225. Magnetic Field Strength: Force on a Moving Charge in a Magnetic FieldLearning Objectives
What is the mechanism by which one magnet exerts a force on another? The answer is related to the fact that all magnetism is caused by current, the flow of charge. Magnetic fields exert forces on moving charges, and so they exert forces on other magnets, all of which have moving charges. Right Hand Rule 1The magnetic force on a moving charge is one of the most fundamental known. Magnetic force is as important as the electrostatic or Coulomb force. Yet the magnetic force is more complex, in both the number of factors that affects it and in its direction, than the relatively simple Coulomb force. The magnitude of the where $\theta $ is the angle between the directions of $\mathbf{\text{v}}$ and $\mathbf{\text{B}}.$ This force is often called the Because $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ is unitless, the tesla is $$\text{1 T}=\frac{\text{1 N}}{\mathrm{C}\cdot \text{m/s}}=\frac{\text{1 N}}{\mathrm{A}\cdot \mathrm{m}}$$(note that C/s = A). Another smaller unit, called the The direction of the magnetic force $\mathbf{\text{F}}$ is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$, as determined by the Making Connections: Charges and MagnetsThere is no magnetic force on static charges. However, there is a magnetic force on moving charges. When charges are stationary, their electric fields do not affect magnets. But, when charges move, they produce magnetic fields that exert forces on other magnets. When there is relative motion, a connection between electric and magnetic fields emerges—each affects the other. Example 1: Calculating Magnetic Force: Earth’s Magnetic Field on a Charged Glass RodWith the exception of compasses, you seldom see or personally experience forces due to the Earth’s small magnetic field. To illustrate this, suppose that in a physics lab you rub a glass rod with silk, placing a 20nC positive charge on it. Calculate the force on the rod due to the Earth’s magnetic field, if you throw it with a horizontal velocity of 10 m/s due west in a place where the Earth’s field is due north parallel to the ground. (The direction of the force is determined with right hand rule 1 as shown in Figure 2.) Strategy We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation $F=\text{qvB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ to find the force. Solution The magnetic force is $$F=\text{qvb}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$$We see that $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$, since the angle between the velocity and the direction of the field is $\text{90\xba}$. Entering the other given quantities yields $$\begin{array}{lll}F& =& \left(\text{20}\times {\text{10}}^{\mathrm{\u20139}}\phantom{\rule{0.25em}{0ex}}\mathrm{C}\right)\left(\text{10 m/s}\right)\left(5\times {\text{10}}^{\mathrm{\u20135}}\phantom{\rule{0.25em}{0ex}}\mathrm{T}\right)\\ & =& 1\times {\text{10}}^{\text{\u201311}}\phantom{\rule{0.25em}{0ex}}\left(\mathrm{C}\cdot \text{m/s}\right)\left(\frac{\mathrm{N}}{\mathrm{C}\cdot \text{m/s}}\right)=1\times {\text{10}}^{\text{\u201311}}\phantom{\rule{0.25em}{0ex}}\mathrm{N.}\end{array}$$Discussion This force is completely negligible on any macroscopic object, consistent with experience. (It is calculated to only one digit, since the Earth’s field varies with location and is given to only one digit.) The Earth’s magnetic field, however, does produce very important effects, particularly on submicroscopic particles. Some of these are explored in Force on a Moving Charge in a Magnetic Field: Examples and Applications. Section Summary
Conceptual QuestionsExercise 1If a charged particle moves in a straight line through some region of space, can you say that the magnetic field in that region is necessarily zero? Problems & ExercisesExercise 1What is the direction of the magnetic force on a positive charge that moves as shown in each of the six cases shown in Figure 3? Figure 3 Show/Hide Solution Solution(a) Left (West) (b) Into the page (c) Up (North) (d) No force (e) Right (East) (f) Down (South) Exercise 2Repeat Exercise 1 for a negative charge. Exercise 3What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in Figure 4, assuming it moves perpendicular to $\mathbf{\text{B}}\mathrm{?}$ Figure 4 Show/Hide Solution Solution(a) East (right) (b) Into page (c) South (down) Exercise 4Repeat Exercise 3 for a positive charge. Exercise 5What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming $\mathbf{\text{B}}$ is perpendicular to $\mathbf{\text{v}}$? Figure 5 Show/Hide Solution Solution(a) Into page (b) West (left) (c) Out of page Exercise 6Repeat Exercise 5 for a negative charge. Exercise 7What is the maximum force on an aluminum rod with a $0\text{.}\text{100}\text{\mu C}$ charge that you pass between the poles of a 1.50T permanent magnet at a speed of 5.00 m/s? In what direction is the force? Show/Hide Solution Solution$7\text{.}\text{50}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N}$ perpendicular to both the magnetic field lines and the velocity Exercise 8(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a $0\text{.}\text{500}\text{\mu C}$ charge and flies due west at a speed of 660 m/s over the Earth’s south magnetic pole, where the $8\text{.}\text{00}\times {\text{10}}^{5}\text{T}$ magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect. Exercise 9(a) A cosmic ray proton moving toward the Earth at $\text{5.00}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ experiences a magnetic force of $1\text{.}\text{70}\times {\text{10}}^{\text{16}}\phantom{\rule{0.25em}{0ex}}\text{N}$. What is the strength of the magnetic field if there is a $\text{45\xba}$ angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface? Discuss. Show/Hide Solution Solution(a) $3\text{.}\text{01}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$ (b) This is slightly less then the magnetic field strength of $5\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$ at the surface of the Earth, so it is consistent. Exercise 10An electron moving at $4\text{.}\text{00}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ in a 1.25T magnetic field experiences a magnetic force of $1\text{.}\text{40}\times {\text{10}}^{\text{16}}\phantom{\rule{0.25em}{0ex}}\text{N}$. What angle does the velocity of the electron make with the magnetic field? There are two answers. Exercise 11(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than $1\text{.}\text{00}\times {\text{10}}^{\text{12}}\phantom{\rule{0.25em}{0ex}}\mathrm{N}$. What is the greatest the charge can be if it moves at a maximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent. Show/Hide Solution Solution(a) $6\text{.}\text{67}\times {\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{C}$ (taking the Earth’s field to be $5\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{T}$) (b) Less than typical static, therefore difficult
