The Large Hadron Collider (LHC) of circumference 27km uses magnetic fields to accelerate a proton repeatedly in a circular path. Calculate the flux density of a uniform magnetic field required for the proton to travel at a tenth of the speed of light.

Firstly, we must clearly set out the information we have. The particle in question is a proton, which has a mass of 1.67e-27 kg, and a charge of 1.6e-19 C. The path it takes has a circumference of 27000m, meaning the radius of its path is (27000/(2pi)), which is 4297m. The speed it is travelling at is c/10, or 3e7 m/s. The particle takes a circular path, meaning there must be a centripetal force acting on it, and this is given by F = (mv^2)/r. In addition the charged particle is moving through a magnetic field, which means it experiences a force perpendicular to its travel, given by F = Bqv, where B is the magnetic flux density, q is the charge of the particle, and v is the velocity it is travelling at. This is the only force that can provide the centripetal force required for the proton to maintain its path, meaning the above two equations must be equal: (mv^2)/r = Bqv. We want to find the value of B, so rearranging the above equation, we find that: B=(mv)/(rq) =(1.67e-27 * 3e7)/(4297 * 1.6e-19) = 7.29e-5 T