# Will increasing the strength of a magnetic field affect the circular motion of a charged particle?

Say you have a charged object in uniform circular motion and increase the magnitude of the magnetic field. By $$Fm = qvB\cdot \sin(t)$$, the $$Fm$$ should increase, as should the centripetal force. My question is, I know the $$Fc = mv^2/r$$ equation describes the minimum centripetal force required for uniform circular motion. So, will an increase in strength of the magnetic field induce a change in $$v$$ or $$r$$ (or does it merely allow for a higher $$v$$ or lower $$r$$)? Thanks!

• The problem is more complicated than that however, as the changing field will induce a circulating electric field (which can accelerate or decelerate the particle in it's orbit). Jan 28, 2021 at 20:51

If we were to simply consider 2 situations with the same initial conditions; and each has different magnetic fields strengths; the change would be in $$r$$, and not in $$v$$; since the magnetic field will not do work on the particle (this claim is often misused however in this situation it is warranted).
However if we just increase/decrease the magnetic field over time, due to Faraday's law we will induce an electric field corresponding to our increase/decrease and the motion of the charge. That Electric field will do work, and change the velocity of the particle, however that change in velocity will impact the magnetic force and so you would end up with a system of 2 Differential Equations. Computationally this is easy to solve, but analytically might be a little involved; I expect however that both $$v$$ and $$r$$ would be affected by this.
Of course after the transient change is complete, our system will return to a steady state orbit that is consistent with the field strength that we ended with, and the steady state velocity; but then again that steady state velocity depends on how the field strength was changed ($$\frac{dB}{dt}$$).