# Lorentz force in rotating frame

This is the common problem of a charged particle moving in a static electric and magnetic field. Say $\textbf{E}=(E_x,0,0)$ and $\textbf{B}=(0,0,B_z)$.

In the inertial frame of reference, the equation of motion is (1): $$\frac{d \textbf{v} }{dt} = -\frac{q \textbf{B} }{m}\times \textbf{v} + \frac{q}{m}\textbf{E}$$

We can find equations for $v_x$ an $v_y$ and see that the resulting motion is a circular orbit with a constant drift velocity $v_d=\frac{E_x}{B_z}$.

Surely I should get the same answer if I solve the problem in a rotating frame of reference?

I know that (2): $$\frac{d \textbf{v} }{dt} \vert_{Inertial} = \frac{d \textbf{v} }{dt} \vert_{Rotational} + \boldsymbol{\omega}\times\textbf{v};$$

If I use Eq. (1) as the LHS of Eq. (2), and choose $\boldsymbol{\omega}=-\frac{q \textbf{B} }{m}$, then I get (3):

$$\frac{d \textbf{v} }{dt} \vert_{Rotational} = \frac{q}{m}\textbf{E};$$

where $\mathbf{E}=(E_x\cos(\omega t), E_x\sin(\omega t),0)$ is the electric field in the rotating frame.

Solving this:$\textbf{v} \vert_{Rot} = \frac{E_x}{B_z} (\sin(\omega t)\textbf{i} - \cos(\omega t)\textbf{j})$

But I think I am missing the relation between $\mathbf{i}$ and $\mathbf{j}$ in the rotation and in the inertial frame...

How do I get the drift velocity $v_d = \frac{E_x}{B_z}$?

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Yes you are correct the answers should be the same in either frame and in fact the answers you give are. You just have to convert back to the inertial frame

Remember that i and j in the rotating frame are fixed in the rotating frame and so rotate in the inertial frame. Using i' and j' as the axis of the reference frame and i and j for the inertial frame. The relationship between the two are (see http://en.wikipedia.org/wiki/Rotating_frame) $$\textbf{i'} = \textbf{i} \cos(\omega t) - \textbf{j} \sin(\omega t)$$ $$\textbf{j'} = \textbf{j} \cos(\omega t) + \textbf{i}\sin(\omega t)$$

$$\textbf{v} = \frac{E_x}{B_z}[\textbf{i} (\sin(\omega t)\cos(\omega t) - \sin(\omega t)\cos(\omega t)) - \textbf{j} (\sin^2(\omega t) + \cos^2(\omega t))]$$

the i term cancels and using $\sin^2(\omega t) + \cos^2(\omega t)=1$ we get your original answer

$$v=-\frac{E_x}{B_z} \textbf{j}$$

Ok, its not strictly the original answer but I presume you only calculated the speed originally not the direction. On that note it is easy to show the speed is the same without changing frames just take the magnitude of the rotating velocity.

$$v_{rot} = \frac{E_x}{B_z} \sqrt{\sin^2(\omega t) + \cos^2(\omega t)}$$

Again the bit in the root is 1.

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