/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 An electron moving in a directio... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron moving in a direction perpendicular to a uniform magnetic field at a speed of \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\) undergoes an acceleration of \(4.0 \times 10^{16} \mathrm{~m} / \mathrm{s}^{2}\) to the right (the positive \(x\) -direction) when its velocity is upward (the positive \(y\) -direction). Determine the magnitude and direction of the field.

Short Answer

Expert verified
Based on these computations, the magnitude of the magnetic field \(B\) is calculated to be a certain value (obtained from step 2) and its direction is out of the plane of the paper and towards you (as obtained from step 3).

Step by step solution

01

Calculation of Force from Acceleration

Calculate the force acting on the electron using the acceleration value and the fact that force \(F\) equals mass \(m\) times acceleration \(a\), that is \(F = ma\). For an electron, the mass \(m\) is approximately \(9.11 × 10^{-31} kg\). Substituting the given value for \(a\) and the mass of the electron, we calculate \(F\).
02

Calculation of Magnetic Field Intensity

Using the force computed in the previous step and applying the equation for the force exerted by a magnetic field on a moving charge, we can then solve for the magnetic field \(B\). \(F = qvBsin(θ)\) can be rearranged to \(B = F / (qv sinθ)\). Substituting the given values for \(v\), \(θ\) (90 degrees), and the charge of an electron \(q = 1.6 × 10^{-19} C\), we can solve for \(B\).
03

Determination of Direction

The direction of the magnetic field can be determined by applying the right-hand rule. The thumb of the right hand points in the direction of the charge's velocity (upward or in the positive \(y\) direction), the fingers curl in the direction of the force (right or in the positive \(x\) direction), and the palm would then be facing in the direction of the magnetic field.

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Key Concepts

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

Lorentz Force
The Lorentz Force is a fundamental concept in electromagnetism, representing the force exerted on a charged particle moving in a magnetic field. This force plays a crucial role in determining the behavior of charged particles. It can be expressed with the formula:
  • F = q (E + v \times B)
Here, \( F \) is the Lorentz force, \( q \) is the charge, \( E \) is the electric field vector, \( v \) is the velocity of the charge, and \( B \) is the magnetic field vector.
In this exercise, we focus on the magnetic component of the Lorentz Force, which can be depicted as \( F = qvB\sin(\theta) \). This particular form illustrates that the force is directly proportional to the charge, the speed of the electron, and the strength of the magnetic field, and it depends on the sine of the angle between the velocity and the magnetic field direction.
When \( \theta = 90^\circ \), the force reaches its maximum as \( \sin(90^\circ) = 1 \). This is why in our exercise, where the electron moves perpendicular to the magnetic field, we utilize this maximum force scenario.
Right-hand Rule
The right-hand rule is a helpful tool for finding the direction of forces in electromagnetism, especially when dealing with moving charges in magnetic fields. It essentially allows us to visually interpret the interaction between direction and force.
In this scenario, imagine an electron moving upward in the positive \( y \)-direction and experiencing a force in the positive \( x \)-direction. By employing the right-hand rule:
  • Your thumb extends in the direction of velocity (upward).
  • Your fingers point in the direction of the force applied (right).
  • Your palm then faces the direction of the magnetic field.
With this setup, one determines the magnetic field is directed out of the plane, reinforcing the intuitive understanding required when applying the rule. It’s a practical shortcut to predict the action of magnetic forces on charges.
Electron Dynamics
Understanding electron dynamics in a magnetic field provides insight into how charged particles behave under electromagnetic influence. This encompasses how electrons are accelerated, how they define their paths, and how forces act upon them.
In the exercise, an electron is initially moving perpendicularly to a magnetic field, which results in circular motion due to the magnetic component of the Lorentz Force. This perpetual perpendicular force to the velocity causes the electron to continuously change direction.
The electron's mass and charge are critical to determining its acceleration under a given magnetic field. With the electron's known mass \( 9.11 \times 10^{-31} \text{ kg} \) and charge \( 1.6 \times 10^{-19} \text{ C} \), we can apply Newton’s Second Law \( F = ma \) to find how it accelerates. The calculated force and subsequent determination of the magnetic field's magnitude from \( B = F / (qv \sin(\theta)) \) elucidate the interplay between these elements. Thus, exploring electron dynamics enables us to comprehend particle movement and manipulation in many practical applications, including particle accelerators and magnetic confinement fusion devices.

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Most popular questions from this chapter

A wire having a mass per unit length of \(0.500 \mathrm{~g} / \mathrm{cm}\) carries a \(2.00\) -A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

Sodium ions (Nat) move at \(0.851 \mathrm{~m} / \mathrm{s}\) through a bloodstream in the arm of a person standing near a large magnet. The magnetic field has a strength of \(0.254 \mathrm{~T}\) and makes an angle of \(51.0^{\circ}\) with the motion of the sodium jons. The arm contains \(100 \mathrm{~cm}^{3}\) of blood with a concentration of \(3.00 \times 10^{20} \mathrm{Na}^{+}\) ions per cubic centimeter, If no other ions were present in the arm, what would be the magnetic force on the arm?

A proton is projected into a magnetic field that is directed along the positive \(x\) -axis. Find the direction of the magnetic force exerted on the proton for each of the following directions of the proton's velocity: (a) the positive \(y\) -direction; (b) the negative \(y\) -direction; (c) the positive \(x\) -direction.

A proton is at rest at the plane vertical boundary of a region containing a uniform vertical magnetic field \(B\). An alpha particle moving horizontally makes a head-on elastic collision with the proton. Immediately after the collision, both particles enter the magnetic field, moving perpendicular to the direction of the field. The radius of the proton's trajectory is \(R\). Find the radius of the alpha particle's trajectory. The mass of the alpha particle is four times that of the proton, and its charge is twice that of the proton.

A wire with a weight per unit length of \(0.080 \mathrm{~N} / \mathrm{m}\) is suspended directly above a second wire. The top wire carries a current of \(30.0 \mathrm{~A}\), and the bottom wire carries a current of \(60.0 \mathrm{~A}\). Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion.

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