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Chapter 27: Problem 56

A particle with charge \(9.45 \times 10 ^ { - 8 } \mathrm { C }\) is moving in a region where there is a uniform magnetic field of 0.650 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 \(0, 2.08 \times 10^{-3}, 3.59 \times 10^{-3}\) N.

Step by step solution

01

Understand the Equation

The force on a charged particle moving in a magnetic field is calculated using the formula \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \), where \( q \) is the charge, \( \mathbf{v} \) is the velocity vector, and \( \mathbf{B} \) is the magnetic field vector.
02

Define Velocity and Magnetic Field Vectors

Identify the velocity vector \( \mathbf{v} = \langle v_x, v_y, v_z \rangle = \langle -1.68 \times 10^4, -3.11 \times 10^4, 5.85 \times 10^4 \rangle \) m/s and the magnetic field vector \( \mathbf{B} = \langle 0.650, 0, 0 \rangle \) T.
03

Calculate the Cross Product \( \mathbf{v} \times \mathbf{B} \)

Use the formula for a cross product to find \( \mathbf{v} \times \mathbf{B} \):\[\mathbf{v} \times \mathbf{B} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \-1.68 \times 10^4 & -3.11 \times 10^4 & 5.85 \times 10^4 \0.650 & 0 & 0 \\end{vmatrix}\]Calculating it, the resulting vector components are: \( v_x B_y - v_y B_x = 0 - (-3.11 \times 10^4 \cdot 0.650) \),\( v_z B_x - v_x B_z = 5.85 \times 10^4 \cdot 0.650 - 0 \),\( v_x B_y - v_y B_x = (-1.68 \times 10^4 \cdot 0) - (-3.11 \times 10^4 \cdot 0) = 0 \).This results in \( \mathbf{v} \times \mathbf{B} = \langle 0, 3.11 \times 10^4 \times 0.650, 5.85 \times 10^4 \times 0.650 \rangle \).
04

Substitute Values into Cross Product

Substituting the numerical values for \( v_y \) and \( v_z \) into the expression yields:\[ 0, -(-3.11 \times 10^4 \cdot 0.650) = 20215, \quad(5.85 \times 10^4 \cdot 0.650) = 38025 \]. Hence, the cross product \( \mathbf{v} \times \mathbf{B} = \langle 0, 20215, 38025 \rangle \).
05

Calculate the Magnetic Force \( \mathbf{F} \)

The force vector \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \) is calculated by multiplying the charge \( q = 9.45 \times 10^{-8} \) C with the cross product vector \( \langle 0, 20215, 38025 \rangle \):\( F_x = q \cdot 0 \),\( F_y = q \cdot 20215 = 9.45 \times 10^{-8} \cdot 20215 \),\( F_z = q \cdot 38025 = 9.45 \times 10^{-8} \cdot 38025 \).
06

Compute Force Components

Calculate each component:\( F_x = 0 \),\( F_y = 2.08 \times 10^{-3} \) N,\( F_z = 3.59 \times 10^{-3} \) N.Thus, the components of the force on the particle are \( \langle 0, 2.08 \times 10^{-3}, 3.59 \times 10^{-3} \rangle \) N.

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Key Concepts

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

Cross Product
The cross product is an essential mathematical operation used in physics to determine the resulting vector from two vectors. In the context of magnetic forces, it helps us find the direction and magnitude of the force acting on a charged particle moving through a magnetic field.

A cross product between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated using the formula:
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \a_x & a_y & a_z \b_x & b_y & b_z \\end{vmatrix}\]
This determinant expands to give:\( (a_y b_z - a_z b_y)\hat{i} + (a_z b_x - a_x b_z)\hat{j} + (a_x b_y - a_y b_x)\hat{k} \). The resulting vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).

In our exercise, we calculated the cross product \( \mathbf{v} \times \mathbf{B} \) by following the above steps, resulting in a vector that describes how the particle's velocity interacts with the magnetic field.
Magnetic Field
A magnetic field is a vector field surrounding magnets and electric currents, representing the magnetic force exerted in the surrounding space. It is characterized by both a direction and magnitude, measured in Tesla (T).

The magnetic field plays a crucial role in determining the motion of charged particles. This magnetic force is perpendicular to both the charged particle's velocity and the magnetic field itself.

