A proton moving with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.40 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.

Short Answer

Expert verified
Question: Calculate the magnitude of the force on a proton moving with a speed of \(4.0 \cdot 10^{5}\mathrm{m}/\mathrm{s}\) in the positive y-direction entering a uniform magnetic field pointing in the positive x-direction with a strength of \(0.40\mathrm{T}\). Answer: The magnitude of the force acting on the proton is \(2.56 \cdot 10^{-13}\mathrm{N}\).

Step by step solution

01

Identify given values

We are given the following values: - The speed of the proton, \(v = 4.0 \cdot 10^{5}\mathrm{~m} / \mathrm{s}\) - The magnetic field strength, \(B = 0.40\mathrm{~T}\) - The charge of a proton, \(q = 1.60 \cdot 10^{-19}\mathrm{C}\)
02

Use the formula for force

According to the analysis, the force acting on the proton can be calculated using the formula \(F = qvB\). We can plug the given values into the formula to find the force: \(F = (1.60 \cdot 10^{-19}\mathrm{C})(4.0 \cdot 10^{5}\mathrm{~m} / \mathrm{s})(0.40\mathrm{~T})\)
03

Calculate the force

Now we can calculate the force: \(F = (1.60 \cdot 10^{-19}\mathrm{C})(4.0 \cdot 10^{5}\mathrm{~m} / \mathrm{s})(0.40\mathrm{~T}) = 2.56 \cdot 10^{-13}\mathrm{N}\) So the magnitude of the force acting on the proton is \(2.56 \cdot 10^{-13}\mathrm{N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lorentz force
The concept of the Lorentz force is fundamental in understanding how charged particles are influenced when they move within magnetic fields. It's an amalgamation of electric and magnetic force on a point charge due to electromagnetic fields.

When a charged particle, such as a proton, moves through a magnetic field, it experiences a force that is perpendicular to both the velocity of the particle and the direction of the magnetic field. This is known as the magnetic component of the Lorentz force, described mathematically by the equation: \( F = qvB \sin(\theta) \), where \( F \) is the force, \( q \) is the electric charge of the particle, \( v \) is the velocity of the particle, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the direction of the velocity and the magnetic field. For a proton moving perpendicular to the magnetic field, as in the exercise provided, the sine function becomes 1 since \( \theta = 90\degree \) and the force can be simplified to \( F = qvB \). Understanding this force is not only crucial for solving physics problems but also for applications like cyclotrons and mass spectrometry.
Magnetic field
A magnetic field is an invisible field that exerts a force on particles that are charged and on materials that are magnetic. It's represented by the symbol \( B \) and in the International System of Units, is measured in tesla (T).

Visualizing this field can be done through imagining lines that run from the north to the south poles of a magnet. These lines are a way to model the direction and strength of the magnetic field: the closer together they are, the stronger the field. In a uniform magnetic field, like the one described in the exercise, these lines are parallel and evenly spaced, which means the strength and direction of the field are the same at every point. This concept is vital in the study of electromagnetism and is key to understanding how a proton's path is changed when entering such a field, causing it to move in a circular path due to the Lorentz force.
Proton charge
The charge of a proton is a fundamental property that plays a critical role in the study of electromagnetism and quantum mechanics. Every proton carries a positive charge of \( 1.60 \times 10^{-19} \) coulombs, symbolically represented as \( q \).

In our exercise, the proton's charge interacts with the magnetic field to create the Lorentz force that influences the proton's trajectory. It's this very charge that causes the proton to experience a force when it moves through magnetic fields. The proton's charge is also equal in magnitude but opposite in sign to the charge of an electron, making the two particles fundamental opposites when it comes to electrical properties.
Uniform magnetic field
A uniform magnetic field is a field that has the same intensity and direction at any given point. This uniformity simplifies calculations and predictions of a charged particle's behavior within the field. As envisaged in our exercise, the uniform magnetic field has a strength of \( 0.40 T \), and it's directed along the positive x-axis.

In a uniform field, charged particles like protons will move in a circular or helical path, depending on their entry angle. This occurs because the Lorentz force acts as a centripetal force, constantly pulling the proton towards the center of its circular path and keeping it moving in that motion. The concept of a uniform magnetic field is not just a theoretical convenience; it approximates the conditions in many practical applications like particle accelerators and magnetic confinement devices in fusion reactors.

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Most popular questions from this chapter

A particle with a charge of \(+10.0 \mu \mathrm{C}\) is moving at \(300 \cdot \mathrm{m} / \mathrm{s}\) in the positive \(z\) -direction. a) Find the minimum magnetic field required to keep it moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(y\) -direction. b) Find the minimum magnetic field required to keep the particle moving in a straight line at constant speed if there is a uniform electric field of magnitude \(100 . \mathrm{V} / \mathrm{m}\) pointing in the positive \(z\) -direction.

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.

A proton, moving in negative \(y\) -direction in a magnetic field, experiences a force of magnitude \(F\), acting in the negative \(x\) -direction. a) What is the direction of the magnetic field producing this force? b) Does your answer change if the word "proton" in the statement is replaced by “electron"?

A straight wire of length \(2.00 \mathrm{~m}\) carries a current of \(24.0 \mathrm{~A} .\) It is placed on a horizontal tabletop in a uniform horizontal magnetic field. The wire makes an angle of \(30.0^{\circ}\) with the magnetic field lines. If the magnitude of the force on the wire is \(0.500 \mathrm{~N}\), what is the magnitude of the magnetic field?

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{i}\) d) \(2 r_{i}\) e) \(4 r_{\mathrm{i}}\)

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