Magnetic Forces do no work.

[Je m'appelle]

It is known that magnetic fields do no work at a moving particle, all they can do is to change the particle's direction.

So, I've been trying to understand a step on the mathematical explanation, but I'm stuck.

I'm using this source: http://www.tutorvista.com/content/physics/physics-iv/moving-charges-magnetism/lorentz-force.php

What I don't get is the step below

$$m \frac{d}{dt} (v \cdot v) = m(v} \cdot \frac{d v}{dt} + \frac{d v}{dt} \cdot v) = 2m v\frac{d v}{dt}$$

Shouldn't it be

$$2m v \cdot \frac{d v}{dt} = 2m v \frac{d v}{dt} cos \theta$$

What happened to the cosine? The dot product simply disappeared, it's like he considered $$cos \theta = 1$$, but as far as I understood it, the cosine is actually zero and not one.

OBS: 'v' is a vector.

 

[The legend]

your latex image seems to be invalid and I can't view it

 

[Je m'appelle]

your latex image seems to be invalid and I can't view it

I've fixed it already, somehow the vector function in the latex wasn't working.

 

[Jolsa]

It looks like a typo to me. He does it correctly the first time when he says

$$m\frac{d\vec{v}}{dt}\cdot \vec{v}=\frac{m}{2}\frac{d}{dt}(\vec{v} \cdot \vec{v})=0$$

Now $\vec{v} \cdot \vec{v}=v^2$ so it follows that the change in kinetic energy is zero.

 

[rwisz]

I can provide some clarification on this step. The dot product is used to calculate the work done by a force, which is defined as the product of the force and the displacement in the direction of the force. In this case, we are dealing with a magnetic force, which does not do any work. Therefore, the dot product is not necessary in this step and can be omitted.

In other words, the dot product is not relevant in this case because the magnetic force does not contribute to the work done on the particle. It only changes the direction of the particle's motion. So, in this step, we are simply calculating the rate of change of the particle's velocity, which is equal to the acceleration.

I hope this helps to clarify the confusion. Keep in mind that in physics, different mathematical approaches can be used to arrive at the same result. The important thing is to understand the underlying concepts and principles.