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 \(l\).

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: \(ma_x = qE - q(v_yB)\).

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.

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 \(E = v\sin{\theta} B\).

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: \(\frac{E}{B} - qB \frac{l}{v_x} = 0\).

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.