Lorentz Force and Larmor radius

Question

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 :(

Answer 1

From the comments

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.

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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.

Answer 2

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 \begin{equation} 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} \end{equation}

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

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

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