Question

A velocity selector is used in a mass spectrometer to produce a beam of charged particles with uniform velocity. Suppose the fields in a selector are given by \(\vec{E}=\left(1.749 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x}\) and \(\vec{B}=(46.23 \mathrm{mT}) \hat{y} .\) Find the speed with which a charged particle can travel through the selector in the \(z\) -direction without being deflected.

Short answer

Answer: The required speed for the charged particle to travel without being deflected is 378 m/s.

Step-by-step solution

Relate electric force and magnetic force

The forces acting on a charged particle in electric and magnetic fields are given by \(\vec{F_e}=q\vec{E}\) and \(\vec{F_m}=q\vec{v} \times \vec{B}\), respectively, where \(q\) is the charge of the particle and \(\vec{v}\) is its velocity. Since the charged particle is not deflected, the electric force and magnetic force must be equal to each other. As a result, we have the equation: \(q\vec{E} = q(\vec{v} \times \vec{B})\).

Simplify the equation

Simplify the equation by dividing both sides by \(q\) and plugging in the given values for \(\vec{E}\) and \(\vec{B}\). We have: \(\vec{E} = \vec{v} \times \vec{B}\) \(\left(1.749 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x} = v_z(\hat{z} \times (46.23 \mathrm{mT}) \hat{y})\)

Calculate the components of the cross product and find \(v_z\)

In order to solve for the z-direction velocity \(v_z\), we need to compute the cross product \(\vec{v} \times \vec{B}\) and compare it with the electric field vector. The only component of the cross product that can potentially match the electric field vector is the x-component. Writing \(\vec{v} = v_z \hat{z}\), we can calculate the x-component of the cross product as follows: \((v_z \hat{z} \times (46.23 \mathrm{mT}) \hat{y})_x = v_z B_y\) Since the x-components must be equal, we can set up the following equation: \(1.749 \cdot 10^{4} \mathrm{~V} / \mathrm{m} = v_z (46.23 \mathrm{mT})\) Now, solve for \(v_z\): \(v_z = \frac{1.749 \cdot 10^{4} \mathrm{~V} / \mathrm{m}}{46.23 \mathrm{mT}}\)

Convert units and calculate final speed

Convert 46.23 mT into tesla (T) by dividing it by 1000: \(B_y = \frac{46.23 \mathrm{mT}}{1000} = 0.04623 \mathrm{T}\) Now, plug the value of \(B_y\) into the equation and solve for \(v_z\): \(v_z = \frac{1.749 \cdot 10^{4} \mathrm{~V} / \mathrm{m}}{0.04623 \mathrm{T}} = 378 \mathrm{m/s}\) Therefore, the speed with which a charged particle can travel through the selector in the z-direction without being deflected is 378 m/s.