Circular trajectory traveled by a charged particle in a magnetic field

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Homework Statement

Derive the general formula for the radius of the trajectory traveled by a charged particle in a magnetic field $\vec{B}$ perpendicular to its velocity $\vec{v}$.

Relevant Equations

Centripetal force: $F_{C}=\frac{mv^2}{r}$

The Lorentz's force acting on a charged particle perpendicularly "hitting" a magnetic field will be directed upwards, and generally directed towards the center of the circumference traveled by this particle, and so will cause a centripetal acceleration to keep it in a circular motion.
By equalling the centripetal force formula which causes this acceleration, to the Lorentz's force acting on the particle(considering $\theta = 90^{\circ}$), will give us:
$$qvB=\frac{mv^2}{r}\rightarrow r=\frac{mv}{qB}$$
Now, my question is: how do we treat negatively charged particles? By this formula, wouldn't their radiuses be $<0$, since $r\propto \frac{1}{q}$?
I can't understand why we don't use the absolute value of $q$, since(correct me if I'm wrong), even in case of a negative charge the trajectory doesn't change, apart from the Lorentz's force direction(which would be inverted).

 

[BvU]

apart from the Lorentz's force direction(which would be inverted).

Right ! So the trajectory curves in the opposite direction !

 

[BvU]

Bear in mind that a force is a vector: it has magnitude and a direction.
The expression for the Lorentz force is a vector equation and all three of $q$, $\vec v$ and $\vec B$ determine the direction. Your $r= {mv\over qB}$ is written as a scalar equation containing only magnitudes (and units).

 

[PeroK]

Homework Statement:: Derive the general formula for the radius of the trajectory traveled by a charged particle in a magnetic field $\vec{B}$ perpendicular to its velocity $\vec{v}$.
Relevant Equations:: Centripetal force: $F_{C}=\frac{mv^2}{r}$

The Lorentz's force acting on a charged particle perpendicularly "hitting" a magnetic field will be directed upwards, and generally directed towards the center of the circumference traveled by this particle, and so will cause a centripetal acceleration to keep it in a circular motion.
By equalling the centripetal force formula which causes this acceleration, to the Lorentz's force acting on the particle(considering $\theta = 90^{\circ}$), will give us:
$$qvB=\frac{mv^2}{r}\rightarrow r=\frac{mv}{qB}$$
Now, my question is: how do we treat negatively charged particles? By this formula, wouldn't their radiuses be $<0$, since $r\propto \frac{1}{q}$?
I can't understand why we don't use the absolute value of $q$, since(correct me if I'm wrong), even in case of a negative charge the trajectory doesn't change, apart from the Lorentz's force direction(which would be inverted).

Yes, it's the magnitude of the whole thing:$$r=\big | \frac{mv}{qB} \big |$$