A magnetic field pushes on a charge only while the charge moves, and only on the part of its motion that cuts across the field. This is the magnetic partner of the electric force you met in The Electric Field, acting in the magnetic fields you have just met, and you will need vector components to split a velocity into parts along and across $\vec{B}$.
Key idea
A charge $q$ moving at speed $v$ at angle $\theta$ to a field $B$ feels $$F = |q|\,vB\sin\theta.$$ The force points at right angles to both $\vec{v}$ and $\vec{B}$. For a positive charge, use the right-hand rule; for a negative charge, the force points the other way.
- Point the fingers of your right hand along $\vec{v}$.
- Curl them toward $\vec{B}$ (through the smaller angle).
- Your thumb points along $\vec{F}$ for a positive charge. Reverse it for a negative charge.
Why it works
This rule is how the field is defined: $B$ is the force per unit charge per unit speed, $B = F/(qv)$ for motion straight across the field. That gives the tesla its size, $1\ \text{T} = 1\ \text{N·s/(C·m)}$. The $\sin\theta$ says that only the part of the velocity across the field, $v\sin\theta$, counts. Motion along the field lines feels nothing.
Common mistake
Two common slips. (1) Thinking the force points along the field lines, as an electric force does. It never does: it is always sideways to $\vec{B}$. (2) Thinking a magnet pushes on a charge at rest. With $v = 0$ there is no magnetic force at all. Check: if your answer has $\vec{F}$ parallel to $\vec{B}$, or a force on a resting charge, start over.
Worked example: A proton crossing a field
A proton moves to the right at $2.0 \times 10^{5}\ \text{m/s}$ through a $0.50\ \text{T}$ field that points into the page. Find the size and direction of the force.
- The velocity is across the field, so $\theta = 90^\circ$ and $\sin\theta = 1$.
- $$\begin{aligned} F &= qvB \\ &= (1.60 \times 10^{-19})(2.0 \times 10^{5})(0.50) \\ &= 1.6 \times 10^{-14}\ \text{N}. \end{aligned}$$
- Direction: fingers to the right, curl them into the page. Your thumb points toward the top of the page.
Answer: $1.6 \times 10^{-14}\ \text{N}$, toward the top of the page
Worked example: An electron at an angle
An electron moves at $3.0 \times 10^{6}\ \text{m/s}$ at $30^\circ$ to a $0.020\ \text{T}$ field. How big is the force?
- Use the size of the charge: $|q| = 1.60 \times 10^{-19}\ \text{C}$.
- $$\begin{aligned} F &= |q|vB\sin\theta \\ &= (1.60 \times 10^{-19})(3.0 \times 10^{6}) \\ &\quad \times (0.020)(0.50) \\ &= 4.8 \times 10^{-15}\ \text{N}. \end{aligned}$$
- If the electron moved along the field ($\theta = 0$), the force would be zero.
Answer: $4.8 \times 10^{-15}\ \text{N}$