Chapter 7: Problem 12
A stream of charged particles with various velocities is projected in a direction at right angles both to an E-field of \(4 \mathrm{~V} \mathrm{~cm}^{-1}\) and a B-ficld of \(10^{-3} \mathrm{~T}\). What is the speed of undeflected particles? What energy have they if they are protons?
Short Answer
Expert verified
If the particles are protons, what is their energy?
Answer: The particles should be projected at a speed of \(4 \times 10^7 \mathrm{m}\mathrm{s}^{-1}\). If the particles are protons, their energy is \(5.6 \times 10^{-13} \mathrm{J}\).
Step by step solution
01
Identify the Forces Acting on the Charged Particles
Since the charged particles are projected at right angles to both the electric (E) and magnetic (B) fields, they experience forces due to both fields. The electric field exerts an electric force given by \(F_e = qE\), and the magnetic field exerts a magnetic force given by \(F_m = qvB\), where \(q\) is the charge of the particle, \(v\) is its velocity, and \(E\) and \(B\) are the magnitudes of the electric and magnetic fields, respectively.
02
Apply the Principle of Equilibrium
For the charged particles to be undeflected, the forces due to the electric and magnetic fields must cancel each other out, or \(F_e = F_m\). From Step 1, we have \(F_e = qE\) and \(F_m = qvB\). Equating the two, we get \(qE = qvB\).
03
Solve for the Speed of the Charged Particles
From the equation in Step 2, we can now solve for the speed \(v\) of the charged particles: \(v = \frac{E}{B}\). Using the given values for E and B, which are \(4 \mathrm{~V}\mathrm{~cm}^{-1} = 4 \times 10^4 \mathrm{~V}\mathrm{~m}^{-1}\) and \(10^{-3} \mathrm{~T}\), respectively, we get \(v = \frac{4 \times 10^4 \mathrm{~V}\mathrm{~m}^{-1}}{10^{-3} \mathrm{~T}} = 4 \times 10^7 \mathrm{m}\mathrm{s}^{-1}\).
04
Calculate the Energy of the Protons
To calculate the energy of the protons, we'll use the formula for kinetic energy: \(E_k = \frac{1}{2}mv^2\), where \(m\) is the mass of a proton, which equals \(m = 1.67 \times 10^{-27} \mathrm{kg}\). Plugging in the values, we get \(E_k = \frac{1}{2}(1.67 \times 10^{-27} \mathrm{kg})(4 \times 10^7 \mathrm{m}\mathrm{s}^{-1})^2 = 5.6 \times 10^{-13} \mathrm{J}\).
So, the particles have a speed of \(4 \times 10^7 \mathrm{m}\mathrm{s}^{-1}\), and if they are protons, they have an energy of \(5.6 \times 10^{-13} \mathrm{J}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electric Field
The electric field is a fundamental concept in physics, indicating the area around a charged particle where forces are exerted on other charges within that field. It’s measured in volts per meter (V/m) and represents the force per unit charge that would act on a positive test charge placed in the field. For example, in the exercise provided, the electric field is set up with a magnitude of \(4 \mathrm{~V} \mathrm{~cm}^{-1}\), which equates to \(4 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}\). When a positively charged particle, such as a proton, is placed within this electric field, it experiences a force in the direction of the field if the field is also positive, and opposite if the field is negative.
This concept plays a crucial role in determining the behavior of charged particles, as it can accelerate or decelerate a charge depending on the direction of the field relative to the particle’s motion. It's the electric field's uniform force that influences the charged particles to move at constant speeds if no other forces are acting upon them.
This concept plays a crucial role in determining the behavior of charged particles, as it can accelerate or decelerate a charge depending on the direction of the field relative to the particle’s motion. It's the electric field's uniform force that influences the charged particles to move at constant speeds if no other forces are acting upon them.
Magnetic Field
The magnetic field, on the other hand, is a vector field surrounding magnets and moving charges and it's described by a vector field where each point in space, and its size is known as magnetic field strength, or magnetic flux density, measured in teslas (T). This exercise features a magnetic field strength of \(10^{-3} \mathrm{~T}\). Magnetic fields can interact with charged particles, but unlike electric fields, the force exerted by a magnetic field on a moving charge is always perpendicular to the direction of the particle's velocity and the magnetic field lines.
This perpendicular force doesn't do work on the particle since it causes no displacement in the direction of the force, but rather it changes the particle's direction. Charged particles in magnetic fields experience a centripetal force that can ultimately lead to circular or helical paths, a fundamental principle that is instrumental in designing devices like cyclotrons and synchrotrons.
This perpendicular force doesn't do work on the particle since it causes no displacement in the direction of the force, but rather it changes the particle's direction. Charged particles in magnetic fields experience a centripetal force that can ultimately lead to circular or helical paths, a fundamental principle that is instrumental in designing devices like cyclotrons and synchrotrons.
Lorentz Force
The Lorentz force is the combined force on a charged particle due to electric and magnetic fields. It is given by the equation \(\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})\), where \(F\) is the force experienced by the particle, \(q\) is its charge, \(E\) is the electric field, \(v\) is the particle's velocity, and \(B\) is the magnetic field. The \(\times\) symbol indicates the cross product, which shows that the force due to the magnetic field is perpendicular both to the velocity of the particle and the magnetic field itself.
In our exercise, charged particles with various velocities moving through perpendicular electric and magnetic fields will remain undeflected when the Lorentz forces cancel each other out. This equilibrium is the principle applied in step 2, which is essential in finding the velocity of particles that aren’t influenced by these fields.
In our exercise, charged particles with various velocities moving through perpendicular electric and magnetic fields will remain undeflected when the Lorentz forces cancel each other out. This equilibrium is the principle applied in step 2, which is essential in finding the velocity of particles that aren’t influenced by these fields.
Kinetic Energy of Protons
Kinetic energy represents the energy that particles possess due to their motion and is given by the equation \(E_k = \frac{1}{2}mv^2\), with \(m\) being mass and \(v\) being velocity. For protons, which are positively charged particles found in the nucleus of an atom, the mass is a constant \(1.67 \times 10^{-27} \mathrm{kg}\).
In the context of our exercise, once the protons reach the speed where the electric and magnetic forces balance each other, they move with constant velocity, maintaining their kinetic energy consistently. The calculation step 4 shows how to find this kinetic energy using the previously found velocity from the equilibrium principle. Here, the kinetic energy is \(5.6 \times 10^{-13} \mathrm{J}\), which is a critical factor when examining the motion of charged particles not just in fields, but also in potential applications like particle accelerators or energy calculations in quantum systems.
In the context of our exercise, once the protons reach the speed where the electric and magnetic forces balance each other, they move with constant velocity, maintaining their kinetic energy consistently. The calculation step 4 shows how to find this kinetic energy using the previously found velocity from the equilibrium principle. Here, the kinetic energy is \(5.6 \times 10^{-13} \mathrm{J}\), which is a critical factor when examining the motion of charged particles not just in fields, but also in potential applications like particle accelerators or energy calculations in quantum systems.