If we interchange velocity with charge and omega that is rotation of a system with $B$, magnetic field, we get the same thing. Is there any deeper meaning to this same mathematical form?
[Physics] Resemblance between Coriolis force and magnetic part of Lorentz force
coriolis-effectelectromagnetismforcesnewtonian-mechanicsreference frames
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Nothing feels a reaction force to the Coriolis force because Newton's laws only apply in an inertial frame. If we imagine a frame with some sort of wild, random acceleration, then in that frame at a given moment it would look like, in a room with completely stationary tables and chairs and things, everything was simultaneously accelerating the same direction, and there would be huge "fictitious forces" needed to explain this. There are no third-law pairs going on, though. The fictitious forces were necessary to make $F=ma$ work out in this accelerating frame, but by introducing them we sacrificed the third law (at least in regards to those kinematic forces).
Imagine dropping a ball off a tower. It gets deflected east.
If we analyze this scenario from a frame rotating along with the Earth, there is no conflict with Newton's third law because it does not hold for the Coriolis force or other kinematic forces.
If we analyze the scenario from an inertial frame in which the Earth is rotating, the trajectory simply is not deflected. The ball lands east of where it was dropped because its velocity in a tangential direction, which comes from the rotation of the Earth, stayed the same, but as it fell its angular velocity $v/r$ increased because $r$ decreased. That led it to "deflect east" when it was really moving in a perfectly ordinary parabola.
As nobody has done this yet, let's try to give an answer to your question in the right framework, i.e. through the formalism of differential geometry (and of action principles, as far as the physics is concerned). This formalism has the advantage of allowing for the use of arbitrary coordinate systems, so that the problem of the arising of "fictitious" force terms such as the Coriolis force can be addressed in a rigorous way. Moreover, it allows to generalize the standard description of electromagnetism in such a way that magnetic monopoles indeed are permitted to exist. I will show and motivate why in my opinion there is no connection between the Coriolis force and the magnetic force, explain what it means for the magnetic force to be a relativistic effect of the electric force and show how to introduce monopole fields in the magnetic field. I'll try to make myself as clear as possible, as I understand that you are not familiar with the formalism.
First of all, the Lagrangian for a point-like, massive particle in an arbitrarily curved spacetime, subject to the electromagnetic field, can be expressed as
$$\mathscr{L}=-m\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}-e\,A_{\mu}(x^{\mu})\frac{dx^{\mu}}{ds}$$
(factors of $c$ missing). Here $g_{\mu\nu}$ is the spacetime metric, the object which encodes the spacetime curvature, $A_{\mu}$ is the covariant form of the electromagnetic four-potential, $A_{\mu}=(\phi,-\vec{A})$, $s$ is an arbitrary parameter, $m$ and $e$ the mass and charge of the particle, $x^{\mu}(s)$, with $\mu=0,1,2,3$, the trajectory of the particle in spacetime. We will be interested in flat spacetimes, i.e. Minkowski spacetime, but one needs the full generalization to extract useful results from the formalism. One finds the equations of motion for the particle by minimizing the action integral $S$, that is
$$ S[x]=\int_{a}^{b}\mathscr{L}(x^{\mu}(s),\dot{x}^{\mu}(s))\ ds $$ where the dot denotes a derivation with respect to the parameter $s$. Finding a minimum for $S$ is completely equivalent to the procedure shown in Frédéric's answer: the curve which minimizes the action is the curve which solves the Euler-Lagrange equations, or equivalently the Hamilton equations (those in the cited answer). Notice that the action
