Special relativity, circular motion

[Adorno]

Homework Statement

A charged particle (mass $m$, charge $q$) is moving with constant speed $v$. A magnetic field $\vec{B}$ is perpendicular to the velocity of the particle. Find the strength of the field required to hold the particle on a circular orbit of radius $R$.

Homework Equations

$\vec{F} = q\vec{v} \times \vec{B}$
$\vec{F} = m\vec{a}_c$

The Attempt at a Solution

Well, I know that in the "classical" case this is fairly easy. One just sets

$qvB = ma$,

and since $a = \frac{v^2}{R}$, one gets

$qvB = m \frac{v^2}{R}$
$\Rightarrow B = \frac{mv}{qR}$

However, I am not sure if I can use this here, because the particle is assumed to be traveling at close to the speed of light. I have read somewhere that I should use the relativistic mass in the calculation of the centripetal force, i.e.

$F = \frac{\gamma m v^2}{R}$,

but I am not sure why this is the case. Could anyone help?

 

[mathfeel]

One way to see this is go into the momentarily comoving frame (not an inertial frame, obviously) of the particle. In this frame, the particle is at rest so the Lorentz force only comes from the electric field in this frame, which is: $q E^{\prime} = \gamma q \mathbf{v} \times \mathbf{B}$
So it experiences a force which is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ with the $\gamma$ as promised.