A field that circles the wire

Around a long straight wire, the field lines are circles centred on the wire. There is no component toward the wire, away from it, or along it.

Get the direction by gripping the wire with your right hand, thumb along the conventional current; your fingers curl the way the field runs. So for a current flowing to the right on the page:

1. Above the wire, the field points out of the page.
2. Below the wire, the field points into the page.

Drawing spokes radiating outward is importing the picture of a charged rod, and it breaks every force and flux problem downstream, because a force computed from a field pointing the wrong way lands in the wrong direction. The falloff is usually the part students get right; the direction is the part that costs them.

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

So $B \propto I/r$. Doubling the current doubles the field; doubling the perpendicular distance halves it. Moving from $4$ cm to $2$ cm multiplies the field by $2$, not by $4$.

Inverse square belongs to sources whose influence spreads over a sphere, so it thins out in two directions at once. A long wire's field spreads around a circle, which costs only one power of $r$. That is worth knowing as a reason rather than as a fact to memorise, because it tells you which sources get which exponent.

It is also directly checkable: measured $B$ plotted against $1/r$ falls on a straight line through the origin, and the slope gives the current. Plotted against $1/r^2$ it does not.

Two parallel wires carrying current in the same direction attract. Antiparallel currents repel. That reverses the rule you already own for charges, so matching the electrostatic pattern flips the answer in every two-wire problem.

The result comes from applying two right-hand rules in sequence, and that is what a free-response answer has to show:

1. Rule one: find the field wire 1 creates at wire 2's location. For two vertical wires with current upward, wire 1's field at wire 2 points into the page.
2. Rule two: find the force that field exerts on wire 2's current, using $\vec{F}$ perpendicular to both the current direction and $\vec{B}$. An upward current in an into-the-page field is pushed back toward wire 1.

Run the same two steps on wire 1 and it is pushed toward wire 2, as Newton's third law requires. Reverse one current and both steps flip once, so the pair repels.

The force on a current-carrying wire in an external field is

$$F = BIL\sin\theta,$$

and both of the last two symbols are constrained by the geometry of the setup.

1. $L$ counts only the length actually inside the field region. A $50$ cm wire passing through a $10$ cm magnet gap contributes $0.10$ m, not $0.50$ m. The leads running outside the magnet feel nothing.
2. $\theta$ is the angle between the current direction and $\vec{B}$. Parallel gives zero force; perpendicular gives the maximum.

Both conditions are geometric, so settle them off the diagram before any numbers go in. A diagram tends to hide the parallel case in particular, since a wire lying along the field looks perfectly well placed and feels nothing at all.