Problem 12 - Classical Mechanics

Question

A beam of particles with velocity \((v, 0,0)\) enters a region containing crossed electric and magnetic fields, as in the example at the end of $$\$S$$ \(5.2.\) Show that if the ratio \(E / B\) is correctly chosen the particles are undeviated, while particles with other speeds follow curved trajectories. Suppose the particles have velocities equal to \(v\) in magnitude, but with a small angular dispersion. Show that if the path length \(l\) is correctly chosen, all such particles are focussed onto a line parallel to the \(z\)-axis. (Thus a slit at that point can be used to select particles with a given speed.) For electrons of velocity \(10^{8} \mathrm{~m} \mathrm{~s}^{-1}\) in a magnetic field of \(0.02 \mathrm{~T}\), find the required electric field, and the correct (smallest possible) choice for \(l\).

Step-by-Step Solution

Answer

Answer: The required electric field is \(2\times10^6\) N/C, and the smallest possible path length for the focusing effect to occur is approximately \(1.25\) meters.

Step 1: Equation of Motion Along the x-axis

We start by writing an equation for the motion of the particles as they enter the region with the crossed electric and magnetic fields. Newton's second law states that the net force acting on a particle equals its mass times its acceleration. For a charged particle, the force due to the electric field \(F_E\) is given by \(F_E = qE\), where \(q\) is the charge of the particle and \(E\) is the electric field strength. The force due to the magnetic field \(F_B\) is given by \(F_B = q(v \times B)\), where \(v\) is the velocity vector of the particle and \(B\) is the magnetic field vector. Since the velocity \(v\) is along the x-axis, and the electric field \(E\) and magnetic field \(B\) are crossed, we can say that E is along the y-axis and B is along the z-axis. Therefore, the magnetic force will be along the y-axis too. The net force acting on the particle along the x-axis can be calculated as the sum of the electric field force and the magnetic field force. Therefore, the equation of motion along the x-axis for the particle becomes: \(ma_x = qE - q(v_yB)\).

Step 2: Condition for Undeflected Particles

For the particles to remain undeflected, their acceleration along the x-axis must be zero. This means that the net force acting on them along this axis should be zero. Therefore, we have \(qE - q(v_yB) = 0\). Dividing both sides by \(q\) and rearranging, we get the condition for the undeflected particles: \(E = v_yB\). This tells us that if the ratio of the electric field to the magnetic field is equal to the y-component of the particles' velocity, they will remain undeflected.

Step 3: Motion of Particles with Angular Dispersion

Now, let's consider the particles with velocities equal to \(v\) in magnitude but with a small angular dispersion in their velocities. Their velocity vectors will make an angle \(\theta\) with the x-axis. These particles can be projected onto the y-axis, forming a right triangle with the x-axis as the base. We can thus write their y-component velocity as \(v_y = v\sin{\theta}\), where \(\theta\) is the angle between the velocity vector and the x-axis. Substituting this expression into our previous result, we get \(E = v\sin{\theta} B\).

Step 4: Focused Particles along the z-axis

To find the condition for the particles to be focused onto a line parallel to the z-axis, we need to ensure that their y-component velocity is zero after traveling a distance \(l\) along the x-axis. The time taken by the particles to traverse the path length \(l\) can be calculated as: \(t = \frac{l}{v_x}\). From step 2, we already know that \(v_y = \frac{E}{B}\). When the particles enter the region with crossed electric and magnetic fields, they experience a magnetic force in the negative y-direction. Therefore, the y-component of their velocity will decrease as they move along the x-axis. After traveling a distance \(l\), this y-component must be zero: \(v_y - Ft = 0\). Substituting \(v_y = \frac{E}{B}\) and \(F = qB\), we get \(\frac{E}{B} - qBt = 0\). Substituting the time expression, we have: \(\frac{E}{B} - qB \frac{l}{v_x} = 0\).

Step 5: Calculating the Required Electric Field and Path Length

Now, we can use the given values for the velocity of electrons (\(v = 10^8\) m/s) and the magnetic field strength (\(B = 0.02\) T) to find the required electric field and the smallest possible path length \(l\). First, we find the electric field by substituting the given values in the ratio \(E / B\): \(E = vB = (10^8) (0.02) = 2\times10^6\) N/C. This is the required electric field for the electrons to be undeflected. Next, we find the smallest possible path length \(l\) using the formula from step 4: \(\frac{E}{B} - qB \frac{l}{v_x} = 0\). Rearranging for \(l\) and using the given values and the calculated electric field strength, we get \(l = \frac{E v_x}{qB^2} = \frac{(2\times10^6)(10^8)}{(1.6\times10^{-19})(0.02)^2}\), which yields \(l \approx 1.25\) meters. This is the smallest possible path length for the focusing effect to occur. In summary, the required electric field for the electrons to be undeflected is \(2\times10^6\) N/C, and the smallest possible path length for the focusing effect to occur is approximately \(1.25\) meters.

Key Concepts

Crossed Electric and Magnetic Fields

Imagine a scenario where a charged particle, such as an electron or a proton, moves into a region where an electric field and a magnetic field cross at right angles. The charged particle will experience forces from both fields. The electric field exerts a force that's dependent on the particle's charge and the strength of the field, while the magnetic field affects the particle in a way that depends on both the charge and its velocity.

When these two fields are crossed, meaning they are perpendicular to each other, the forces they exert on a particle can have interesting consequences. If the fields are properly adjusted, a particle moving through this region can essentially 'ignore' these forces, continuing on its path undisturbed, which is known as the undeflected motion condition—a vital concept for specific applications like mass spectrometry or the operation of a cyclotron.

Charged Particle Trajectory

The path of a charged particle through crossed electric and magnetic fields is fascinating. The particle's trajectory isn't just a matter of its initial velocity but also its charge and the strengths of the fields it encounters. In the absence of fields, the particle would simply travel in a straight line. However, the electric field tends to push it in a direction perpendicular to the field lines, and the magnetic field, curiously, forces it to rotate perpendicular to both its velocity and the field lines.

Understanding this trajectory allows scientists and engineers to manipulate the paths of particles for various uses. For example, in a television tube, adjusting the fields can manipulate the electron beams to hit the right spots on the screen to generate an image. Similarly, in particle accelerators, controlling the trajectories is crucial for accelerating particles to high speeds and for steering them along the desired paths within the accelerator structures.

Undeflected Motion Condition

Achieving an undeflected motion condition is akin to finding a perfect balance. For a particle moving through crossed electric and magnetic fields to continue in a straight line, untouched by the forces around it, a precise relationship between the electric field strength (E), the magnetic field strength (B), and the velocity component of the particle perpendicular to these fields must exist. As shown in the textbook solution, the condition for their balance, leading to no deflection, is expressed by the equation
\[ E = v_yB \].

Using this equation, one can calculate the speed of particles that will pass through the region without deviation, which is immensely useful, for example, in the design of velocity filters in physics research labs. By selecting for particles with a specific velocity, experts can filter out unwanted particles and gather only those with the desired qualities.