sPHENIX discussion of the superconducting solenoid

[Kin Yip <kinyip AT bnl.gov>]

Hi Dave,

Below is what Uncle Meng has told/taught me ... which reminded me of what you said more than one time, ie. the magnetic forces don't change even if you flip the polarity.
I now think/understand, since force is the vector product of  I current (or electron velocity in the wire)  and B-field, when you flip the polarity, you  only
change both signs (but not magnitude) of I and B (in Biot-Savart Law)  and    -B*-I = B*I  => same force (magnitude and direction).

I like the story of the magnetized steel.    I'll go to test Uncle Meng's  "prediction"  after the High-Field Test   :-) 

Kin

-------- Forwarded Message -------- Subject: RE: polarity Date: Thu, 22 Jun 2017 15:55:39 -0400 From: Meng, Wuzheng <meng AT bnl.gov> To: Yip, Kin <kinyip AT bnl.gov>

Kin,

 

If you change the polarity of the current, you actually reverse the direction of the flux lines; but the forces calculated  will remain the same: magnitude and direction.

 

If you care about tiny changes due to the magnetizations on the steel part, there will be some residual fields even when you turn the current to zero: you may test after you power down the coil to see the steel can attract small iron piece (a paper clip, or a nail). If we ignore this effect, then everything will be the same regardless the polarity changes. I guess if you ask old folks from STAR or PHENIX experiments, they might tell you that after they flip the polarity, they needed to increase the operation current by little bit to compensate such residual field effect (produced by the previous polarity).

 

When I first time did the simulations, I just pick one polarity in sPhenix. In old PHENIX, I was told all the coils should be in the same polarity (meaning, to produce the same direction in Bz, in central, north and south muon magnets). So, yes, randomly.

 

Wuzheng

 

 

From: Kin Yip [mailto:kinyip AT bnl.gov]
Sent: Thursday, June 22, 2017 2:57 PM
To: Meng, Wuzheng
Subject: polarity

 

Hi,

Just an question without doing calculation/simulation ...

For the Magnet (either in High-Field or final sPHENIX configuration), if we change the polarity of the current, I guess the magnetic field strength or the forces on the coil
(etc.) won't be identical due to our lead-end and non-lead-end asymmetry.

But apart from the signs, do you expect that the differences in magnitude (of forces or magetici field strength) will be small ?

In fact, how did you choose the polarity when you first did the simulation  ??    If you have seen it in some literature, could you please share with me ?

When they asked me this question, I found a page (attached) in the Power Supply section which says Lead A => +ve.   So I could tell them what to set.   But
I wonder how you assign the polarity ??   Randomly ??

Kin