The field only moving charge can feel

The magnetic field $\vec{B}$ is defined by what it does to a moving charge:

$$\vec{F} = q\,\vec{v}\times\vec{B}, \qquad |\vec{F}| = |q|\,v\,B\sin\theta,$$

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. The unit of $B$ is the tesla, $1\ \text{T} = 1\ \text{N}/(\text{A}\cdot\text{m})$, which is enormous: a strong lab magnet is a few tesla, the Earth's field is about $5\times10^{-5}$ T.

Read the cross product literally. $\vec{v}\times\vec{B}$ is perpendicular to $\vec{v}$ AND perpendicular to $\vec{B}$, so:

1. A magnetic force along $\vec{B}$ is impossible. Not unlikely, not rare: the cross product cannot produce it.
2. A magnetic force along $\vec{v}$ is impossible too, which is why the magnetic force never speeds anything up.
3. When $\vec{v}$ is parallel or antiparallel to $\vec{B}$, $\sin\theta = 0$ and the force is zero, not aligned with anything.

The right-hand rule executes the cross product: point the fingers along $\vec{v}$, curl them toward $\vec{B}$, and the thumb gives $\vec{v}\times\vec{B}$. Multiply by $q$ at the end, keeping its sign. Order matters, because $\vec{B}\times\vec{v} = -\vec{v}\times\vec{B}$; running the rule backward gets the direction exactly reversed every single time.

Set $v = 0$ in $\vec{F} = q\vec{v}\times\vec{B}$ and the force is zero, however strong the field. A proton parked between the poles of a $2$ T magnet sits there. This is the cleanest break from electrostatics, where $\vec{F} = q\vec{E}$ acts on a charge whether it moves or not.

Only the component of $\vec{v}$ perpendicular to $\vec{B}$ buys any force. Split the velocity:

$$v_\perp = v\sin\theta \quad \text{(does the work of producing force)}, \qquad v_\parallel = v\cos\theta \quad \text{(produces nothing)}.$$

The reverse of this error is just as common. A current-carrying wire is electrically neutral, and students conclude it feels no magnetic force. It does: the carriers inside it are moving, and summing $q\vec{v}\times\vec{B}$ over all of them gives $d\vec{F} = I\,d\vec{\ell}\times\vec{B}$. Neutrality kills the ELECTRIC force on the wire, not the magnetic one.

What about a paperclip stuck to a magnet, with nothing obviously moving? That is induced magnetization: atomic current loops in the iron aligning with the field and then being pulled by the field's gradient. It is not a force on free charge, and it is not $q\vec{v}\times\vec{B}$ with $v = 0$.

Electric field lines start on positive charge and end on negative charge, because charge is a source. Magnetic field lines do neither. Outside a bar magnet they run from north to south; inside the magnet they continue from south back to north, closing the loop. Every line is a closed curve.

The formal statement is Gauss's law for magnetism:

$$\oint \vec{B}\cdot d\vec{A} = 0 \quad \text{through ANY closed surface.}$$

Compare the electric version, $\oint\vec{E}\cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0$. The zero on the right is the statement that there is no magnetic charge to enclose. Draw a closed surface around one pole of a bar magnet and the net flux is still zero, because every line that enters through the surface leaves through it somewhere else.

The experimental face of this: cut a bar magnet in half and you do not get a loose north and a loose south. You get two shorter magnets, each with both poles. Cut again and again, all the way down to a single electron, and there is still no isolated pole.

Two different readings of a field diagram both go wrong, and they are worth separating.

1. The line direction is not the force direction. The arrow on a field line tells you where $\vec{B}$ points. The force on a moving charge is perpendicular to that arrow. A charge released on a field line does not slide along it toward a pole.
2. Line spacing IS the field strength. Where the lines crowd together, $B$ is large. That part of the electric-field vocabulary carries over unchanged.

What does line direction predict directly? The orientation of a compass needle. A small magnetic dipole in a uniform field feels a torque that rotates it into alignment with $\vec{B}$, and zero net force. That is what "the field points this way" means operationally: it is where the needle settles, not where a charge gets dragged.

Page notation, since almost every problem uses it: a dot means "out of the page toward you", an $\times$ means "into the page away from you". They are the tip and the tail of an arrow. Getting these backward flips every answer in the problem set, so check the convention once at the top of the page and then trust it.