D3-3. Charged Particles in Magnetic Fields
1. Circular Motion in Magnetic Fields
The Law
- Because the magnetic force is always perpendicular to motion, it acts as a centripetal force, causing the particle to move in a circle.
- Radius of path: $r = \frac{mv}{qB}$
- Charge-to-mass ratio: $\frac{q}{m} = \frac{v}{rB}$
Constant Kinetic Energy (IB Exam Trap)
- The magnetic force $\vec{F}$ is always perpendicular to velocity $\vec{v}$.
- Because the force is perpendicular to motion, it does zero work ($W = Fd\cos(90^\circ) = 0$).
- Crucial Rule: A uniform magnetic field can change a particle's direction, but its speed and kinetic energy stay perfectly constant.
2. Examples
Example 1
Problem: [N18, P1, Q22] A particle of mass $m$ and charge of magnitude $q$ enters a region of uniform magnetic field $B$ that is directed into the page. The particle follows a circular path of radius $R$. What is the sign of the charge of the particle and the speed of the particle?
| Charge of the particle | Speed of the particle | |
|---|---|---|
| A. | positive | $\frac{qBR}{m}$ |
| B. | negative | $\frac{qBR}{m}$ |
| C. | negative | $\sqrt{\frac{qBR}{m}}$ |
| D. | positive | $\sqrt{\frac{qBR}{m}}$ |
Solution:
Step 1: Using the Right-Hand Rule, with fingers pointing into the page and thumb (force) pointing towards the center of the curve, the velocity vector must point opposite to the actual motion for this to work. This means the charge is negative.
Step 2: Equating centripetal and magnetic forces: $qvB = \frac{mv^2}{R}$. Solving for speed $v$ gives $v = \frac{qBR}{m}$.
The correct answer is B.
Example 2
Problem: [N19, P2, Q4] A proton is moving in a region of uniform magnetic field. The magnetic field is directed into the plane of the paper. The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton.
(a) Explain why the path of the proton is a circle.
(b) The speed of the proton is $2.0 \times 10^6\text{ m s}^{-1}$ and the magnetic field strength $B$ is $0.35\text{ T}$.
(i) Show that the radius of the path is about $6\text{ cm}$.
(ii) Calculate the time for one complete revolution.
(c) Explain why the kinetic energy of the proton is constant.
Solution:
(a) The magnetic force is always perpendicular to the velocity of the proton. This provides a constant centripetal force directed towards the center, resulting in circular motion.
(b) (i) Using $r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(1.60 \times 10^{-19})(0.35)} = 0.0596\text{ m} \approx 6\text{ cm}$.
(b) (ii) Time for one revolution is $T = \frac{2\pi r}{v} = \frac{2\pi (0.0596)}{2.0 \times 10^6} = 1.87 \times 10^{-7}\text{ s}$.
(c) The magnetic force acts at $90^\circ$ to the displacement. Since Work = $F d \cos\theta$ and $\cos(90^\circ) = 0$, zero work is done on the proton. With no work done, its kinetic energy ($1/2 mv^2$) remains constant.