Question

The velocity selector described in Solved Problem 27.2 is used in a variety of devices to produce a beam of charged particles of uniform velocity. Suppose the fields in such a selector are given by \(\vec{E}=\left(1.00 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x}\) and \(\vec{B}=(50.0 \mathrm{mT}) \hat{y} .\) Find the velocity in the \(z\) -direction with which a charged particle can travel through the selector without being deflected.

Short answer

Answer: The velocity in the \(z\)-direction required for a charged particle to travel through the velocity selector without being deflected is \(-200 \mathrm{ m/s}\).

Step-by-step solution

Write the Lorentz force equation

The Lorentz force equation is given by \(\vec{F}=q(\vec{E}+\vec{v} \times \vec{B})\), where \(q\) is the charge of the particle, \(\vec{E}\) is the electric field, \(\vec{v}\) is the velocity of the particle, and \(\vec{B}\) is the magnetic field.

Find the condition for no deflection

The particle will not be deflected if the net force acting on it is zero. Therefore, we need to find the condition for which the Lorentz force is zero: \(q(\vec{E}+\vec{v} \times \vec{B}) = 0\).

Expand the cross product and solve for the velocity in the \(z\)-direction

We are given the electric field \(\vec{E}=\left(1.00\cdot 10^{4}\mathrm{~V} / \mathrm{m}\right) \hat{x}\) and magnetic field \(\vec{B}=(50.0\mathrm{ mT})\hat{y}\). So, we have: \(\vec{v} = v_x \hat{x} + v_y \hat{y} + v_z \hat{z}\) Now, let's compute the cross product \(\vec{v} \times \vec{B}\): \(\vec{v} \times \vec{B} = (50.0\mathrm{ mT})(v_z \hat{x} - v_x \hat{z})\) Now, we can substitute \(\vec{E}\) and \(\vec{v} \times \vec{B}\) into the Lorentz force equation: \(q(\left(1.00\cdot 10^{4}\mathrm{ V}\mathrm{/ m}\right) \hat{x} + (50.0\mathrm{ mT})(v_z \hat{x} - v_x \hat{z})) = 0\) We can equate the components along the \(x\) and \(z\) directions to solve for \(v_z\): \(q(1.00\cdot 10^{4}\mathrm{ V}\mathrm{/ m}) + q(50.0\mathrm{ mT})v_z = 0 \Rightarrow v_z = -\frac{1.00\cdot 10^{4} \mathrm{ V} \mathrm{/ m}}{50.0\mathrm{ mT}}\)

Calculate the value of \(v_z\)

Using the given values of the electric and magnetic fields: \(v_z = -\frac{1.00\cdot 10^{4} \mathrm{ V} \mathrm{/ m}}{50.0\mathrm{ mT}} = -\frac{1.00\cdot 10^{4} \mathrm{ V} \mathrm{/ m}}{50.0\cdot 10^{-3} \mathrm{ T}}\) \(v_z = -200 \mathrm{ m/s}\) So, the velocity in the \(z\)-direction with which a charged particle can travel through the selector without being deflected is \(-200 \mathrm{ m/s}\).

Key concepts

Velocity Selector

A velocity selector is a device used to filter out charged particles based on their velocities, widely used in physics experiments and applications such as mass spectrometry. It consists of perpendicular electric and magnetic fields, typically represented as \(\vec{E}\) and \(\vec{B}\), respectively. The fundamental operation of a velocity selector rests on the principle that only charged particles with a specific velocity, which is the result of the ratio of these fields, can pass through without deflection.

When a charged particle with velocity \(\vec{v}\) enters the velocity selector, it experiences the Lorentz force due to both fields. If the velocity of the particle is such that the electric force \(q\vec{E}\) is exactly balanced by the magnetic force \(q\vec{v} \times \vec{B}\), the particle continues in a straight path. Any charged particle moving slower or faster will be deflected and filtered out. Therefore, a velocity selector can be used to select particles with the desired velocity from a beam of particles with mixed velocities.

Electric Field

An electric field is a region of space around a charged particle or object within which another charged particle will experience a force. It is a vector field, typically represented by \(\vec{E}\), and can be visualized as imaginary lines extending outward from a positive charge and converging in toward a negative charge. The direction of the electric field at any point is the direction of the force that a positive test charge would experience at that point.

The strength of the electric field is measured in volts per meter (V/m) and is a measure of how strong the force is on a charge within that field. In the context of the velocity selector, the electric field is responsible for exerting a force on the charged particles that is proportional to their charge and the strength of the electric field. Thus, a particle with charge \(q\) in an electric field \(\vec{E}\), is subject to a force \(q\vec{E}\).

Magnetic Field

A magnetic field is a vector field that exerts a force on moving charges, such as electrons or other charged particles. It is often represented by the symbol \(\vec{B}\). This field is characterized by both a magnitude and a direction, and it can be visualized by the pattern of iron filings that align themselves along the field lines around a magnet.

The strength of a magnetic field is measured in teslas (T) or, in smaller units, milliteslas (mT). Importantly, the force a charged particle experiences in a magnetic field is always perpendicular to both the direction of the magnetic field and the velocity of the particle. This is known as the magnetic force and can be mathematically expressed using the cross product \(q\vec{v} \times \vec{B}\). The perpendicular nature of this force is key to many applications, including the velocity selector, where it works in conjunction with the electric field to filter particles by velocity.

Charged Particle Motion

The motion of charged particles in electric and magnetic fields is a fundamental concept in electromagnetism, governed by the Lorentz force law. According to this law, a charged particle with charge \(q\) moving at velocity \(\vec{v}\) in an electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) will experience a force, \(\vec{F}\), given by \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\).

For a particle to pass through a velocity selector without deflection, this force must be zero. That happens when the electric force, \(q\vec{E}\), and the magnetic force, \(q\vec{v} \times \vec{B}\), are equal in magnitude but opposite in direction, thus canceling each other out. This precise balance depends on the velocity component of the particle in the direction perpendicular to both fields. In many applications, understanding and predicting the motion of charged particles enables the development of various instruments and technologies, ranging from particle accelerators to everyday electronics.