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I was solving the following homework problem:

A force $\vec{F} = \vec{k} \times \vec{v}$ is applied to a particle of mass $m$. Here $\vec{k}$ is a fixed vector and $\vec{v}$ is the velocity of the particle. Show that the kinetic energy of the particle is constant.

My attempt at a solution was as follows. We know that the work done by a particle being subjected to some force $\vec{F}$ from $a$ to $b$ is the change in kinetic energy. Using this, if I show that the work is $0$ for any starting time $a$ and ending time $b$, then the kinetic energy must be constant. We thus see that $$ W_{ab} = \int_{a}^{b} \vec{F} \cdot \vec{v} \ dt = \int_{a}^{b} (\vec{k} \times \vec{v})\cdot \vec{v} \ dt = \int_{a}^{b} \underbrace{(\vec{v} \times \vec{v})}_{\color{purple}{\vec{0}}}\cdot \vec{k} \ dt = 0 $$ and since $a$ and $b$ are arbitrary, the kinetic energy is constant.


While thinking about this solution, I noticed that I didn't use the fact that the vector $\vec{k}$ was fixed. So does this mean that the kinetic energy is still constant even if $\vec{k}$ isn't constant? Or did I make a mistake in my analysis? Thank you very much!

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    $\begingroup$ Your conclusion is correct, if the vector k where not constant, it would need to be cast as a path dependent quantity in the integrand, but the argument would still hold. Adding to this, the force used in this exercise is analogous to the magnetic field force on a charged particle, so this tells you the magnetic field doesnt do work on charged particles. $\endgroup$ May 25, 2021 at 5:03
  • $\begingroup$ from Newton law $ m\dfrac{d^{2}x}{dt^{2}}=F $,you obtain this equation $\dfrac{mv^{2}}{2}=\int \overrightarrow{F}\cdot \overrightarrow{v}dt=0$ thus v is zero or constant ( Initial condition) $\endgroup$
    – Eli
    May 25, 2021 at 10:10

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The work done is zero because the force is perpendicular to the velocity, it will be zero even if $\vec{k}$ is time-dependent. Think of it like this: Work done by a Lorentz force is zero even if the magnetic field is time-dependent.

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