$$ S=\int_{a}^{b}\bigg\{-m\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}-e\,A_{\mu}\frac{dx^{\mu}}{ds}\bigg\}\ ds $$ is invariant with respect to three different kinds of transformation. The first one is a monotone, increasing change of parametrization $s\to s'$ (i.e. one with $ds'/ds>0$), as the transformation gets absorbed in the measure of integration $ds$. The second one is an arbitrary change of coordinates: every time you see two indices contracted, as the two objects involved in the contraction transform in opposite ways (this is symbolically expressed by the positioning of the indices), the overall object remains invariant under a change of coordinates. The last one is the transformation
$$ A_{\mu}\to A_{\mu}+\partial_{\mu}\chi $$ where $\chi$ is an arbitrary function of the variables $x^{\mu}$, known as a gauge transformation of the electromagnetic potential. Under such a transformation, the action gains the additional term
$$ \delta S=\int_{a}^{b}-e\ \partial_{\mu}\chi\ \frac{dx^{\mu}}{ds}\ ds=\int_{a}^{b}-e\ \frac{d\chi}{ds}\ ds=-e\ \bigg[\chi(b)-\chi(a)\bigg] $$ which is a constant. So the action may not actually be invariant under such a transformation, but as $S$ only gets shifted by a constant amount, its minima are preserved by the transformation. The property of $S$ being invariant under such transformations has one important consequence: the general form of the dynamical equations one gets from the minimization of $S$ is valid with respect to any parametrization of the curve of the particle, which in turn can be expressed in any coordinate system you like, and the equations are not changed by a gauge transformation. To give the equations a simpler look, I will use the parametrization in which
$$ \sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}=1 $$
Please notice that this condition does not affect the choice of the coordinates through which you express the dynamical curve: the condition itself is unaffected by a change of coordinates. From the above action $S$ and given the former condition on $s$, it can be shown that the dynamical equations have the following form:
$$ \frac{d^{2} x^{\mu}}{ds^{2}}=-\Gamma^{\mu}_{\nu\tau}\ \frac{dx^{\nu}}{ds}\frac{dx^{\tau}}{ds}+\frac{e}{m}\ F^{\mu}_{\ \nu}\ \frac{dx^{\nu}}{ds} $$
Here $$ F^{\mu}_{\ \nu}=g^{\mu\sigma}(\partial_{\sigma}A_{\nu}-\partial_{\nu} A_{\sigma}) $$ with $g^{\mu\sigma}$ the inverse of the matrix $g_{\mu\sigma}$, is the electromagnetic field tensor and the functions $$ \Gamma^{\mu}_{\nu\tau}=\frac{1}{2}g^{\mu\sigma}\ \big[\partial_{\nu}g_{\sigma\tau}+\partial_{\tau}g_{\sigma\nu}-\partial_{\sigma}g_{\nu\tau}\big] $$ are called the "Christoffel symbols" related to the metric $g$ in the $x^{\mu}$ coordinate system. The Christoffel symbols encode two kinds of information: first of all, if the metric $g$ describes a curved spacetime, they encode the effect of curvature on the particle, i.e. they encode the gravitational field; second of all, they encode the fictitious forces due to the choice of a specific coordinate system. It can be shown that, when spacetime is not curved, there exist coordinates with respect to which every $\Gamma$ is zero. These are the (in)famous intertial frames, in which $\eta_{\mu\nu}=\text{diag}(1,-1,-1,-1)$ and the equations take the form
$$ \frac{d^{2} x^{\mu}}{ds^{2}}=\frac{e}{m}\ F^{\mu}_{\ \nu}\ \frac{dx^{\nu}}{ds} $$ $$ F^{i}_{\ 0}=-\vec{\nabla}\phi-\partial_{0}\vec{A}=\vec{E} $$ $$ F^{i}_{\ j}=-\partial_{i}A_{j}+\partial_{j}A_{i}=\epsilon_{ijk}(\vec{B})_{k} $$ If we substitute $ds$ with $dt ds/dt$, where $t$ is the time coordinate of the inertial observer, i.e. $t=x^{0}$, we find for the $\mu=1,2,3$ equations $$ \frac{d}{dt}\bigg(\frac{dt}{ds}\frac{d\vec{x}}{dt}\bigg)=\frac{e}{m}\ \left(\vec{E}+\vec{v}\times\vec{B}\right) $$ As $ds/dt$ turns out to be $\sqrt{1-v^{2}/c^{2}}$, the former is exactly the Lorentz equation for a relativistic particle. Now let's work in more general coordinates. Let's call these coordinates $y^{\mu}$, with equations of motion $$ \frac{d^{2} y^{\mu}}{ds^{2}}=-\Gamma^{\mu}_{\nu\tau}\ \frac{dy^{\nu}}{ds}\frac{dy^{\tau}}{ds}+\frac{e}{m}\ F^{\mu}_{\ \nu}\ \frac{dy^{\nu}}{ds} $$ It can be shown that the general relation between the Christoffel symbols wrt the $y$ and the $x$ coordinates is
