# Lorentz Force and Larmor radius

I have the following exercise: I have a charge (q) accelerated by a potential difference ($\Delta V$) which enters a region in which a magnetic uniform field is present (the field is not know). I have the distance FO and I have to determinate the distance OS, where 'S' is the point of impact on the "Schermo". The magnetic field B and the particle speed v forms an angle of 90°

That's my idea: The electrostatic field is conservative so I can write this: $$\Delta K = W_T = - \Delta U_e = q\Delta V$$ so: $1/2 m v^2 = q\Delta V => v^2 = 2q\Delta V / m$

Now when the charge reaches the "magnetic region" moves in circular motion because of the Lorentz Force with this radius: $qvB = mv^2 / R => R = mv/qB$

I can't go on with the exercise because I haven't got the magnetic field B. Can you please help me completing this? Thank you so much :(

• Which given value have you not used yet and what does the given value represent? – Sanya Aug 10 '16 at 22:12
• You would need to know either the angle it impacts, B, or the distance OS to solve everything. Without any of the three there is no way to solve it. Did they give you anything? Or do they want it solved in terms of B? – Kevin Kostlan Aug 10 '16 at 22:13
• I have to determinate the distance OS; B is not given, I just know it's 90° with v and makes q go down – PeppeDAlterio Aug 10 '16 at 22:15
• Isn't this just a kinematics question? You have an initial velocity, the particle experiences a constant acceleration, and you have the final velocity (i.e., the magnetic field won't change the speed, just the direction and you know the final direction, thus the final velocity). So I don't think you need the magnetic field... – honeste_vivere Aug 12 '16 at 13:41
• Can you please explain this better? :( @honeste_vivere – PeppeDAlterio Aug 12 '16 at 16:39

I just know it's 90° with v and makes q go down.

I conclude that the angle of impact on the bottom wall is $90°$. If this is not the case then my answer would not address the question.

Answer: In a uniform magnetic field with $\vec{B} \perp \vec{v}$ the particle trajectory is a circle. Hence since we know the final angle the solution is

$$\overline{OF} = \overline{OS}.$$

The magnetic field is not needed since the geometry already specifies the trajectory, which is probably what honeste_vivere meant in his comment.

• How can you tell that the center of the circle is the O point? – PeppeDAlterio Aug 13 '16 at 17:47
• @PeppeDAlterio if the walls are perpendicular and the velocity is perpendicular to the wall at F and at S then this is a geometric requirement (since the centre point of a circle lies on the lines perpendicular to its tangents). – Wolpertinger Aug 13 '16 at 17:51
• But what happens when at the point of impact v is not $90°$ with the schermo as the question doesn't seem to mention that? – Weezy Aug 13 '16 at 18:16
• @Weezy Then the problem is underdetermined. – Wolpertinger Aug 13 '16 at 20:31

From the first Figure

\begin{align} OS^{2} & =KS^{2}-OK^{2}=KS^{2}- \left( KF-OF \right)^{2} \tag{001}\\ & =R^{2}-\left(R-OF\right)^{2}=OF\cdot\left(2R-OF\right) \end{align} so $$OS = \begin{cases} \sqrt{OF\cdot\left(2R-OF\right)} & \text{for 0 < OF \le 2R } \\ \text{no impact} & \text{otherwise} \end{cases} \tag{002}$$

$$R\equiv \dfrac{mv}{qB}=\sqrt{\dfrac{2m\Delta V}{qB^2}} \tag{003}$$

We must have the value of $\:B\:$.
If we want impact normal to Schermo then $\:OF=R\:$ that is

$$OF\cdot B =\dfrac{mv}{q}=\sqrt{\dfrac{2m\Delta V}{q}} \tag{004}$$

• Can you explain how you got the first equation of (001)? – Weezy Aug 14 '16 at 8:54