- #1
greg_rack
Gold Member
- 363
- 79
- 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).
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).