$$ \Gamma^{\mu (y)}_{\nu\tau}=\frac{\partial x^{\sigma}}{\partial y^{\nu}}\frac{\partial x^{\lambda}}{\partial y^{\tau}}\Gamma^{\alpha\ (x)}_{\sigma\lambda}\frac{\partial y^{\sigma}}{\partial x^{\alpha}}+\frac{\partial y^{\mu}}{\partial x^{\alpha}}\frac{\partial^{2} x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}} $$ where $\partial y/\partial x$ and its inverse are the matrices of the change of coordinates. In our specific case ($\Gamma^{\alpha\ (x)}_{\sigma\lambda}=0$), $$ \Gamma^{\mu}_{\nu\tau}=\frac{\partial y^{\mu}}{\partial x^{\alpha}}\frac{\partial^{2} x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}} $$ So the dynamical equations can be written as $$ \frac{d^{2} y^{\mu}}{ds^{2}}=-\frac{\partial y^{\mu}}{\partial x^{\alpha}}\frac{\partial^{2} x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}}\ \frac{dy^{\nu}}{ds}\frac{dy^{\tau}}{ds}+\frac{e}{m}\ F^{\mu\ (y)}_{\ \nu}\ \frac{dy^{\nu}}{ds} $$ Now let's focus on each term of the equation. $\frac{d^{2} y^{\mu}}{ds^{2}}$ plays the role of an acceleration wrt to the parameter $s$ (which does not need to be time). $\frac{e}{m}\ F^{\mu (y)}_{\ \nu}\ \frac{dy^{\nu}}{ds}$ is the ordinary electromagnetic acceleration, expressed in an arbitrary coordinate system and wrt to the parameter $s$. But what about the functions $ -\frac{\partial x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}}\ \frac{dy^{\nu}}{ds}\frac{dy^{\tau}}{ds}$? It is useful to notice that these functions depend on second derivatives, i.e. they do not appear if the relation between the coordinates $y$ and $x$ is linear, of the form
$$ x^{\mu}=\Lambda^{\mu}_{\ \nu}y^{\nu}+a^{\mu} $$ It is easy to realize that the former is the correct relation between two inertial coordinate systems. In a special-relativistic setting, we choose $\Lambda$ to be a Lorentz transformation, in order to keep the metric $g_{\mu\nu}=\text{diag}(1,-1,-1,-1)$, and in turn the $g^{\mu\nu}$ in the definition of tensor $F^{\mu}_{\nu}$, invariant. Following a Lorentz transformation then, the equations of motion do not change, as advised by the theory of special relativity. But if we were to make different coordinate changes, the equations would indeed be different. In particular, new terms proportional to the velocities $dy^{\mu}/ds$ would arise. This is how Coriolis and fictitious forces arise: $\vec{v}\times\vec{\omega}$ is none other than the product between a velocity and a reference frame parameter $\vec{\omega}$, which you can see in the general equation given above. The other velocity disappears when you choose $s$ to be time.
Now that we have the needed machinery and conceptual rigour, let's go back to your questions. First of all, what does it mean for the magnetic force to be a relativistic effect of the electric force? The electric and magnetic fields are related by coordinate transformations through the equations
$$ F_{\mu\nu}^{(y)}=\frac{\partial x^{\sigma}}{\partial y^{\mu}}\frac{\partial x^{\lambda}}{\partial y^{\nu}}\ F_{\sigma\lambda}^{(x)} $$ On the right side of the equation, the electric and magnetic fields get mixed due to the change of coordinates. Are there coordinate systems in which the field is entirely electric or magnetic? Yes (and in principle they can even be inertial frames). This is because if the electric source is static in some reference frame, then in that frame the field is entirely electric. Thus, for example, in any frame related to this frame by a space rotation or translation the field will remain totally electric (Lorentz boosts, on the other hand, mix the two). On the other hand, imagine setting the source into motion. Then the field produced by the force in that frame is both electric and magnetic, but the source itself hasn't changed a bit! This means that the splitting of the EM field between an electric component and a magnetic component is not an intrinsic property of the source, but rather an artifact of the choice of a coordinate system, relative to the state of motion of the source itself. This is seen by the very definition of the E and B fields, which are deduced from the tensor $F_{\mu\nu}$, which in turn depends on the choice of coordinates. So in this case "relativistic effect" means "relative wrt the state of motion of the source, wrt to a chosen coordinate system". (The question of magnetic fields due to spin currents, on the other hand, must be addressed in a different, quantum-mechanical, setting).
