/*! 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 58 The force on a charged particle ... [FREE SOLUTION] | 91Ó°ÊÓ

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The force on a charged particle moving in a magnetic field can be computed as the vector sunt of the forces due to each separate component of the magnetic field. As an example, a particle with charge \(q\) is moving with speed \(v\) in the \(-y\) -direction. It is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=\overrightarrow{\boldsymbol{B}}_{\boldsymbol{x}} \hat{\boldsymbol{i}}+\boldsymbol{B}_{\boldsymbol{y}} \hat{\boldsymbol{j}}+\boldsymbol{B}_{\boldsymbol{z}} \hat{\boldsymbol{k}}\) (a) What are the components of the force \(\overrightarrow{\boldsymbol{F}}\) exerted on the particle by the magnetic field? (b) If \(q>0,\) what must the signs of the components of \(\overrightarrow{\boldsymbol{B}}\) be if the components of \(\overrightarrow{\boldsymbol{F}}\) are all nonnegative? (c) If \(q<0\) and \(B_{x}=B_{y}=B_{z}>0,\) find the direction of \(\vec{F}\) and find the magnitude of \(\vec{F}\) in terms of \(|q|, v,\) and \(B_{x}\) .

Short Answer

Expert verified
(a) \(F_x = qvB_z\), \(F_y = 0\), \(F_z = qvB_x\); (b) \(B_x, B_z \geq 0\); (c) Direction: \(-\hat{i} - \hat{k}\), Magnitude: \(|q|vB_x\sqrt{2}\).

Step by step solution

01

Understanding the Force on a Charge in a Magnetic Field

The magnetic force \( \overrightarrow{\boldsymbol{F}} \) on a charged particle moving in a magnetic field is given by the cross product \( \overrightarrow{\boldsymbol{F}} = q \cdot (\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \). Here, \( \overrightarrow{\boldsymbol{v}} = -v \hat{\boldsymbol{j}} \) because the particle moves in the \(-y\)-direction and \( \overrightarrow{\boldsymbol{B}} = B_x \hat{\boldsymbol{i}} + B_y \hat{\boldsymbol{j}} + B_z \hat{\boldsymbol{k}} \). We calculate the components separately.
02

Calculating the Cross Product

To calculate \( \overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}} \), we use the cross product formula:\[\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & -v & 0 \B_x & B_y & B_z \\end{vmatrix} = (vB_z)\hat{i} - (0)\hat{j} + (vB_x)\hat{k} = vB_z \hat{i} + vB_x \hat{k}\]
03

Write the Force Expression

From the cross product, the magnetic force is:\[ \overrightarrow{\boldsymbol{F}} = q(vB_z \hat{i} + vB_x \hat{k}) = qvB_z \hat{i} + qvB_x \hat{k} \]This gives us the force components:\(F_x = qvB_z\), \(F_y = 0\), \(F_z = qvB_x\).
04

Determine Conditions for Nonnegative Force Components (q > 0)

If \( q > 0 \), then both \( F_x = qvB_z \) and \( F_z = qvB_x \) need to be nonnegative. This implies that both \( B_z \geq 0 \) and \( B_x \geq 0 \). Therefore, the signs of the components of \( \overrightarrow{\boldsymbol{B}} \) must be nonnegative for the components of \( \overrightarrow{\boldsymbol{F}} \) to be nonnegative.
05

Force Direction and Magnitude when q < 0 and B_x = B_y = B_z > 0

When \( q < 0 \) and \( B_x = B_y = B_z > 0 \), the direction of \( \overrightarrow{\boldsymbol{F}} \) can be determined by the negative of the cross product found earlier, pointing in the direction \(-\hat{i} - \hat{k}\). The magnitude is:\[ |\overrightarrow{\boldsymbol{F}}| = |-qvB_z \hat{i} - qvB_x \hat{k}| = |q| \sqrt{(vB_z)^2 + (vB_x)^2} = |q|vB_x \sqrt{2} \]

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Key Concepts

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

charged particle
When we say a "charged particle," we are referring to any tiny object that carries an electric charge. This charge can be positive, like a proton, or negative, like an electron. In a magnetic field, a charged particle experiences a force that depends on its velocity and the magnetic characteristics of the field. The particle's charge and its motion influence the strength and the direction of this force.

In the case given in the exercise, the charged particle is moving in the "-y" direction. This movement means it has a velocity vector pointing along the negative y-axis. It's essential to know the sign of the charge because it determines the direction in which the magnetic force will act. A positive charge would experience a force in one direction, while a negative charge would experience it in the opposite direction.

The motion of charged particles in magnetic fields is a fundamental phenomenon in physics, crucial for understanding concepts like electromagnetism, and it plays a significant role in a range of applications, from electric motors to particle accelerators.
cross product
The cross product is a method used in vector mathematics to find a vector that is perpendicular to two other vectors. This operation is crucial when calculating the magnetic force acting on a moving charged particle. The magnetic force is given by the cross product formula: \[ \overrightarrow{\boldsymbol{F}} = q (\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \]

Here, \( \overrightarrow{\boldsymbol{v}} \) is the velocity vector of the particle, and \( \overrightarrow{\boldsymbol{B}} \) represents the magnetic field vector. This calculation involves using determinants to solve for each component of the force vector. The cross product will yield a new vector that points in a perpendicular direction to both \( \overrightarrow{\boldsymbol{v}} \) and \( \overrightarrow{\boldsymbol{B}} \).

In our exercise, the cross product produces components in the \( \hat{i} \) and \( \hat{k} \) directions because the velocity vector is along \( \hat{j} \) and the magnetic field vector has components in all three directions. This specific arrangement maximizes certain force components while nullifying others. Understanding the cross product is essential for visualizing and predicting the behavior of charged particles in magnetic fields.
magnetic field components
The magnetic field is defined as a vector field, meaning it has both a magnitude and a direction in space. It can be decomposed into components along the x, y, and z axes, often represented by \( B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \). These components help in understanding the influence of the magnetic field on a charged particle.

In problems like the one given, knowing the components is crucial because the interaction with the velocity vector of a charged particle is not uniform in all directions. Each component of the magnetic field can influence a different aspect of a particle's movement, contributing to the overall magnetic force experienced by the particle.

For example, in the exercise, it's observed that certain field components contribute to nonzero force components, especially if both the velocity vector and magnetic field have component overlap. This helps explain why \( F_y = 0 \) in our problem: the y-component of both force vectors cancels out. Hence, students must identify and understand these components to solve such physics problems.
vector sum
Vector sum refers to the process of adding two or more vectors together, creating a new vector. In physics, the magnetic force on a charged particle often results from summing the effects due to each magnetic field component. You perform this operation component-wise, meaning you add corresponding components from different vectors.

In our exercise, we sum the contributions from each component of the magnetic field. This vector sum helps determine the total force acting on the particle in three-dimensional space, represented as \( F_x \hat{i} + F_y \hat{j} + F_z \hat{k} \).

Because force is a vector, its direction and magnitude depend on all contributing factors—velocity, magnetic field orientation, and charge sign. If the field has both positive and negative influences, the vector sum effectively shows how these compete or complement each other, leading to the final force experienced by the particle.

Understanding vectors and their summation is essential in physics because it allows one to predict accurately how various forces combine and interact with one another.

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Most popular questions from this chapter

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2\(\Omega\) . When the motor is running at full load on a \(120-\mathrm{V}\) line, the emf in the rotor is 105 \(\mathrm{V}\) . (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

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