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Chapter 27: Problem 4

A particle with mass \(1.81 \times 10^{-3} \mathrm{~kg}\) and a charge of \(1.22 \times 10^{-8} \mathrm{C}\) has, at a given instant, a velocity \(\overrightarrow{\boldsymbol{v}}=\left(3.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath}\) What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field \(\vec{B}=(1.63 \mathrm{~T}) \hat{\imath}+(0.980 T) \hat{\jmath} ?\)

Short Answer

Expert verified
The magnitude of the particle's acceleration is 33.15 m/s虏, and the direction is along the negative z-axis.

Step by step solution

01

Identify Known Quantities

From the task, we can identify the particle's charge \(q = 1.22 \times 10^{-8} C\), velocity \(\vec{v}=(3.00 \times 10^{4} m/s) \hat{\jmath}\) and mass \(m = 1.81 \times 10^{-3} kg\). The magnetic field is \(\vec{B}=(1.63 T) \hat{\imath}+(0.980 T) \hat{\jmath}\).
02

Calculate Magnetic Force

Using the formula \( \vec{F} = q \vec{v} \times \vec{B} \), only the cross product is nonzero on the i-component because \(\hat{\jmath} \times \hat{\jmath} = 0 \) and \(\hat{\jmath} \times \hat{\imath} = -\hat{k}\). Therefore \(\vec{F} = -q \cdot v \cdot B\_{xi} \hat{k} = - (1.22 \times 10^{-8} C) \cdot (3.00 \times 10^{4} m/s) \cdot (1.63 T) \hat{k} = -6.00 \times 10^{-5} N \hat{k}\).
03

Calculate Acceleration

Acceleration \(\vec{a}\) is the force divided by mass, so \(\vec{a} = \vec{F}/m = \frac{-6.00 \times 10^{-5} N}{1.81 \times 10^{-3} kg} \hat{k} = -33.15 m/s^2 \hat{k}\). The negative sign indicates the direction is along the negative z-axis.

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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 Lorentz force is crucial when studying the effects of electromagnetic fields on charged particles. It represents the force experienced by a charged particle due to electromagnetic fields. The force is given by the equation \( \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \) where \( q \) is the electric 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. Since the exercise only mentions a magnetic field and no electric field, the electric field component \( \vec{E} \) can be ignored, simplifying the equation to \( \vec{F} = q \vec{v} \times \vec{B} \).
In the given problem, applying the Lorentz force equation, we see that the charged particle's acceleration is purely the result of the magnetic component of the Lorentz force. There is no electric field acting on the particle, thus the force and resulting acceleration are solely due to the magnetic field.
Cross product in magnetic force
Understanding the cross product is vital when calculating the magnetic force on a charged particle moving in a magnetic field. The cross product, represented by \( \times \), is a mathematical operation between two vectors resulting in a third vector that is perpendicular to the plane formed by the initial vectors. Its magnitude is equal to the product of the magnitudes of the two vectors and the sine of the angle between them.
In the context of magnetic force, the cross product occurs between the velocity \( \vec{v} \) of the particle and the magnetic field \( \vec{B} \), as seen in the formula \( \vec{F} = q \vec{v} \times \vec{B} \). In our exercise, because the velocity vector is directed along \( \hat{j} \) and the magnetic field has components along both \( \hat{i} \) and \( \hat{j} \), we find the force to be in the direction of \( \hat{k} \), which is perpendicular to the plane formed by \( \hat{i} \) and \( \hat{j} \). The vector nature of this product results in a force (and thus acceleration) that is not along the direction of either the velocity or the magnetic field, but orthogonal to both.
Motion of charged particles in a magnetic field
The motion of charged particles in a magnetic field is profoundly affected by the magnetic force, and that motion can be complex due to the force being perpendicular to the particle鈥檚 trajectory. A charged particle moving in a uniform magnetic field experiences a centripetal force which causes it to move in a circular path if its velocity is perpendicular to the field lines, or in a spiral path if it also has a velocity component parallel to the field lines.
In our exercise, the charged particle moving through the uniform magnetic field experiences a force perpendicular to its motion, causing it to accelerate in the direction opposite to \( \hat{k} \) due to the specific directions of \( \vec{v} \) and \( \vec{B} \). The acceleration being perpendicular to the velocity suggests that the particle will not speed up or slow down; instead, it will change direction, resulting in circular or helical motion. This outcome, predicted by the right-hand rule, demonstrates the complex interplay between charge, magnetic fields, and motion in the absence of other forces.

