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If the velocity of a particle and magnetic field are perpendicular, force on the particle is also perpendicular to velocity. Suppose the force is along $x$ direction and velocity along $y$.

Then the initial component of velocity in the $y$ direction must be unchanged because acceleration is along $x$. So $v_x$ increases and $v_y$ is constant. But in circular motion, both $v_x$, $v_y$ change continuously. It's also said that charged particle moves in a circle in a magnetic field perpendicular to initial velocity.

Can someone resolve this confusion?

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  • $\begingroup$ The same question would apply to any circular motion at constant speed - a planet in orbit, a car driving around a curve, or a rock on a string. The rock on a string example may be easier to visualize because the force is parallel to the string. $\endgroup$
    – mmesser314
    Commented Dec 11, 2017 at 14:26
  • $\begingroup$ The force will always be perpendicular the velocity, as velocity changes so will the force $\endgroup$
    – Rick
    Commented Dec 11, 2017 at 14:27
  • $\begingroup$ I just got an idea, please confirm somebody: since the particle is accelerated in x direction, the velocity vector changes which implies the force vector also changes, it's because force by a magnetic field depends upon the velocity vector. So a changing force vector changes initial vy also $\endgroup$
    – Neeladri Reddy
    Commented Dec 11, 2017 at 15:52

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But in circular motion, both vx, vy change continuously

That is not true. If you think about sinusoidal motion, there is a point during the motion that velocity does not change (that is, that the derivative is zero). That is exactly what happens at the point in the circular motion that you are talking about: If we imagine the motion to be counterclockwise, then as the particle arrives at the y=0 point, the y velocity has been increasing; after that point, it will be decreasing. But at that precise moment, it is neither increasing nor decreasing.

This can be shown mathematically as well: if we assume the orbit to have a radius R centered on the origin, and reaching (R,0) at t=0, we can write the position as:

$$x(t)=R\cos\omega t\\ y(t)=R\sin\omega t$$

Differentiating with respect to time, we find the velocity; differentiating again, we find the acceleration. And you will see the acceleration (change in velocity) in Y is zero at that precise point:

$$\begin{align}v_y(t)&=\omega R \cos\omega t\\ a_y(t)&=-\omega^2\sin\omega t\\ a_y(0)&=0\end{align}$$

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