The 4.25 km radius tunnel that is used to house the magnets for the Large Hadron

Question

The 4.25 km radius tunnel that is used to house the magnets for the Large Hadron Collider (LHC) calls for proton beams of energy around 3.9 TeV. What magnetic field is required (in T)? (Hint: This proton is moving so close to the speed of light that its kinetic energy is not equal to (1/2)mv2. Instead, you have to treat the proton with special relativity so that p=mv=E/c. This is kind of like a massless photon in which the momentum is related to the energy by p=E/c.) Please show steps with answer!

Explanation / Answer

=E/C

p=(3.9*1012/3*108 )*1.602*10-19

p=1.3*104*1.602*10-19

r=p/qB

4250=(1.3*104*1.602*10-19)/(1.602*10-19*B)

B=3.05T

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