In Magnetic Force on a Moving Charge you found that the magnetic force is always at right angles to the velocity. In Forces in Circular Motion you met a force like that before: a sideways pull that bends a path into a circle without changing the speed.
Why it works
A force at right angles to the motion does no work, $W = Fd\cos 90^\circ = 0$, so the kinetic energy and the speed never change. The force only turns the velocity. A constant-size force that keeps turning with the velocity is a centripetal force, so the charge moves in a circle: $$|q|vB = \frac{mv^{2}}{r} \quad\Rightarrow\quad r = \frac{mv}{|q|B}.$$
Key idea
A charge moving across a uniform field circles at constant speed with radius and period $$r = \frac{mv}{|q|B}, \qquad T = \frac{2\pi m}{|q|B}.$$ The period does not depend on the speed.
Why is the period fixed? The time for one lap is $T = 2\pi r/v$. Put in $r = mv/(|q|B)$ and the $v$ cancels. A faster particle runs a longer lap at a higher speed. Cyclotrons use this: they kick particles in step with a period that never changes.
Common mistake
Thinking the magnetic force speeds the particle up as it circles. It cannot: it is always at right angles to the motion, so it does no work. Check: the kinetic energy at the end must equal the kinetic energy at the start. Only an electric field can change a charge's speed.
Worked example: A proton in a field
A proton moves at $3.0 \times 10^{6}\ \text{m/s}$ at right angles to a $0.20\ \text{T}$ field. Find the radius of its path.
- $m_p = 1.67 \times 10^{-27}\ \text{kg}$ and $q = 1.60 \times 10^{-19}\ \text{C}$.
- $$\begin{aligned} r &= \frac{mv}{qB} \\ &= \frac{(1.67 \times 10^{-27})(3.0 \times 10^{6})}{(1.60 \times 10^{-19})(0.20)} \\ &= 0.157\ \text{m}. \end{aligned}$$
- About 16 cm. An electron at the same speed has about $\frac{1}{1800}$ of the mass, so its circle would be about 1800 times smaller.
Answer: $r \approx 0.16\ \text{m}$
Worked example: An electron from an accelerator
An electron is accelerated from rest through $100\ \text{V}$ and then enters a $1.0\ \text{mT}$ field at right angles. Find the radius.
- Energy first: it gains $$K = eV = (1.60 \times 10^{-19})(100) = 1.60 \times 10^{-17}\ \text{J}.$$
- Speed: $$v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(1.60 \times 10^{-17})}{9.11 \times 10^{-31}}} = 5.93 \times 10^{6}\ \text{m/s}.$$
- Radius: $$r = \frac{mv}{eB} = \frac{(9.11 \times 10^{-31})(5.93 \times 10^{6})}{(1.60 \times 10^{-19})(1.0 \times 10^{-3})} = 0.0337\ \text{m}.$$
Answer: $r \approx 3.4\ \text{cm}$
Which way does it circle? With $\vec{B}$ into the page, a positive charge goes counterclockwise and a negative charge clockwise. If the velocity has a part along $\vec{B}$, that part is untouched, and the circle stretches into a helix that winds along the field lines. That winding is how magnetic "bottles" hold hot plasma for fusion: try it in the Magnetic Bottles lab.