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You can find the symbol of love, the heart, by looking at the motion of a charged particle that's emitted from a hot wire in a specific electromagnetic field. The electric field is directed radially from the wire and has strength $1V/m$ everywhere. The magnetic field $B=1T$ is parallel with the wire axis. If the charged particle has a very small radial velocity with absolutely no velocity parallel to the wire direction when it escapes the wire, find how long the total length it has to travel to come back to the wire in meters?

If you plot the orbit of the particle you'll find that it looks like a heart!

Neglect the size of the wire. The electric charge of the particle is $1C$ and the mass is $1kg$.

I am stuck on how to analyze the motion of the object.

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the solution must be one of these heart shapes – anna v May 26 '14 at 8:53

Fun problem! In cylindrical coordinates you'll have the forces $$ \begin{align*} \hat r \cdot \vec F_e &= q E \\ \hat r \cdot \vec F_m &= -q r \dot \theta B & \hat \theta \cdot \vec F_m &= q \dot r B \end{align*} $$ Part of the trick is that the unit vectors $\hat r,\hat\theta$ for cylindrical coordinates change with time. I think it's \begin{align*} \dot{\hat r} &= \hat\theta\dot\theta & \dot{\hat\theta} &= -\hat r\dot\theta \end{align*} which gives $$ \begin{align*} \frac d{dt} \hat rr &= \hat r \dot r + \hat\theta\dot\theta r \\ \frac {d^2}{dt^2} \hat rr &= \hat r \left( \ddot r - r\dot\theta^2\right) + \hat\theta \left( r\ddot\theta + 2\dot r\dot\theta \right) \end{align*} $$ Since $\vec F = m\ddot{\vec r}$, now you have two coupled second-order differential equations to work with, which will give you a framework to find the curve. There may be another cute trick to find an analytic solution, or you may need to solve numerically.

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I checked, it can be done with out numerical analysis, can u pls clarify how you got value for the unit vectors. – Derg May 26 '14 at 9:45
@Derg You can start with the transformation $\hat r = \hat x \cos\theta + \hat y \sin\theta$, and its complement for $\hat\theta$. Then $\frac d{dt}\hat x =0$, and $\frac d{dt} \cos\theta = -\frac{d\theta}{dt}\sin\theta$, and so on. From that starting point everything I've written is just applications of the chain rule and product rule. – rob May 26 '14 at 15:50
@Derg Yes, I get separable differential equations from $F=ma$ in both the $\hat r$ and $\hat\theta$ directions. A good take-home exam question. – rob May 26 '14 at 16:13
Thanks, I'll see if I could develop on this – Derg May 26 '14 at 18:04

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