magnetic pressure, like pressure in a hose, will tend to split the solenoid open. If the tensile strength of the wire is 2*10$^{9}$ dynes/cm$^{2}$, what is the strongest magnetic field that could be contained in that solenoid

The answer is given as 35 kilogauss.

2. Relevant equations

For the field along the axis of a solenoid: $\frac{2\pi*In}{10}$(cos$\theta_{1}-cos\theta_{2}$)

Force on a current-carrying wire: $\frac{I\vec{dl}\times\vec{B}}{10}$ = $\vec{dF}$

3. The attempt at a solution

First, I decided to express current in terms of the magnetic field, since current is not given.

I = $\frac{10B}{2\pi*n(cos\theta_{1}-cos\theta_{2})}$

I'm using the field along the axis of the solenoid, although I know it will be weaker at the rim where the wires are, because that's all I know how to do, and all that's ever been done in the book I'm using (Purcell).

Then I plugged in the above for I in the force equation, also noting that the current and magnetic field are perpendicular, assuming that the loops are perfectly circumferential (they're obviously not, but I checked the angle, and the error introduced to the cross product seems to be less than .2%). I also divided the expression by 2 since the force on the wire will be due to the average field acting on the wire. Since the field is approximately zero outside the solenoid, I used half the axial field.

$\frac{B^{2}(dl)}{4\pi*n(cos\theta_{1}-cos\theta_{2})}$ = dF

So this is the radial force acting on an infinitesimal length of wire (I think).

And then I have no clue what to do next. I've been trying to derive an expression for tension in terms of radial force on a loop, but I'm not making much progress. Am I totally approaching this wrong?

Last edited: Jun 26, 2012
2. Jun 26, 2012

### tiny-tim

Hi Opus_723!
(i haven't checked the rest of your work, but …)

to find the tension, do a free body diagram for a short arc of angle dθ …

there'll be two tension forces at either end, and the radial force (which you can treat as constant over the arc) in the middle

3. Jun 26, 2012

### Opus_723

Of course! Thanks. I'll try the numbers when I get home, but let me check the reasoning with you now.

Let's call the line perpendicular to the outward magnetic force and tangent to the loop in the plane of the loop the horizontal. The tension forces at each end of a segment of wire, due to the curvature of the wire, will make an angle with this horizontal. Geometrically, I think that this angle is half the angle subtended by the wire segment. By symmetry, the non-inward components cancel out when the tension at each end is added together. So the net force due to the tension is 2Tsin(dθ/2), where T is the tension force at one end. Since dθ = dl/r, this becomes 2Tsin(dl/2r). And since it is a very small angle, we can approximate this as $\frac{T*dl}{r}$. I approximate it so that when I set this expression equal to my outward force expression (Until the maximum tension is reached, the wire does not move, so the net force is zero) the infinitesimal dl will drop out.

Does that all seem good? Like I said, I'll check the numbers in a few hours to see if I have the right answer.

Last edited: Jun 26, 2012
4. Jun 27, 2012

### tiny-tim

(just got up :zzz:)

yes, that's fine!