D3-4. Charged Particles in Electric and Magnetic Fields

Example 1: Deflection in an Electric Field

Problem: A single proton travelling with a constant horizontal velocity enters a uniform electric field between two parallel charged plates. Describe the path taken by a Boron nucleus (Atomic number = 5, Mass number = 11) that enters the electric field at the same point and with the same velocity as the proton.

Solution:

Step 1 (Compare Charges): Boron has 5 protons, meaning its charge is $5\times$ greater than the proton. The electrical force $F$ pulling it will be $5\times$ greater ($F = qE$).
Step 2 (Compare Masses): The Boron nucleus has 11 nucleons, meaning its mass $m$ is roughly $11\times$ greater than the proton.
Conclusion: The acceleration is $a = \frac{qE}{m}$. Since the mass factor ($11\times$) is much greater than the force factor ($5\times$), its vertical acceleration will be roughly half that of the proton. Because it accelerates slower vertically, the Boron nucleus will be less deflected and hit the plate further down the line.

Example 2: Balanced Crossed Fields

Problem: An electron passes between two parallel metal plates moving with a constant velocity of $2.1 \times 10^7\text{ m s}^{-1}$. The potential difference between the plates is $3100\text{ V}$. A uniform magnetic field of magnitude $0.054\text{ T}$ acts perpendicular to the electric field and the movement of the electron to perfectly balance the forces. Calculate the separation of the plates.

Solution:

Step 1: Because the forces are balanced and velocity is constant, $E = vB$.
$$E = (2.1 \times 10^7\text{ m s}^{-1}) \times (0.054\text{ T}) = 1.134 \times 10^6\text{ N/C}$$
Step 2: Use the electric field between parallel plates equation $E = \frac{V}{d}$ to solve for $d$.
$$d = \frac{V}{E} = \frac{3100\text{ V}}{1.134 \times 10^6\text{ N/C}} = \mathbf{2.73 \times 10^{-3}\text{ m}}$$

Example 3: Particle in a Magnetic Field

Problem: 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?

Solution:

Step 1 (Speed): The magnetic force provides the necessary centripetal force for the particle to move in a circle. We can equate the two formulas:
$$qvB = \frac{mv^2}{R}$$
Canceling one $v$ from both sides and rearranging for speed ($v$) gives:
$$v = \frac{qBR}{m}$$
This immediately eliminates options C and D.

Step 2 (Sign of Charge): Apply the Right-Hand Rule. Point your fingers into the page (the direction of the magnetic field, $B$) and your thumb to the right (the initial direction of the velocity vector $v$). Your palm indicates the direction of the magnetic force for a positive charge, which would be upwards.

Because the diagram clearly shows the particle curving downwards—in the opposite direction out the back of your hand—the charge must be negative. Therefore, the correct combination is a negative sign and speed $\frac{qBR}{m}$. The answer is B.

Example 4: Proton in a Uniform Magnetic Field

Problem: 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.

1. Explain why the path of the proton is a circle.
2. The speed of the proton is $2.0 \times 10^6\text{ m s}^{-1}$ and the magnetic field strength $B$ is $0.35\text{ T}$.
   1. Show that the radius of the path is about $6\text{ cm}$.
   2. Calculate the time for one complete revolution.
3. Explain why the kinetic energy of the proton is constant.

Solution:

(a) The magnetic force acting on the moving proton is always perpendicular to its velocity vector. Because the force is perpendicular to motion, it changes the direction of the proton but not its speed. A constant force acting perpendicular to a constant speed provides a centripetal force, resulting in uniform circular motion.

(b)(i) Start by equating the magnetic force to the centripetal force: $qvB = \frac{mv^2}{r}$. Rearranging for the radius $r$ gives: $$r = \frac{mv}{qB}$$ Substitute the known values for a proton (mass $m = 1.67 \times 10^{-27}\text{ kg}$, charge $q = 1.60 \times 10^{-19}\text{ C}$): $$r = \frac{(1.67 \times 10^{-27}\text{ kg})(2.0 \times 10^6\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C})(0.35\text{ T})}$$ $$r \approx 0.0596\text{ m}$$ Converting to centimeters, $r \approx 5.96\text{ cm}$, which is approximately $\mathbf{6\text{ cm}}$.

(b)(ii) The time for one complete revolution is the period ($T$). Using the formula for constant speed in circular motion ($v = \frac{2\pi r}{T}$), we can rearrange for $T$: $$T = \frac{2\pi r}{v}$$ $$T = \frac{2\pi (0.0596\text{ m})}{2.0 \times 10^6\text{ m s}^{-1}} \approx \mathbf{1.87 \times 10^{-7}\text{ s}}$$

(c) Work done by a force is given by $W = Fd \cos(\theta)$. Because the magnetic force is always at a $90^\circ$ angle to the displacement of the proton, $\cos(90^\circ) = 0$, meaning the magnetic field does zero work on the proton. According to the work-energy theorem ($W = \Delta KE$), if no work is done, the change in kinetic energy is zero, meaning the kinetic energy remains constant.