Forces

Taking a closer look at LHC

CENTRIPETAL  FORCE

The necessary force on each proton travelling at almost the speed of light will be:

Con v~c :    Fc = mc2/r      Fc = E/r

To generate this force there are 1232 magnetic dipoles located on the eight archs, each one having a magnetic length of 14,3 m, giving a total implied length:

1232 × 14,3 = 17618 m

Precisely, we can calculate the so-called "bending radius":

rb =17618/(2π)   ⇒  rb =  2804 m

 

So,

Fc = 7 TeV /2804 m     Fc = 1.12·10-6J /2804 m

Fc = 4·10-10 N
on each proton.


ACTION  AND  REACTION

But the accelerator will suffer the same force from each proton.

Considering the two beams with 2808 bunches and the number of protons inside (1,15·1011):

FT  = 2 · 2808 × 1,15·1011 × 4·10-10

FT    260000 N ≈ 26 Tm_Force

So the reaction force over the curved part of the accelerator will be equivalent to 26 tonnes of force.

And the most incredible is that this force is created in interaction with 6·1014 protons with a rest mass of only 1 ng (one nanogram!).

An other more impresive force acts over the beam pipe because the superconductive currents. which is explained in Magnetic Dipoles.

The designers have taken these forcee into account because very high precision is required.

AUTHORS


Xabier Cid Vidal, PhD in experimental Particle Physics for Santiago University (USC). Research Fellow in experimental Particle Physics at CERN from January 2013 to Decembre 2015. Currently, he is in USC Particle Physics Department (Spanish Postdoctoral Junior Grants Programme).

Ramon Cid Manzano, secondary school Physics Teacher at IES de SAR (Santiago - Spain), and part-time Lecturer (Profesor Asociado) in Faculty of Education at the University of Santiago (Spain). He has a Degree in Physics and in Chemistry, and is PhD for Santiago University (USC).

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© Xabier Cid Vidal & Ramon Cid - rcid@lhc-closer.es  | SANTIAGO (SPAIN) | Template based on the design of the CERN website

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