# Magnetic force on a charged particle doesn't affect total energy

So there's this sentence in a book I am reading. It states "When a charged particle is placed within a magnetic field, despite the magnetic force acting on the particle, there will be no change in the total energy content of the particle. As the force acts along the radius, it is a non-effective force. So, no work is done by the particle. Hence, the change in total energy content is zero."

• I'd really like to understand what this means. If someone could explain what they're trying to say here.
• My first thought was that this is nothing complicated and a simple iteration of the force times component of displacement theorem. Since the magnetic field acts perpendicular to the charge the force also acts perpendicular to the electric field. So there's no resultant work.
• Yes. You are correct. Since magnetic force can't do any work on the particle, it cannot change its energy, – Ari May 6 '16 at 5:44
• Situation is different for time dependent magnetic fields – jim May 7 '16 at 7:59

The magnetic force acting on a charged particle doesn't affect the particle's energy. Otherwise magnetic forces cannot do work. It's because the magnetic force equation is given by

$$\vec{F}=q\vec{v}\times\vec{B}$$

where $q$ is the charge, $\vec{v}$ is the velocity and $\vec{B}$ is the magnetic flux density.

So, it is clear that by virtue of the cross-product term, the magnetic force on the charge will be perpendicular to the plane containing the vectors $\vec{v}$ and $\vec{B}$. If the charge enters perpendicular to the field then the magnetic force will be maximum and it's magnitude will be $F=qvB$. This means the force acts in a direction perpendicular to the motion of the charge (or the velocity vector $\vec{v}$). This means the displacement and force are in different directions. So the work done by the magnetic force in displacing the charge will be zero, since a work is said to be done if displacement is in the same direction of the applied force. This means that the magnetic force has nothing to do with the speed of the particle since the displacement ($\vec{v}dt$) is not produced by the magnetic field. So the magnetic force cannot increase the speed (or kinetic energy) of the particle.

This can be proved as follows:

Let the charge covers a displacement of $\vec{v}dt$ in a time $dt$. Then the small work done is

$$dW=\vec{F}\cdot \vec{v}dt=q(\vec{v}\times\vec{B})\cdot \vec{v}dt=0$$

since $\vec{v}\times\vec{B}$ is perpendicular to $\vec{v}$ and hence their dot product vanishes ($\cos 90^0=0$). Hence magnetic force has done no work on the charge. this means the kinetic energy of the charge has not increased or decreased.

But the magnetic force could accelerate the charged particle by deflecting it's path, without affecting it's speed $v$. So the charged particle moves in a circular path. Remember that you cannot have uniform motion along a circular path. Even if the speed is a constant, the direction of velocity changes which causes an acceleration.

So, in effect, what the magnetic force do is just deflect the charge in a circular path and provide the necessary centripetal force corresponding to the field value. This is why the total energy of the charged particle is unaffected by a magnetic field. The things change if you had electric field.