# Intuition for why the motion of a charged particle in a uniform magnetic field is circular

Suppose a positive point charge is moving in a plane and there is a uniform magnetic field directed into the plane. My textbook (Young and Freedman) claims that because the Lorentz force exerted on the point charge is always perpendicular to its motion, it follows a circular path (or helical in the general case). The textbook continues on with that assumption and uses centripetal acceleration to derive the radius of curvature.

I understand that mathematically, Newton's laws give equations of motion that are consistent with circular (helical) motion (Helical motion of charged particle in external magnetic field). However, why is that the case?

In my understanding, the fact that the Lorentz force is perpendicular to motion should only guarantee a constant speed curvilinear motion. For it to be a circular motion, the provided centripetal force ($$q\vec{v}\times\vec{B}$$) must exactly equal the required centripetal force ($$mv^2/r$$) which is not immediately obvious to me why that should be the case.

• That match is satisfied for one value of r, isn't it? Apr 8, 2023 at 0:24
• Have you tried to solve Newton's equations in the presence of Lorentz force? What else is there to understand? Apr 8, 2023 at 0:33
• hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movchg.html#c1 Apr 8, 2023 at 4:57

You have quite correctly said that:

the fact that the Lorentz force is perpendicular to motion should only guarantee a constant speed curvilinear motion

But if the speed is constant the Lorentz force is constant because it is proportional to the speed. That means the particle is moving with a constant centripetal force and therefore its path has a constant curvature. And the only way to move with constant curvature (in 2D) is to move in a circle.

• Why does constant centripetal force mean constant curvature? Doesn't it just mean that $\ddot r-r\dot\theta^2$ is constant? Apr 9, 2023 at 5:58

The Magnetic force given by qv x B is (by definition of vector product) perpendicular to v and B, hence, if the Force is perpendicular the acceleration is also perpendicular. In sum, the only motion that satisfies this kind of relation between a and v is the circular and helical motion.

Not a bad question! Mostly mathy in answer though.

# Recharacterizing circles

So a circle is defined in the upper half-plane by the curve $$y(x) = \sqrt{R^2 - x^2}$$ And since for small k, $$\sqrt{1+k}\approx 1+\frac12 k,$$ we have that for small $$x$$ a circle can be approximated by the parabola $$y(x) \approx R - \frac{x^2}{2R}.$$ Now you can invert your perspective on all of this, you can say that a curve has a radius of curvature given by $$[y''(x)]^{-1}$$ and that circles are precisely those planar curves (they must be stuck in a 2D plane) which keep a constant radius of curvature (caveat: also not flip-flopping in whether it curves clockwise then anticlockwise in the plane, but you could say that the radius of curvature becomes undefined at such points).

Essentially, the radius of curvature tells us, if we want to approximate a curve with a little segment of circle, how big should that circle be.

# Perpendicular acceleration

If we now graduate to $$x(t), y(t)$$ we have parametric motion in a plane, without loss of generality we can choose coordinate axes such that at some time $$t_0$$ the particle happens to be travelling in just the $$x$$-direction, $$\dot x(t_0)=v, ~~\dot y(t_0) =0.$$ If $$\ddot x = a_\parallel$$ and $$\ddot y=a_\bot$$ are nontrivial then we can expand the motion about $$t=t_0$$ as approximately, $$x(t) =x_0+ v~(t-t_0) + \frac12 a_\parallel (t-t_0)^2,\\ y(t) = y_0+ \frac12 a_\bot (t-t_0)^2.$$ One can do a quadratic formula for $$t-t_0$$ in terms of $$x-x_0$$ then substitute into the $$y$$-equation and take two derivatives to find, I believe, $$y''(x) = {a_\bot ~ v\over (v^2 + 2 a_\parallel(x-x_0))^{3/2}},$$ so when we look for the radius of curvature at $$x=x_0$$ the tangential acceleration $$a_\parallel$$ is irrelevant and only the perpendicular component $$a_\bot$$ remains, moreover we find, $$R = \frac{v^3}{v~a_\bot} = \frac {v^2}{a_\bot}.$$ So this familiar formula for circular motion, if you are careful to apply it ONLY to the perpendicular acceleration component $$a_\bot$$, actually can be used in reverse to calculate a radius of curvature for a parametric path at some time $$t_0$$.

# Assembling the argument

The argument is now the previous two sections put together:

1. If $$\mathbf B$$ is in the $$z$$-direction and $$v_z=0$$ is an initial condition, we never generate a $$z$$-component of acceleration and motion remains confined to the $$xy$$-plane.

2. There is never a tangential acceleration from a magnetic force, the cross product forces it to be perpendicular to velocity, so the velocity maintains an overall constant magnitude.

3. As a result $$v^2$$ is constant and $$a_\bot$$ is constant, so the radius of curvature is constant.

4. This finally implies that the motion is circular.

The statement that a moving subatomic particle in a magnetic field describes a circular motion - or helical motion for the case of the angle between the direction of motion and the magnetic field being less than 90° and greater than 0° - is not precise.

1. the deflection is demonstrably accompanied by the emission of electromagnetic radiation.
2. the emission of energy can only come from the kinetic energy of the particle. This is because the magnetic field can come from a permanent magnet, and its field is proven not to weaken with time.
3. because of the reduction of the kinetic energy with further photon emission, the particle describes a spiral path and comes to a standstill in its centre.

Point 1 deserves special attention, because if one considers the photon emission as the cause of the deflection (the Lorentz force), the search for a centripetal force is no longer necessary.