A proof that magnetic forces do no work?

[Dale]

What is actual displacement? That \Delta \theta ?

It is the actual path that the object travels. It is a vector, so no, it is not $\Delta \theta$. It is the vector $\Delta \vec r$ which can be written as $\Delta \vec r = \Delta x \hat x + \Delta y \hat y = (\Delta x, \Delta y)$ in Cartesian coordinates or as $\Delta \vec r = \Delta r \hat r + \Delta \theta \hat \theta = (\Delta r, \Delta \theta)$ in polar coordinates. So in polar coordinates you can write it $(0,\Delta \theta)$ and as $\Delta t \to 0$ we get $(0,\Delta \theta/\Delta t) \to (0,d\theta/dt)=(0,v_{\theta})$ which is always perpendicular to the force $(F_B,0)$ and therefore the work is always 0.

 

[Adesh]

Causation doesn't enter the definition of work.

Please explain me a little further, I really want to learn it.

 

[Adesh]

It is the actual path that the object travels. It is a vector, so no, it is not $\Delta \theta$. It is the vector $\Delta \vec r$ which can be written as $\Delta \vec r = \Delta x \hat x + \Delta y \hat y = (\Delta x, \Delta y)$ in Cartesian coordinates or as $\Delta \vec r = \Delta r \hat r + \Delta \theta \hat \theta = (\Delta r, \Delta \theta)$ in polar coordinates. So in polar coordinates you can write it $(0,\Delta \theta)$ and as $\Delta t \to 0$ we get $(0,\Delta \theta/\Delta t) \to (0,d\theta/dt)=(0,v_{\theta})$ which is always perpendicular to the force $(F_B,0)$ and therefore the work is always 0.

But we want to prove that the displacement is perpendicular to the force not the velocity.

 

[Dale]

But we want to prove that the displacement is perpendicular to the force not the velocity.

The velocity is the instantaneous displacement. You have studied calculus.

 

[dsanz]

Please explain me a little further, I really want to learn it.

Forget about what causes what. To calculate work done by a particular force on a given object, you only need to know what that force is and what the displacement of the object is. It doesn't matter whether "only part of the displacement was caused by the force" or anything of that sort. You look at the force and take the dot product with displacement (both at the same instant).

 

[PeroK]

But we want to prove that the displacement is perpendicular to the force not the velocity.

In one sense there is no instantaneous displacement. There is an instantaneous displacement from the origin, which is the instantaneous position (vector). But, displacement is the difference of two position measurements. As the time increment tends to zero, the displacement over that time increment tends to zero.

The instantaneous quantities must be the time derivative of something. Velocity is the derivative of position and acceleration is the derivative of velocity. Neither of these need tend to zero as the time increment tends to zero.

 

[alantheastronomer]

In this picture, I have drawn the violet arrow which represents the velocity after a time \Delta t from where the + sign has been made. I have drawn the components of violet velocity, the components are in yellow color and there you see we have a component inwards. What has caused this inward component of velocity? I think it is \mathbf{F_{B}} which acted at + sign has caused this, correct me if I'm wrong.
View attachment 254197

I see what you're saying - the displacement is not parallel to the velocity, and we can clearly see that there is a component of the displacement parallel to the force vector, so their dot product clearly isn't zero, so what gives? Well, this would be true if the force were impulsive - occurring over a finite amount of time - so that the force vector was fixed in that direction. But this force arises out of that magnetic field interacting with the particle's motion, perpendicular to them both. The force vector is not static, it's continually changing in lockstep with the particle's position, always perpendicular.