# Does a force perpendicular to the direction of movement not do work?

I have read that since the magnetic force acting on moving charge is always perpendicular to the velocity of moving charge, the field is not doing any work on the charge.

However, if a ball rolling on the X axis collides with a heavier ball rolling on the Y axis, the first ball will get some of the kinetic energy of the second ball, even though their velocities are perpendicular. (At least, that's what I assume that happens. I imagine a truck hitting a person crossing the road).

Why can kinetic energy transfer to the ball in the second example, but the charges from the first example cannot get any (even though both cases deal with perpendicular forces)? Is energy transferred in the second case because the collision is not instant, and we could say that the second ball first "rotates" the first ball and only then "pushes" it in the direction of its velocity? If so, can there be an instant collision where the heavier object only gets to rotate the lighter object?

• Also note that a force against the direction of movement does not do work either. For example, if you compress a spring, the spring pushes you back with a certain force, but the spring does not do work, you do. However, when you release the pressure and allow the spring uncompress, it will push you with a force and do work, because now its force is in the direction of movement. – safesphere Oct 9 '17 at 4:17

First, there certainly could be collisions where no energy is transferred, only direction changes. But you're right, we could construct a case where the initial contact is perpendicular, but energy is transferred.

Is energy transferred in the second case because the collision is not instant, and we could say that the second ball first "rotates" the first ball and only then "pushes" it in the direction of its velocity?

Correct. In a large collision, there's a significant change in velocity during the interaction. But the angle of the delivered force remains mostly the same. That means it cannot be perpendicular to the velocity during the entire time.

If so, can there be an instant collision where the heavier object only gets to rotate the lighter object?

Not if the collision is large. In the limit where the collision is small, the momentum change and therefore the velocity change are also small. In that case, the force can remain almost completely perpendicular. Instead of a big ball hitting, imagine instead we have a machine-gun ping-pong ball shooter. Each individual ping-pong ball transfers very little momentum, and we can line up each new one as nearly perpendicular as possible. The total result would be to change the direction of the target, not to change the energy.

In your second example, the collision is not instantaneous, and the velocity of the lighter ball changes throughout the collision.

It's easiest to think of this velocity at the instants just before the balls part, at which the velocity of the lighter ball is already close or equal to its final velocity, with a nonzero component along $y$. During those last instants of the collision, then, the force exerted by the heavier ball on the lighter one does perform work. And there is therefore a continuous (in fact linear) ramp-up of the work performed by the heavier ball on the lighter one, as the latter acquires more and more speed along the collision.

I have read that since the magnetic force acting on moving charge is always perpendicular to the velocity of moving charge,...

To clear your mind I want to rephrase your statement and then tell you about Lorentz force.

Under the influence of a magnetic field a moving - not parallel to this field - charge gets deflected in a plane which is perpendicular to both the direction of the electrons movement and the direction of the magnetic field. (Note that the text between the hyphens “-“ is not necessary because in the case of the same directions of the magnetic field and the electrons movement the plane between them is zero.)

The phenomenon of this deflection is called the Lorentz force.

... the field is not doing any work on the charge.

To explained this phenomenon first one has to recognize that the charge gets deflected permanently. Telling you this you perhaps can conclude that the charge will moves in a circle. But carrying out the experiment with electrons in a Bubble chamber you will observe that the electron moves in a spiral path until in comes to rest.

Furthermore the charge loses permanently kinetic energy and this energy loss is realized in the form of emission of photons from the electron.

The next paragraph is a little bit tricky. Magnetic fields don’t interact with the electric field of a charge. So how the deflection happens? Please realize that the electron has not only an electric field, it has a permanent magnetic dipole too.

Moving into an external magnetic field the electrons magnetic dipole gets aligned and the electrons gets a little bit deflected. This deflection is an acceleration and as you know an acceleration of a charge is accompanied by the emission of electromagnetic radiation (photons). By this emission the electron gets disaligned again and the process starts again, but this time on a lower energy level until the kinetic energy is exhausted.

The magnetic field works like spring. You can do the Lorentz force experiment with an electromagnet as well as with a permanent magnet. The permanent magnet will have the same strength before and after the experiment.The electrons temporarely and periodically influence the shape of the external magnetic field and the external magnetic field align temporarily and periodic the electrons.

Does a force perpendicular to the direction of movement not do work?

Driving your bicycle and coming under the influence of a heavy crosswind you very fast will realize that this perpendicular force will do work. For the Lorentz force the emitted photons do the work. The magnetic field is the catalyst only.