1
$\begingroup$

I have recently studied about the movement of a particle in a basic electromagnetic field. It circulates perpendicular to the field lines.

And the equation is then, $F= Bqv$

$B$ is the value of the magnetic field

$q$ is the charge of the particle

$v$ is the linear velocity of the particle

Now if the motion of the particle is not perpendicular to the field lines but rather creates an angle $\theta$ with the field lines, Then to calculate the force exerted on it by the magnetic field, we would take the vertical component of the velocity $v\sin\theta$.

But it does have a horizontal component too. So it's movement would be helical.

Now let's come to the fact that's confusing me. In case of pure circular movement with no horizontal component of velocity, The centripetal force is the same force that we calculate by the equation $F=Bqv$. So the force exerted on the particle is nothing but the centripetal force. I am not sure if I am right about this.

So we can write $F= m \frac {v^2} {r} = Bqv$

And it makes sense actually. If we think about the direction of these forces they are the same.

But in case of helical movement we take the vertical component of velocity and not the actual velocity. But I think the force that i should consider when calculating the centripetal force is the actual velocity. So the $F= Bqv\sin\theta$ and the $F$ denoting centripetal force will have different direction and will not be the same.

Then how do we calculate centripetal force in this case?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Look at it in a frame of reference that is moving at velocity $v\ \text{cos}\ \theta$ in the direction of the magnetic field, and so keeping pace with the motion of the charged particle parallel to the axis of the helix. In this frame of reference the particle is moving in a circle with speed $v\ \text{sin}\ \theta$ and it should then be clear that $v\ \text{sin}\ \theta$ replaces $v$ on both sides of the equation $Bqv=\frac{mv^2}{r}$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Do u mean that in helical movement, the centripetal force is caused by the vertical component of the velocity only? and the vcosθ is just causing the charge to move forward? $\endgroup$
    – Sami
    Commented Jul 25, 2017 at 10:16
  • $\begingroup$ I don't know what you mean by 'vertical'. But I'd agree that the component of velocity parallel to the field lines remains unaltered and gives the motion along the axis of the helix, whereas the velocity component at right angles to the field gives rise to the force and is continuously altered in direction by the force, causing the circular motion. I find that looking at it from the new reference frame (see my answer above) is an aid to clear thinking. $\endgroup$
    – Philip Wood
    Commented Jul 25, 2017 at 17:36
0
$\begingroup$

Your formulae are correct. The movements are comletely independent like in an inclined trajactory of a thrown ball where $v_x$ and $v_y$ are independent: one is constant ,the other affected by acceleration due to gravity. Here you have a $v_\perp$ affected by the magnetic field and a $v_\parallel$ that remains just constant. It is probably more obvious if you write down the full vector expression for the force: $$\vec F= q \vec v \times \vec B$$

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.