# Magnetic force doesn't do work and therefore can't change the KE of a particle?

I am having a real hard time understanding the principles behind a question I've come across during test prep. Here is the question:

Q: A proton and an electron are traveling in the uniform magnetic field with identical velocities. If the movement of both particles is perpendicular to the magnetic field lines, is the following true?

The proton will experience a greater kinetic energy change than the electron.

On one hand I think if there is a $\mathrm{KE}$ change for the particles, they would have to be both changed equally, since an input of force to each particle would be converted into $\mathrm{KE}$. On the other hand, since $\mathrm{KE} = \frac{1}{2}mv^2$, and both particles are traveling at the same velocity, it seems the proton's much large mass would mean it has a greater $\mathrm{KE}$ change.

However, neither of my thoughts were on target, as the answer reads as follows:

False, because the magnetic force does not do work, so it cannot change the kinetic energy of either particle.

How does this make sense? Won't the particles change velocity when exposed to the magnetic field, and therefore change $\mathrm{KE}$?

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You basically just need to be careful about the distinction between velocity and speed. In particular, you say that

Won't the particles change velocity when exposed to the magnetic field, and therefore change KE?

A change in velocity is not necessarily accompanied by a change in speed, and it's the speed that determines the kinetic energy. The magnetic field can change the directions of the motions of the charged particles, but it will not change their speeds.

The mathematical details of this are as follows. The force on a charged particle of charge $q$ moving in a magnetic field $\mathbf B$ is $$\mathbf F = q\mathbf v\times\mathbf B$$ where $\mathbf v$ is the velocity of the particle. Now, note that Newton's second law says that the force on the particle equals the rate of change of its momentum; $\mathbf F = \dot{\mathbf p}$. So we get $$\dot{\mathbf p} = q\mathbf v\times\mathbf B$$ Now take the dot product of both sides with $\mathbf p$. Note that the (non-relativistic) momentum of the particle is $\mathbf p = m\mathbf v$, so when we take the dot product of the right hand side with $\mathbf p$, we get zero. Putting this all together gives $\mathbf p \cdot\dot{\mathbf p} = 0$ which implies that $$\frac{d}{dt}\frac{\mathbf p^2}{2m} = 0$$ In other words, the kinetic energy is constant in time.

The details of the argument above assumed the non-relativistic expression for the momentum of a mass particle. However, the argument still carries through in a relativistic context. To see this, note that the relativistic expression for the momentum of a particle is $$\mathbf p = \gamma m\mathbf v, \qquad \gamma = (1-\mathbf v^2/c^2)^{-1/2}$$ and using this definition, Newton's second law can in this context still be written as $\mathbf F = \dot{ \mathbf p}$, so the equation of motion for a massive particle moving in a magnetic field is still $\dot {\mathbf p} = q\mathbf v\times \mathbf B$. Moreover, since the velocity and momentum are still proportional to one another, dotting both sides of this equation of motion still gives $\mathbf p\cdot\dot{\mathbf p} = 0$ and therefore that $$\frac{d}{dt}\mathbf p^2 = 0$$ Now recall that the kinetic energy of a relativistic particle is given by its total energy $E=\gamma mc^2$ minus its rest energy $mc^2$; $$K = E - mc^2$$ In particular, the time derivative of its total energy equals the time derivative of its rest energy; $\dot K = \dot E$. On the other hand, recall the relation $$m^2c^4 = E^2 - c^2\mathbf p^2$$ Combining this relation, and the fact that the time derivatives of the kinetic and total energies are equal, we find that the desired result $$\dot K = 0$$ holds in a relativistic context as well.