Current makes a field that circles it

For a long straight wire carrying current $I$, the field at perpendicular distance $r$ has magnitude

$$B = \frac{\mu_0 I}{2\pi r}, \qquad \mu_0 = 4\pi\times10^{-7}\ \text{T}\cdot\text{m}/\text{A},$$

and its direction comes from the right-hand grip rule: point the thumb along the conventional current, and the curled fingers trace the field lines. Those lines are circles centered on the wire, and $\vec{B}$ is tangent to them at every point. There is no radial component anywhere.

Two habits to break here:

1. Drawing spokes pointing away from the wire, Coulomb-style. The $1/r$ falloff is real, so the magnitude looks familiar; the direction is nothing like a point charge's.
2. Drawing the field lines along the wire, parallel to the current. That kills every force in the problem, because $I\vec{\ell}\times\vec{B}$ vanishes when $\vec{B}$ is parallel to $\vec{\ell}$.

Worth memorizing one concrete case so you can rebuild the rest: for a wire running left to right across the page with current to the right, the field above the wire points out of the page, and below it points into the page.

Note also that this is $1/r$, not $1/r^2$. Coulomb's inverse square is a point-source law; a straight wire is an infinite line of sources, and integrating along it costs one power of $r$. Both fields come from an inverse-square element law, and they land on different exponents because of the geometry.

Every short piece of current contributes

$$d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\hat{r}}{r^2},$$

where $\hat{r}$ points from the element to the field point and $r$ is that element's own distance. Setting one of these up correctly means respecting three things at once.

1. $r$ varies element to element. The closest approach is one element's distance, not the whole wire's. Writing $\mu_0 I L/(4\pi d^2)$, with the total length at the nearest distance, is the point-charge move and it is wrong by a geometry factor that depends on the wire's extent.
2. The cross product aims each contribution. $d\vec{\ell}\times\hat{r}$ is perpendicular to both. It is largest when the element is perpendicular to the line of sight and exactly zero when the element points straight at the field point, since parallel vectors have zero cross product.
3. Add as vectors. For a straight wire every element happens to give $d\vec{B}$ in the same direction, so the integral reduces to a scalar one. For a bent wire it does not, and components come first.

A finite straight segment gives $B = \dfrac{\mu_0 I}{4\pi d}(\sin\theta_2 - \sin\theta_1)$, with the angles measured to the perpendicular from the field point. Let the wire run to infinity in both directions and the bracket becomes $2$, recovering $\mu_0 I/(2\pi d)$. Point 2 above is why a segment gives nothing at a point on its own extended axis: there, every $d\vec{\ell}$ lies along $\hat{r}$.

At the center of a circular loop of radius $R$ carrying current $I$, every element is perpendicular to its own $\hat{r}$ (which points inward along a radius), so every $|d\vec{\ell}\times\hat{r}| = d\ell$, and every contribution points the same way, along the axis. They add:

$$B = \frac{\mu_0}{4\pi}\frac{I}{R^2}\oint d\ell = \frac{\mu_0}{4\pi}\frac{I}{R^2}(2\pi R) = \frac{\mu_0 I}{2R}.$$

Nothing cancels. Cancellation is what happens between two antiparallel straight wires, and a loop is the opposite arrangement: it is the shape you build precisely because its contributions reinforce, which is why coils are wound and not laid flat. An arc subtending angle $\phi$ gives the same integral over a shorter path, $B = \mu_0 I\phi/(4\pi R)$.

Off center, on the axis at distance $z$:

$$B_z = \frac{\mu_0 I R^2}{2\,(R^2 + z^2)^{3/2}}.$$

Set $z = 0$ and it returns $\mu_0 I/(2R)$. Let $z \gg R$ and it goes as $1/z^3$, the signature of a magnetic dipole with moment $\mu = IA$: not $1/r^2$ like a point charge, and not $1/r$ like a wire. The direction of the axial field comes from a second use of the right hand: curl the fingers along the current around the loop, and the thumb points along $\vec{B}$ on the axis.

Two wires, two fields, one point: get each field's direction from its own grip rule before adding anything. At the midpoint between two parallel wires carrying current the same way, the two circular fields point opposite ways and cancel for equal currents. Carrying current in opposite directions, they point the same way and add to twice one of them. Adding the magnitudes $\mu_0 I/(2\pi r)$ as plain numbers gets one of those two cases exactly backward every time.

The force between the wires chains two rules. Wire 1's field at wire 2 is perpendicular to wire 2, and the force on wire 2 is $\vec{F} = I_2\vec{L}\times\vec{B}_1$, which comes out pointing toward wire 1 when the currents run the same way:

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}, \qquad \text{same direction: ATTRACT, opposite: REPEL.}$$

This is the reverse of the charge rule, and pattern-matching to "like repels like" inverts every two-wire and coil-coil problem you will meet. When the reflex fires, run the two-step instead: grip rule for the field, then $I\vec{L}\times\vec{B}$ for the force. It takes fifteen seconds and it is never wrong.

By Newton's third law the two wires pull on each other equally, which the formula already encodes: $I_1 I_2$ is symmetric, so there is no way for it to favor the wire with the larger current.