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Chapter 27: Problem 54

A particle with charge \(9.45 \times 10^{-8} \mathrm{C}\) is moving in a region where there is a uniform magnetic field of 0.450 \(\mathrm{T}\) in the \(+x\) - direction. At a particular instant of time the velocity of the particle has components \(v_{x}=-1.68 \times 10^{4} \mathrm{m} / \mathrm{s}, v_{y}=-3.11 \times 10^{4} \mathrm{m} / \mathrm{s}\) and \(v_{z}=5.85 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the components of the force on the particle at this time?

Short Answer

Expert verified
The force components are \(0\) N, \(-2.487 \times 10^{-3}\) N, and \(-1.32225 \times 10^{-3}\) N in the \(x, y,\) and \(z\) directions, respectively.

Step by step solution

01

Understanding the Given Values

The problem gives us the following values: charge \(q = 9.45 \times 10^{-8} \text{ C}\), magnetic field \(\mathbf{B} = 0.450 \text{ T}\) in the \(+x\)-direction, and velocity components \(v_x = -1.68 \times 10^4 \text{ m/s}\), \(v_y = -3.11 \times 10^4 \text{ m/s}\), and \(v_z = 5.85 \times 10^4 \text{ m/s}\). We need to find the force on the particle using these values.
02

Formula for Magnetic Force

The magnetic force \(\mathbf{F}\) on a charged particle is given by the formula \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\), where \(\times\) denotes the cross product of the velocity \(\mathbf{v}\) and the magnetic field \(\mathbf{B}\).
03

Determine the Cross Product

To calculate \(\mathbf{v} \times \mathbf{B}\), set \(\mathbf{v} = (-1.68 \times 10^4, -3.11 \times 10^4, 5.85 \times 10^4)\) and \(\mathbf{B} = (0.450, 0, 0)\). Using the determinant formula for the cross product:\[\mathbf{v} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1.68 \times 10^4 & -3.11 \times 10^4 & 5.85 \times 10^4 \ 0.450 & 0 & 0 \end{vmatrix}\]
04

Calculation of the Cross Product

Calculate each component of \(\mathbf{v} \times \mathbf{B}\):- The \(x\)-component is \(0\) because both vectors' components in the \(y\) and \(z\) directions are multiplied to get zero.- The \(y\)-component is \((0 - 0.450 \times 5.85 \times 10^4) = -2.6325 \times 10^4\).- The \(z\)-component is \(0.450 \times -3.11 \times 10^4 - 0 = -1.3995 \times 10^4\).Thus, \(\mathbf{v} \times \mathbf{B} = (0, -2.6325 \times 10^4, -1.3995 \times 10^4)\).
05

Calculate Force Components

Using \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\), we find the force:- The \(x\)-component is \(9.45 \times 10^{-8} \times 0 = 0\).- The \(y\)-component is \(9.45 \times 10^{-8} \times -2.6325 \times 10^4 = -2.487\times 10^{-3} \text{ N}\).- The \(z\)-component is \(9.45 \times 10^{-8} \times -1.3995 \times 10^4 = -1.32225 \times 10^{-3} \text{ N}\).

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Key Concepts

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

Charged Particle Motion
When a charged particle, like the one described in the exercise, moves within a magnetic field, it experiences a force that influences its motion. This phenomenon is quite common in both natural and technological processes.

Key points to consider include:
  • The motion of such a particle is dictated entirely by its charge, speed, direction, and the magnetic field it traverses.
  • The resultant path tends to be curved rather than linear, allowing scientists to predict its trajectory based on specific inputs. A prime example is the circular or helical path particles typically follow under uniform magnetic fields.
In this scenario, the magnetic force doesn't speed up or slow down the charged particle but merely changes its direction.
Cross Product Calculation
The cross product is an essential mathematical operation when dealing with vectors in physics. Specifically, for a charge moving in a magnetic field, the cross product helps calculate the magnetic force direction and magnitude.

