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Why did we need a 8.3 T magnetic field in the LHC beam pipe?

  1. Nov 10, 2015 #1
    I tried to work backwards and verify the magnetic field required by the LHC beam pipes. A circumference of 27km implies a radius R=4300 m. The beam energy E is 7 TeV per proton beam. Considering approximation E approximately equal to pc to get momentum (ultra-relativistic case). Plugging into p=qBr, I am getting a magnetic field of 5.4T as opposed to 8.3T. Clearly, I am doing something wrong. Could someone please point out where I am faltering? Thank you very much.
     
  2. jcsd
  3. Nov 10, 2015 #2

    mfb

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    The LHC tunnel is not a perfect circle. It was built for LEP, where long straight sections are needed for electron/positron acceleration (to counter synchrotron radiation losses). And even in the curves, the dipole field is not everywhere as you need quadrupoles and a few more elements in the ring.
     
  4. Nov 10, 2015 #3
    Thank you so much for your reply. :) Actually, I was wondering more along the lines of why 8.3, it seems like a weird number. Is there some calculation after which we arrive at this figure?
     
  5. Nov 10, 2015 #4

    mfb

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    Count the number of dipole magnets in the ring (1232), look up their effective length (the field is not present at full strength along the whole length), take the product as effective circumference, and you should get some number close to 8.3 T. Even with the full length of 15 meters per magnet you get a reasonable agreement.
     
  6. Nov 10, 2015 #5
    Thank you so much!!
     
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