# Effect of changing magnetic field on charge velocity

Consider a charged particle with velocity $v$ moving in a loop of radius $r$ about a magnetic field $B$ pointing in the $\hat{z}$ direction. I'm trying to figure out how the velocity of the particle will change if $B$ increases.

I haven't done E&M for a little bit now, so I'm a little rusty. I know that the particle will move clockwise in the $xy$ plane looking from above due to Lenz's Law. So if $B$ increases, shouldn't the velocity of the particle increase to counter?

I'm struggling to find the equations to show this. I want to use $\mathcal{E}=-\frac{d\Phi}{dt}=-\pi r^2\frac{dB}{dt}$, and $\mathcal{E}=(2\pi r)Bv$. But wouldn't this imply that $\frac{dB}{dt}\propto-v$, so the velocity would decrease? This doesn't make sense to me.

• The velocity in the Lenz's law, is the velocity of the closed loop in the magnetic field and not the charged particle. Anyway, it depends on whether the particle is moving clockwise or anticlockwise in the wire. There will be emf due to charged particle and the another emf will be induced due to the change in magnetic flux through the wire. In your case, the wire isn't moving but the magnetic field strength is changing and that changes the magnetic flux ($\phi=\vec{B}.\vec{A}$). The net of these should give you the resultant emf. – Mitchell Apr 29 '17 at 17:47

I'm going to assume that the (positively charged) particle was initially given a velocity in the x-y plane, that is at right angles to the field. Provided the magnetic flux density (field strength) isn't changing rapidly, the particle will move in an almost circular path. The sense of revolution will be opposite to that in which an ordinary (right-handed) screw would have to be turned in order to advance in the +z direction (the field direction). I think we're in agreement here, except that I'd attribute this sense of motion to Fleming's Left hand rule, or if you prefer, to the magnetic part of the Lorentz force, $\vec {F}= q \ \vec{v} \times \vec{B}$.