Coriolis forces are too an effect of the choice of a specific coordinate system. As seen, they arise from second derivatives of the coordinate transformation from inertial coordinates. Their "coordinate dependence" status, however, is quite different from that of the E/M splitting. First of all, they do not depend upon the state of motion of any kind of source relative to a specific coordinate system (here we don't regard spacetime itself as one). Second of all, they only appear in non-inertial frames, whereas the E/M splitting is an issue even in inertial frames. Last but not least, Coriolis forces depend upon the mass of the particle, whereas the magnetic force does not. This means that particles with different masses whose motion is described in the same coordinate system will experience different Coriolis forces, but they will experience the very same E/M splitting (the opposite happens with respect to the accelerations). So magnetic forces and Coriolis forces should not be compared to one another: they are two very different objects, and as such no deep connection can exist between them. You noticed, though, that their mathematical form is similar. The mathematical form of a dynamical equation (apart from general covariant equations such as those I wrote above, but this is not the case), though, depends very much on the choice a coordinate system, so one must be careful when comparing force terms of dynamical equations, especially when one is going from one coordinate system to another. The choice of coordinates is unphysical, in the sense that it needn't be connected to underlying physical principles, nor it affects the physics of the system. In this case, though, a comparison can be made on a solid basis and turns out to be useful to answer your question. Now, a Coriolis force term of the form $2m\vec{v}\times\vec{\omega}$ appears in coordinate systems (rotating coordinates wrt a given inertial frame) where the magnetic force can as well not be of the form $e\,\vec{v}\times \vec{B}$. Consider a non-relativistic charged particle in a uniform, constant magnetic field (description given in an inertial frame). The particle will spin around some axis parallel to the magnetic field with frequency $\omega=\frac{eB}{m}$ (factors of $c$ missing, depending on convention on the definition of the magnetic field). If you go to a frame which uniformly rotates around that axis with angular frequency $\omega=\frac{eB}{m}$, there will seem to be no magnetic field at all acting on the particle. This is because the frame is moving together with the component of the motion of the particle which changes due to the magnetic field, thus no motion induced by the magnetic field can be observed in that frame. This does not signal a deep connection between the Coriolis and the magnetic force, it only confirms that there exist specific frames in which specific forces do not appear to act on specific systems. This, of course, is valid for any kind of force, if you don't restrain yourself to simple inertial frames. In this case, magnetic fields make electrically charged particles spin, so it is obvious that the coordinate systems in which the magnetic field does not appear must be a rotating coordinate system. Let us see this for the case of interest. I'll use the same formulas as in Wikipedia's "Rotating reference frame" article (with $\vec{\omega}$ taken in the opposite direction). The acceleration for a charged particle in a uniformly rotating frame about the center of the trajectory, subject to a uniform constant magnetic field, takes the form: $$ \vec{a}_{r}=-\vec{\omega}\times\vec{\omega}\times \vec{r}_{i}+2\vec{\omega}\times\vec{v}_{r}+\frac{e}{m}\ \vec{v}_{i}\times \vec{B} $$ where the subscript $r$ denotes a quantity in the rotating reference frame, while the subscript $i$ denotes the same quantity in an inertial frame. We have: $$ \vec{v}_{i}=\vec{v}_{r}-\vec{\omega}\times\vec{r}_{i} $$ so that $$ \frac{e}{m}\ \vec{v}_{i}\times \vec{B}=\frac{e}{m}\ \vec{v}_{r}\times \vec{B}-\frac{e}{m}\ \vec{\omega}\times\vec{r}_{i}\times \vec{B} $$ As you can see, the mathematical form of the magnetic force changes in a uniformly rotating reference frame. Moreover, as $$ -\vec{\omega}\times\vec{\omega}\times \vec{r}_{i}=-\vec{\omega}(\vec{\omega}\cdot\vec{r}_{i})+\vec{r}_{i}(\vec{\omega}\cdot\vec{\omega})=\frac{e^{2}B^{2}}{m^{2}}\ \vec{r}_{i} $$ where the last identity follows from
$$ \vec{\omega}=\frac{e}{m}\ \vec{B}\ ,\qquad\quad \vec{\omega}\cdot\vec{r}_{i}=0 $$ and $$ -\frac{e}{m}\ \vec{\omega}\times\vec{r}_{i}\times \vec{B}=-\frac{e}{m}\ \vec{r}_{i}(\vec{\omega}\cdot\vec{B})+\frac{e}{m}\ \vec{B}(\omega\cdot \vec{r}_{i})=-\frac{e^{2}B^{2}}{m^{2}}\ \vec{r}_{i} $$ we have $$ \vec{a}_{r}=-2\vec{v}_{r}\times \vec{\omega}+\frac{e}{m}\ \vec{v}_{r}\times\vec{B}=-2\vec{v}_{r}\times \vec{\omega}+\vec{v}_{r}\times\vec{\omega}=-\vec{v}_{r}\times \vec{\omega} $$ Again, $$ \vec{a}_{r}=-\vec{v}_{r}\times \vec{\omega}=-\vec{v}_{i}\times\vec{\omega}-\vec{\omega}\times\vec{r}_{i}\times\vec{\omega} $$ which is zero due to the fact that, having solved the equation in the inertial system, $\vec{v}_{i}=\vec{r}_{i}\times\vec{\omega}$. Hence no uniform constant magnetic field nor Coriolis forces appear in the equations expressed in the uniformly rotating reference frame. This is why the form of the magnetic force term in the inertial frame coincides with the form of the Coriolis force term in the rotating frame: one must compensate the other (together with the centrifugal force) in the transition between the two coordinate systems, given the right rotation frequency.
