Why does a charged particle in a cyclotron moves in circular path?

Question

Force on the particle is charge*(velocity×magnetic field intensity). The magnetic field intensity is perpendicular to velocity of charged particle. What is it that makes it move in a circular path?

Answer 1

I think there is no need of putting cyclotron in this question. If you are interested in working of cyclotron the question is largely ill framed.

From a simplistic point of view consider that a particle enters in a uniform magnetic field in x direction with velocity $v_x$ and the direction of magnetic field is along z axis($B_z$). Then the particle will experience perpendicular force and in an infinitesimal time dt it will gain the velocity $v_y$. Now the vector addition will give final velocity.

$v_f=\sqrt{(v_x^2+{\frac{ev_xB_zdt}{m}}^2)}$

Now as we make dt small the second term tends to zero much faster (due to $dt^2$ dependence) and the magnitude of the final velocity is same as initial velocity, but this interaction give a definite infinitesimal angle change which is non zero as it has first order terms of dt.

Now if you rotate your frame of reference to align your x axis with the new angle you will find yourself in same situation and hence the particle will experience only rotation and not the acceleration. This type of motion will only result in circular movement.

The answer may not be mathematical because we all know mathematics behind this problem but it is my point of view about this problem.

I hope this will help

Answer 2

As you know, the force on a charge by uniform magnetic field is given by

$\vec{F}=q\vec{v}\times\vec{B}$

This force is perpendicular to both velocity and the magnetic field.

During the charge's motion, it experiences a magnetic force just like an other particle experiences centripetal force undergoing uniform circular motion.

When the particle enters the magnetic field, it deviates from its path, which changes the direction of it's velocity (not it's magnitude). After this deviation, the force's direction also changes but this force points towards a same point (the centre of the circle).

So, in its motion, a force acts on the particle that always points towards a single point which is the centre of the circle. This circle is the trajectory of the particle when it moves under the influence of a uniform magnetic field.

The magnetic force is responsible for providing the particle the centripetal acceleration which is the basic requirement for a particle to execute uniform circular motion.

Centripetal force is given by,

$\frac{mv^2}{r}=qvB$

$r=\frac{mv}{qB}$

This gives us the radius of the circle.

Answer 3

Firstly, free charged particles in magnetic fields don't always move in circles - in general they move in helices.

What is always circular is the path of the head of the velocity vector.

You can write down the equations, even get a solution of the form $\mathbf{v} = \exp(B\,t)\mathbf{v}(0)$ where $B$ is a $3\times 3$ skew-symmmetric matrix modelling the magnetic field and integrate to get the path.

Or you can think of it a little more basically and insightfully (but no less rigorously or less mathematically).

Fundamental to this problem is that not only (1) that the magnetic field orthogonal to the velocity, but (2) it is also orthogonal to the force and (3) the force is also orthogonal to the velocity.

From (3), the force does no work. Therefore the length of the velocity vector is constant, and the head of the velocity vector must move on the sphere $|\mathbf{v}|^2=const$. From (2), if the magnetic field is orthogonal to the force, then the force is constrained to a plane normal to the magnetic field. Therefore, by Newton's second law, the time derivative of the velocity is also confined to such a plane. The velocity vector must move on a path that is in the threefold intersection of (1) some plane normal to the the magnetic field, (2) the sphere defined by the velocity's initial length and (3) which contains the head of the initial velocity vector.

Sketch this; these three conditions define a circle.

Now assume that the action of the magnetic field is linear (as it is with the cross product). So the action of the magnetic field is the sum of its actions on the components of the velocity.

Split the initial velocity into components parallel and orthogonal to the magnetic field.

Now reason as above for these two components separately. You'll find that the circular path followed by the head of the parallel component is degenerate - of zero radius, and so the action of the magnetic field on this component is to leave it constant.

The action of the magnetic field on the orthogonal velocity component is to make it move along a circular path in plane containing the origin.

In fact, for constant magnetic field magnitude, this circular motion is uniform. Therefore the orthogonal component results in uniform circular motion in this plane.

Now sum the two effects by linearity. You've got a helical path.