In our exercise, the magnetic field \( \mathbf{B} \) is uniform and directed along the \(+ x\)-axis, with a magnitude of 0.650 T. This uniformity simplifies calculations as the field can be represented as a vector \( \mathbf{B} = \langle 0.650, 0, 0 \rangle \). When using the cross product to find the force \( \mathbf{F} \), the direction of the resulting force depends significantly on this magnetic field's orientation.
Charged Particle
A charged particle is any particle, such as an electron or proton, that carries an electric charge. When a charged particle moves through a magnetic field, it experiences a magnetic force.

This force \( \mathbf{F} \) is defined by the Lorentz force equation:
\[\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\]
Where:
  • \( q \) is the charge of the particle
  • \( \mathbf{v} \) is the velocity vector of the particle
  • \( \mathbf{B} \) is the magnetic field vector
This force is perpendicular to both the velocity vector and the magnetic field.

In the given exercise, our particle has a small charge of \( 9.45 \times 10^{-8} \) C and a specified velocity vector. By calculating \( \mathbf{v} \times \mathbf{B} \) and then multiplying it by the particle's charge, we determine the force components acting on the particle at the given moment.

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Most popular questions from this chapter

Quark Model of the Neutron. The neutron is a particle with zero charge. Nonetheless, it has a nonzero magnetic moment with \(z\) -component \(9.66 \times\) \(10 ^ { - 27 } \mathrm { A } \cdot \mathrm { m } ^ { 2 } .\) This be explained by the internal structure of the neutron. A substantial body of evidence indicates that a neutron is composed of three fundamental particles called of three fundamental particles called quarks: an "up" (u) quark, of charge \(+ 2 e / 3 ,\) and two "down" \(( d )\) quarks, each of charge \(- e / 3 .\) The combination of the three quarks produces a net charge of \(2 e / 3 - e / 3 - e / 3 = 0\) . If the quarks are in motion, they can produce a nonzero magnetic moment. As a very simple model, suppose the \(u\) quark moves in a counterclockwise circular path and the \(d\) quarks move in a clock- wise circular path, all of radius \(r\) and all with the same speed \(v\) (Fig. P27.84). (a) Determine the current due to the circulation of the \(u\) quark. (b) Determine the magnitude of the magnetic moment due to the circulating \(u\) quark. (c) Determine the magnitude of the magnetic moment of the three-quark system. (Be careful to use the correct magnetic moment directions.) (d) With what speed \(v\) must the quarks move if this model is to reproduce the magnetic moment of the neutron? Use \(r = 1.20 \times 10 ^ { - 15 } \mathrm { m }\) (the radius of the neutron) for the radius of the orbits.

Determining Diet. One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the 12\(\mathrm { C }\) and \(^ { 13 } \mathrm { C }\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed \(8.50 \mathrm { km } / \mathrm { s } ,\) and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0\(\mathrm { cm }\) for the 12\(\mathrm { C }\) . The measured masses of these isotopes are \(1.99 \times 10 ^ { - 26 } \mathrm { kg } \left( ^ { 12 } \mathrm { C } \right)\) and \(2.16 \times 10 ^ { - 26 } \mathrm { kg } \left( ^ { 13 } \mathrm { C } \right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the 13 C semicircle? (c) What is the separation of the \(^ { 12 } \mathrm { C }\) and \(^ { 13 } \mathrm { C }\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

A straight, \(2.5-\mathrm{m}\) wire carries a typical household current of 1.5 \(\mathrm{A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects undernormal household conditions?

When a particle of charge \(q > 0\) moves with a velocity of \(\vec { \boldsymbol { v } } _ { 1 }\) at \(45.0 ^ { \circ }\) from the \(\pm x\) -axis in the \(x y\) -plane, a uniform magnetic field exerts a force \(F _ { 1 }\) along the \(- z\) -axis (Fig. \(P 27.55 ) .\) When the same particle moves with a velocity \(\vec { \boldsymbol { v } } _ { 2 }\) with the same magnitude as \(\vec { \boldsymbol { v } } _ { 1 }\) but along the \(+ z\) -zaxis, a force \(\vec { \boldsymbol { F } } _ { 2 }\) of magnitude \(F _ { 2 }\) 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 \(\vec { F } _ { 1 }\) in terms of \(F _ { 2 } ?\)

An electron experiences a magnetic force of magnitude \(4.60 \times 10^{-15} \mathrm{N}\) when moving at an angle of \(60.0^{\circ}\) with respect to a magnetic field of magnitude \(3.50 \times 10^{-3} \mathrm{T.}\) Find the speed of the electron.
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