Consider these details:
  • The cross product is denoted by \( \mathbf{v} \times \mathbf{B} \), where \( \mathbf{v} \) represents the velocity vector and \( \mathbf{B} \) the magnetic field vector.
  • This operation results in a new vector that is perpendicular to both initial vectors. Its direction is determined by the right-hand rule, a handy tool where you use your fingers to guide the vectors' interaction.
  • In this exercise, the calculation uses determinant method, indicating how each component of the resulting vector arises from \ the other components of \( \mathbf{v} \) and \( \mathbf{B} \).
Understanding the cross product's role is fundamental in solving such physics problems accurately.
Magnetic Field Direction
Understanding how the direction of the magnetic field affects a charged particle is crucial. The magnetic field in this exercise points along the \(+x\)-axis, directly influencing the particle鈥檚 vector motion.

Here鈥檚 why the direction matters:
  • Magnetic fields exert forces only on moving charges perpendicular to the field lines. Thus, the particle's velocity components perpendicular to the \(+x\)-axis will interact, influencing motion.
  • Only these perpendicular components (\(v_{y}\) and \(v_{z}\)) will result in non-zero cross products, causing force components along the \(y\) and \(z\)-axes.
  • This interplay between direction and forces finds widespread use in various applications, from MRI machines to particle accelerators.
Through this directional interplay, one can predict the two-dimensional motion resulting from the three-dimensional vector setup.
Vector Components in Physics
Analyzing vector components is a vital step in simplifying and understanding complex physical phenomena. In this context, it helps break down the velocity and force into understandable parts.

Consider these points:
  • Each vector in physics typically has components aligned with the chosen coordinate system; here, it involves \(x\), \(y\), and \(z\) directions.
  • By dissecting vector quantities like velocity, we can easily analyze how each direction contributes to the final vector or subsequent physical outcomes like force.
  • These components allow the application of vector mathematics, including operations such as the cross product, which can derive new vector quantities essential for describing force and motion.
Consequently, understanding vector components enables clearer visualization and manipulation of complex interactions in physics problems.

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Most popular questions from this chapter

A circular area with a radius of 6.50 \(\mathrm{cm}\) lies in the \(x y\) -plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{T}\) (a) in the \(+z\) -direction; \((b)\) at an angle of \(53.1^{\circ}\) from the \(+z\) -direction; \((\mathrm{c})\) in the \(+y\) -direction?

A shunt-wound de motor with the field coils and rotor connected in parallel (Fig. 27.56 ) operates from a \(120-\mathrm{V}\) dc power linc. The resistance of the ficld windings, \(R_{f},\) is 218\(\Omega\) . The resistance of the rotor, \(R_{r}\) is 5.9 . When the motor is running, the rotor develops an emf \(\mathcal{E}\) . The motor draws a current of 4.82 A from the line. Friction losses amount to 45.0 W. Compute (a) the field current; \((b)\) the rotor current; \((c)\) the emf \(\mathcal{E} ;\) (d) the rate of development of thermal energy in the field windings; (e) the rate of development of thermal energy in the rotor; \((f)\) the power input to the motor; (g) the efficiency of the motor.

A wire 25.0 \(\mathrm{cm}\) long lies along the \(z\) -axis and carries a current of 9.00 \(\mathrm{A}\) in the \(+z\) -direction. The magnetic field is uniform and has components \(B_{x}=-0.242 \mathrm{T}, B_{y}=-0.985 \mathrm{T}\) , and \(B_{z}=\) \(-0.336 \mathrm{T} .\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?

When a particle of charge \(q>0\) moves with a velocity of \(\overrightarrow{\boldsymbol{v}}_{1}\) at \(45.0^{\circ}\) from the \(+x\) -axis in the \(x y\) -plane, a uniform magnetic field exerts a force \(\overrightarrow{\boldsymbol{F}}_{1}\) along the \(-z\) -axis (Fig. \(27.58 ) .\) When the same particle moves with a velocity \(\overrightarrow{\boldsymbol{v}}_{2}\) with the same magnitude as \(\overrightarrow{\boldsymbol{v}}_{1}\) but along the \(+z\) -axis, a force \(\overrightarrow{\boldsymbol{F}}_{2}\) of magnitude \(\boldsymbol{F}_{2}\) is exerted on it along the \(+x\) -axis. (a) What are the magnitude (in terms of \(q\) , \(v_{1},\) and \(F_{2} )\) and direction of the magnetic field? (b) What is the magnitude of \(\overrightarrow{\boldsymbol{F}}_{1}\) in terms of \(F_{2} ?\)

In a 1.25 - T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50\(\mu \mathrm{C}\) and initially moving northward at 4.75 \(\mathrm{km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.
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