Why is there no change in speed of a charged particle in a uniform perpendicular magnetic field? [duplicate]

When the initial velocity is perpendicular to the magnetic field, the force acting on the charged particle

F = qvBsin(theta)

F = qvBsin(90)

F = qvB

In my textbook, it is given that the force will continually deflect the particle making it move in a circle. It also says that as the magnetic field acts on a particle perpendicular to its velocity, it does not do any work on the particle, it doesn't change the kinetic energy of the speed or the particle.

However, how come that it doesn't change the speed or the kinetic energy? We do have force acting on the particle which means that there will certainly be some accerleration.

marked as duplicate by Bill N, user36790, John Rennie, Kyle Kanos, ACuriousMind♦Oct 9 '15 at 12:48

• This is similar to another question, for which I gave an answer here. – honeste_vivere Oct 8 '15 at 23:48
• Forces which act perpendicular to the velocity only produce and instantaneous change in the direction of the velocity, not the speed. Change in velocity means there is an acceleration. The magnetic field only produces a force which is instantaneously perpendicular to the velocity of the charge. Your text book is correct. – Bill N Oct 9 '15 at 0:33

The power (work per unit time) of a force $\vec F$ on a particle with velocity $v$ is given by: $$P = \vec F \cdot \vec v.$$ If this number is positive, then the force tends to speed the particle up. If it is negative, it tends to slow the particle down. You can see this in the Work-Energy theorem, the differential form of which consists of applying Newton's law to this as if it's the only force acting on the particle: $$P = m {d\vec v \over dt} \cdot \vec v = m~~\frac{1}{2}~{d \over dt}\left(\vec v \cdot \vec v\right)$$We see that $\vec v\cdot\vec v$ is the speed squared and has no dependencies on anything other than the speed, moreover this is increasing with time when the power is positive, or decreasing when it is negative, or zero when it is zero.
Now, the thing about that dot product, $\vec F \cdot \vec v$, is that it is zero whenever those two vectors are perpendicular. So a force has to have some component in the direction of your velocity.
Magnetic forces do not have this: their Lorentz force is $\vec F = q \vec v\times\vec B$, always perpendicular to velocity. Magnetic forces cannot do work directly. To calculate, say, two magnets attracting each other, you always need to see what sorts of electric fields are getting induced, or at least calculate the energy density in the fields before and after.
The velocity vector is changing all the time. The magnitude is constant, but the direction is changing all the time (circular motion). So, the Lorentz force, which is perpendicular changes $\vec v$ without changing its magnitude $v$, which is prohibited as the energy increase is $\vec F . \vec v =0$.