$F = mv^2/r$ compared to $F=qBv$ how this two connected to each other


Well, since lorentz force is perpendicular to $\vec{B}$ and $\vec{v}$

                                                                        $\vec{F}=q\vec{v}\times \vec{B}$

This force is directed towards the center, and as it is the only force acting on the particle, it equals the centripetal force. Setting the modules equal (and also taking into account relativistic effects):


This is the basic idea behind the working of a cyclotrone, although you would still need to accelerate the particle through a difference in voltage to increase it's velocity, as centripetal acceleration just changes it's direction. The LHC is much more complicated (it is in fact a synchrotron, a kind of accelerator improved from the cyclotron for relativistic velocities and high energies) but the basic underlying principles are the same.


The formula derived by SaudiBombsYemen is often rewritten by accelerator and particle physicists as:

$$Br = \frac{P}{q}$$

where $P=\gamma m v$ is the particle momentum. This is indeed a very general result which applies in all the practical cases: from the mass spectrometers for isotope separation, to the particle tracks in the high energy detectors, to the bending magnets of the LHC.

  • 1
    $\begingroup$ Bringing one term of the Lorentz force vector cross product to the other side of the equation would be $$q_e \vec v_e = ( \vec B \times \vec F) / || \vec B|| ^2$$ and $$\vec B = (\vec F \times q_e \vec v_e ) / || q_e \vec v_e|| ^2$$ $\endgroup$ – HolgerFiedler Oct 13 '16 at 19:23
  • $\begingroup$ @HolgerFiedler when the dynamics is simple enough (uniform B field), so that the directions are obvious, you can forget about vectors and do all the related simplifications. Better not to add complexity where there is no need for it. $\endgroup$ – DarioP Oct 14 '16 at 7:37

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