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Most popular questions from this chapter

You measure the charge-to-mass ratio \(q / m\) for a particle with positive charge in the following way: The particle starts from rest, is accelerated through a potential difference \(\Delta V,\) and attains a velocity with magnitude \(v .\) It then enters a region of uniform magnetic field \(B=0.200 \mathrm{~T}\) that is directed perpendicular to the velocity; the particle moves in a path that is an arc of a circle of radius \(R .\) You measure \(R\) as a function of \(\Delta V\). You plot your data as \(R^{2}\) (in units of \(\mathrm{m}^{2}\) ) versus \(\Delta V\) (in \(\mathrm{V}\) ) and find that the values lie close to a straight line that has slope \(1.04 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{V} .\) What is the value of \(q / m\) for this particle?

A small particle with positive charge \(q=+3.75 \times 10^{-4} \mathrm{C}\) and mass \(m=5.00 \times 10^{-5} \mathrm{~kg}\) is moving in a region of uniform electric and magnetic fields. The magnetic field is \(B=4.00 \mathrm{~T}\) in the \(+z\) -direction. The electric field is also in the \(+z\) -direction and has magnitude \(E=60.0 \mathrm{~N} / \mathrm{C}\). At time \(t=0\) the particle is on the \(y\) -axis at \(y=+1.00 \mathrm{~m}\) and has velocity \(v=30.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction. Neglect gravity. (a) What are the \(x\) -, \(y\) and \(z\) -coordinates of the particle at \(t=0.0200 \mathrm{~s} ?\) (b) What is the speed of the particle at \(t=0.0200 \mathrm{~s} ?\)

Determine the magnetic moment \(\overrightarrow{\boldsymbol{\mu}}\) of a spherical shell with radius \(R\) and uniform charge \(Q\) rotating with angular speed \(\vec{\Omega}=\omega \hat{k} .\) Use the following steps: (a) Consider a coordinate system with the origin at the center of the sphere. Parameterize each latitude on the sphere with the angle \(\theta\) measured from the positive \(z\) -axis. There is a circular current loop at each value of \(\theta\) for \(0 \leq \theta \leq \pi .\) What is the radius of the loop at latitude \(\theta ?\) (b) The differential current carried by that loop is \(d I=\sigma v d W,\) where \(\sigma\) is the charge density of the sphere, \(v\) is the tangential speed of the loop, and \(d W=R d \theta\) is its differential width. Express \(d I\) in terms of \(\sigma, R, \omega, \theta,\) and \(d \theta .\) (c) The differential magnetic moment of the loop is \(d \mu=A d I,\) where \(A\) is the area enclosed by the loop. Express \(d \mu\) in terms of \(R, \omega, \theta,\) and \(d \theta\) (d) Integrate over the sphere to determine the magnetic moment. Express your result as a vector; use the total charge \(Q\) rather than the charge density \(\sigma\). (e) If the sphere is in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(\sin \alpha \hat{\imath}+\cos \alpha \hat{\jmath}) B,\) what is the torque on the sphere?

A particle of mass \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C}\). The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

A Cycloidal Path. A particle with mass \(m\) and positive charge \(q\) starts from rest at the origin shown in Fig. \(\mathbf{P 2 7 . 8 2 .}\). There is a uniform electric field \(\vec{E}\) in the \(+y\) -direction and a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\) directed out of the page. It is shown in more advanced books that the path is a cycloid whose radius of curvature at the top points is twice the \(y\) -coordinate at that level. (a) Explain why the path has this general shape and why it is repetitive. (b) Prove that the speed at any point is equal to \(\sqrt{2 q E y / m}\). (Hint: Use energy conservation.) (c) Applying Newton's second law at the top point and taking as given that the radius of curvature here equals \(2 y,\) prove that the speed at this point is \(2 E / B\)
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