The same can be done for the electric force: it can be made to disappear in simple accelerating coordinate systems, as the force term has the form of a simple acceleration (in contrast with the $\vec{v}\times$ form of the magnetic force).
A magnetic force is not fictitious in the following sense. Magnetic and electric forces have very different effects on the motion of point-like charges. This statement is supported by the very form of the Lorentz equation. Thus, as previously stated, one may find a coordinate system such that the magnetic force does not appear in the equations, but one cannot remove the effect of the magnetic field on the trajectories of charged particles: the trajectories themselves do not depend on the description you give of them. The curve $x^{\mu}(s)$ is an actual collection of points in spacetime, and the location of these points does not depend on the way in which you parametrize it. The equations will change in such ways as to produce the very same effect on the dynamical trajectories, only through a different coordinate description of the dynamics. This is the statement of the coordinate-transformation invariance of the action $S$. It is the description which changes, not the physics. Thus a magnetic field will always have an influence on the system, whether you call it magnetic or (in a different description, i.e. in a different coordinate system) not. One should though keep in mind that magnetic fields arise from the perturbation of the EM field by non static (wrt to some inertial frame) electric charges, with emphasis on the electric (i.e. non-magnetic) nature of the charges. (Again, we are totally underlooking the role of spin currents on the production of magnetic fields).
As a bonus, here is the potential for a monopole magnetic field. Define a coordinate system that covers the entire $\Bbb{R}^{4}$ spacetime except for the non-positive $z$ axis and take $\vec{A}$ to be
$$ \vec{A}^{N}=\left(g\ \frac{y}{r(r+z)},-g\ \frac{x}{r(r+z)},0\right) $$ Then define a coordinate system that covers $\Bbb{R}^{4}$ except for the non-negative $z$ axis and take $\vec{A}$ to be $$ \vec{A}^{S}=\left(-g\ \frac{y}{r(r-z)},g\ \frac{x}{r(r-z)},0\right) $$
$g$ is the magnetic coupling constant, and the magnetic field which corresponds to $\vec{A}^{N}$ and $\vec{A}^{S}$ is the same; it equals
$$ \vec{B}=\vec{\nabla}\times\vec{A}=-g\frac{\vec{r}}{r^{3}} $$ and it is a monopole field. Differential geometry teaches us that it is ok to choose local coordinate systems, i.e. coordinate systems which cover spacetime only in part. It also teaches us that, thanks to the gauge invariance of the action, we can choose potentials to be different in those local coordinate systems, as long as they are related by a gauge transformation in the overlapping regions. In this case, we have
$$ \vec{A}^{N}=\vec{A}^{S}+2g\vec{\nabla} \arctan\frac{y}{x} $$
so that an overall gauge potential is well defined, and it correctly reproduces a monopole field. So no, magnetic monopoles are not forbidden, not even in a classical setting.
Best Answer
I believe you want to replace mass with charge and angular velocity with the magnetic induction.
The Coriolis effect is an apparent force due to the fact that the observer is measuring with respect to a rotating frame of reference. There is no actual force acting on the body, so this can be made to disappear by changing the frame of reference. Classical electrodynamics is best described in 4-vector notation. Using Lorentz transformations (that is, changing frame of reference) it is possible to put an observer in a situation where only an electric field or just a magnetic induction is present.
From a mathematical point of view the two formulae $2m\mathbf v\times\boldsymbol\omega$ and $q\mathbf v\times\mathbf B$ have deep analogies. Both $\boldsymbol\omega$ and $\mathbf B$ are not vectors, but rather pseudo-vector (this notion stems from the fact that these two objects don't change under parity, contrary to a genuine vector). More precisely I'd say that they are 2-forms that can be represented as "vectors" because of the Hodge duality between $\bigwedge^2 V$ and $V$, where $V$ is a Euclidean space of dimension 3 (here I'm also identifying $V$ with its dual $V^*$).