Direction of magnetic field around a straight current carrying wire I am currently studying for O Level and I have been learning about electromagnetism for the past year. The topic has never clicked for me because I've never been taught the true relationship between electricity and magnetism. After looking online, I understand how electromagnetism and magnetism occur on their own (or at least enough to find some closure).
The thing that is still bugging is: why does the magnetic field around a straight current carrying wire 'rotate' a certain direction and not the other? To be specific, I want to understand its direction, not why it's circular. Is it just convention? I know it has something to do with cross-products but not how it applies here. 
I know a lot of the topic is beyond my scope but I want to understand why this apparent 'asymmetry' occurs for the time being to rest my curiosity. Thanks for the any help.
 A: The “right-hand rule” for cross products is just a convention. Physics works fine if you adopt a “left-hand rule” instead. The direction of the magnetic field around a straight wire cannot be directly measured and is dependent on the right-hand convention. All that can be measured is how that field makes charged particles accelerate, and this involves two cross products (the other being in the Lorentz force), so it is independent of the convention.
A: There is no asymmetry!
Or, the asymmetry is a result of using the magnetic vector $\mathbf b$, instead of the magnetic bivector $B$. The former is a directed line segment; the latter is a directed plane segment; they're perpendicular to each other.
$$\mathbf b=b_x\mathbf e_x+b_y\mathbf e_y+b_z\mathbf e_z$$
$$B=-b_x\mathbf e_y\mathbf e_z-b_y\mathbf e_z\mathbf e_x-b_z\mathbf e_x\mathbf e_y$$
Here $\mathbf e_x$ is the unit vector along the $x$-axis, and similarly for $y$ and $z$. And $\mathbf e_x\mathbf e_y$ is the unit bivector in the $x,y$-plane. (Actually there are two; the other is $\mathbf e_y\mathbf e_x=-\mathbf e_x\mathbf e_y$.) The vectors are being multiplied with the geometric product, which is associative, and satisfies $\mathbf v\mathbf v=\lVert\mathbf v\rVert^2$ for any vector $\mathbf v$. Multiplying a vector with a bivector produces two parts: another vector, called $\mathbf v\cdot B$; and a trivector, called $\mathbf v\wedge B$:
$$\mathbf vB=(v_x\mathbf e_x+v_y\mathbf e_y+v_z\mathbf e_z)(-b_x\mathbf e_y\mathbf e_z-b_y\mathbf e_z\mathbf e_x-b_z\mathbf e_x\mathbf e_y)$$
$$=(v_yb_z-v_zb_y)\mathbf e_x+(v_zb_x-v_xb_z)\mathbf e_y+(v_xb_y-v_yb_x)\mathbf e_z\\+(-v_xb_x-v_yb_y-v_zb_z)\mathbf e_x\mathbf e_y\mathbf e_z$$
$$\mathbf v\cdot B=(v_yb_z-v_zb_y)\mathbf e_x+(v_zb_x-v_xb_z)\mathbf e_y+(v_xb_y-v_yb_x)\mathbf e_z$$
$$=\mathbf v\times_3\mathbf b$$
$$\mathbf v\wedge B=(-v_xb_x-v_yb_y-v_zb_z)\mathbf e_x\mathbf e_y\mathbf e_z$$
$$=(-\mathbf v\cdot\mathbf b)I$$
Here $I=\mathbf e_x\mathbf e_y\mathbf e_z$ is the right-handed unit trivector (the left-handed one is $-I$), and $\times_3$ is the vector cross product which only works in 3D. (In contrast, geometric algebra works in any number of dimensions.) We also have $IB=\mathbf b$, an instance of Hodge duality.
So the Lorentz force $q\,\mathbf v\times_3\mathbf b$ can be replaced with $q\,\mathbf v\cdot B$. Geometrically, this is the projection of $\mathbf v$ onto the $B$ plane, rotated $90^\circ$ in that plane, and scaled by $q\lVert B\rVert$.
Here's a 3D illustration of the magnetic bivector field around a wire with current flowing downward. The magnitude of $B$ is represented by the area of the disk, which decreases with distance from the wire. The plane of $B$ always contains the wire. There is rotational symmetry, and mirror symmetry.


More information: http://www.av8n.com/physics/clifford-intro.htm , https://en.wikipedia.org/wiki/Geometric_algebra
