# Bitter magnet calc.

1. Apr 12, 2010

### lamonster

I understand how to figure the magnetic field of an air core electromagnet, but apparently bitter magnets are slightly different. I've looked around and couldn't find equations on the site for the lab in Florida.
If anyone could help me out I would be grateful :D

2. Apr 12, 2010

### marcusl

If a single layer solenoid of inner radius r1 has n turns along its length, a typical thick solenoid of outer radius r2 is made of m layers of n turns each or m*n total turns. Each carries the same current, so the field is found by integrating the B field equation over the winding volume, using a constant current density.

A Bitter magnet uses n annular copper disks separated by insulating disks, instead of wires. The inner and outer radii r1 and r2 of the annulae are the same as before, that is, the number of turns along the solenoid length is the same but there's only one very thick winding instead of m layers of windings. Axial holes in the disks pass cooling water, cleverly solving one of the difficulties faced in building very high current solenoids. This is still an air core magnet (unless they've added iron) so the usual equations will work if you know the current density in the disks.

I've seen treatments of Bitter magnets in books on high field techniques, but I can't remember any titles right now. Sorry! We can probably figure it out, though. I'd expect the potential drop to be constant across the radius of each disk, in which case the current density is inversely proportional to the length of the path carrying the current. Since path length is $$2\pi r$$, I'd expect the current density to be inversely proportional to radius. You should be able to integrate the usual solenoid equations with that current density to get the field.

3. Apr 12, 2010

### lamonster

Well I'm looking to figure out a practical number of turns to achieve 5T. I have access to 200 amps, but could grab a spot welder from he junk yard or a surplus store if needed. You could plausably run anti freeze through as an alternative to water or air since I'm not trying to break a world record :P
Also, I've heard that at a certain amp peak you have to allow for a higher voltage. Anyone know if this Is true?

4. Apr 12, 2010

### marcusl

A 5T magnet is something of a beast, so please take the time to understand fully what you're doing. When energized, your magnet will resemble a small bomb and explosion is a real possibility. The insulation and joints are as important as the conductors. You'll need to design for strength and safety under the considerable Lorentz forces, and slowly ramp the current up and down rather than turn off and on. Do you know why?

Last edited: Apr 12, 2010
5. Apr 12, 2010

### lamonster

The forces Ill be dealing with are comparable to that of a propane tank, but it will get hot as well... rapidly. I don't want to be near a heat fractured bitter magnet pumping 5T. I'll make sure I have It looked over by my two buddies, one is a mechanical engineer and the other an electrician, beforehand.
You make a good point about slowly bringing it to full strength. Heat is going to be my biggest Issue, Other than finding a vacuum pump.... /cry

6. Apr 12, 2010

### marcusl

Why a vacuum pump?

Do you understand why you need to ramp the current? Do you understand dI/dt?

7. Apr 13, 2010

### lamonster

I'm making a verry large penning trap. The 5T field is just a good place to start. Ps what's di/dt

8. Apr 14, 2010

### marcusl

dI/dt is the current rate of change. The voltage developed across the windings is
$$V=-L \frac{dI}{dt}$$
which you need to control to avoid flashover or dielectric breakdown.

9. Apr 14, 2010

### lamonster

Well I'm using teflon washers that are the size of the conducting plates themselves. For the magnets I can't go much higher than 1kv with the resources at hand. Now for the penning trap itself... we'll see >_>. Honestly I knew everything had a dielectric breakdown, but never saw the formula for It.