11.2 Magnetic Fields and Lines

Defining the Magnetic Field

A magnetic field is defined by the force that a charged particle experiences moving in this field, after we account for the gravitational and any additional electric forces possible on the charge. The magnitude of this force is proportional to the amount of charge q, the speed of the charged particle v, and the magnitude of the applied magnetic field. The direction of this force is perpendicular to both the direction of the moving charged particle and the direction of the applied magnetic field. Based on these observations, we define the magnetic field strength B based on the magnetic force [latex]\stackrel{\to }{\textbf{F}}[/latex] on a charge q moving at velocity [latex]\stackrel{\to }{\textbf{v}}[/latex] as the cross product of the velocity and magnetic field, that is,

In fact, this is how we define the magnetic field [latex]\stackrel{\to }{\textbf{B}}[/latex]—in terms of the force on a charged particle moving in a magnetic field. The magnitude of the force is determined from the definition of the cross product as it relates to the magnitudes of each of the vectors. In other words, the magnitude of the force satisfies

where θ is the angle between the velocity and the magnetic field.

The SI unit for magnetic field strength B is called the tesla (T) after the eccentric but brilliant inventor Nikola Tesla (1856–1943), where

A smaller unit, called the gauss (G), where [latex]1\phantom{\rule{0.2em}{0ex}}\text{G}={10}^{-4}\text{T},[/latex] is sometimes used. The strongest permanent magnets have fields near 2 T; superconducting electromagnets may attain 10 T or more. Earth’s magnetic field on its surface is only about [latex]5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{T},[/latex] or 0.5 G.

There is no magnetic force on static charges. However, there is a magnetic force on charges moving at an angle to a magnetic field. When charges are stationary, their electric fields do not affect magnets. However, when charges move, they produce magnetic fields that exert forces on other magnets. When there is relative motion, a connection between electric and magnetic forces emerges—each affects the other.

Example

An Alpha-Particle Moving in a Magnetic Field

An alpha-particle [latex]\left(q=3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)[/latex] moves through a uniform magnetic field whose magnitude is 1.5 T. The field is directly parallel to the positive z-axis of the rectangular coordinate system of Figure 11.5. What is the magnetic force on the alpha-particle when it is moving (a) in the positive x-direction with a speed of [latex]5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s?}[/latex] (b) in the negative y-direction with a speed of [latex]5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s?}[/latex] (c) in the positive z-direction with a speed of [latex]5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s?}[/latex] (d) with a velocity [latex]\stackrel{\to }{\textbf{v}}=\left(2.0\hat{\textbf{i}}-3.0\hat{\textbf{j}}+1.0\hat{\textbf{k}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s?}[/latex]

Strategy

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation [latex]\stackrel{\to }{\textbf{F}}=q\stackrel{\to }{\textbf{v}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{\textbf{B}}[/latex] or [latex]F=qvB\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta[/latex] to calculate the force. The direction of the force is determined by RHR-1.

Solution

Show Answer

1. First, to determine the direction, start with your fingers pointing in the positive x-direction. Sweep your fingers upward in the direction of magnetic field. Your thumb should point in the negative y-direction. This should match the mathematical answer. To calculate the force, we use the given charge, velocity, and magnetic field and the definition of the magnetic force in cross-product form to calculate:
   
   [latex]\stackrel{\to }{\textbf{F}}=q\stackrel{\to }{\textbf{v}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{\textbf{B}}=\left(3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{i}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(1.5\phantom{\rule{0.2em}{0ex}}\text{T}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{k}}\right)=-2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{N}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{j}}.[/latex]
2. First, to determine the directionality, start with your fingers pointing in the negative y-direction. Sweep your fingers upward in the direction of magnetic field as in the previous problem. Your thumb should be open in the negative x-direction. This should match the mathematical answer. To calculate the force, we use the given charge, velocity, and magnetic field and the definition of the magnetic force in cross-product form to calculate:
   
   [latex]\stackrel{\to }{\textbf{F}}=q\stackrel{\to }{\textbf{v}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{\textbf{B}}=\left(3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)\left(-5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{j}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(1.5\phantom{\rule{0.2em}{0ex}}\text{T}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{k}}\right)=-2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{N}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{i}}.[/latex]

   
   An alternative approach is to use Equation 11.2 to find the magnitude of the force. This applies for both parts (a) and (b). Since the velocity is perpendicular to the magnetic field, the angle between them is 90 degrees. Therefore, the magnitude of the force is:
   

   [latex]F=qvB\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =\left(3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m}\text{/}\text{s}\right)\left(1.5\phantom{\rule{0.2em}{0ex}}\text{T}\right)\text{sin}\left(90\text{°}\right)=2.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{N.}[/latex]
3. Since the velocity and magnetic field are parallel to each other, there is no orientation of your hand that will result in a force direction. Therefore, the force on this moving charge is zero. This is confirmed by the cross product. When you cross two vectors pointing in the same direction, the result is equal to zero.
4. First, to determine the direction, your fingers could point in any orientation; however, you must sweep your fingers upward in the direction of the magnetic field. As you rotate your hand, notice that the thumb can point in any x– or y-direction possible, but not in the z-direction. This should match the mathematical answer. To calculate the force, we use the given charge, velocity, and magnetic field and the definition of the magnetic force in cross-product form to calculate:
   
   [latex]\begin{array}{cc}\hfill \stackrel{\to }{\textbf{F}}& =q\stackrel{\to }{\textbf{v}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{\textbf{B}}=\left(3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)\left(\left(2.0\hat{\textbf{i}}-3.0\hat{\textbf{j}}+1.0\hat{\textbf{k}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(1.5\phantom{\rule{0.2em}{0ex}}\text{T}\phantom{\rule{0.2em}{0ex}}\hat{\textbf{k}}\right)\hfill \\ & =\left(-14.4\hat{\textbf{i}}-9.6\hat{\textbf{j}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\text{N.}\hfill \end{array}[/latex]

   
   This solution can be rewritten in terms of a magnitude and angle in the xy-plane:
   

   [latex]\begin{array}{ccc}\hfill |\stackrel{\to }{\textbf{F}}|& =\hfill & \sqrt{{F}_{x}^{2}+{F}_{y}^{2}}=\sqrt{{\left(-14.4\right)}^{2}+{\left(-9.6\right)}^{2}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\text{N}=1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{N}\hfill \\ \hfill \theta & =\hfill & {\text{tan}}^{-1}\left(\frac{{F}_{y}}{{F}_{x}}\right)={\text{tan}}^{-1}\left(\frac{-9.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\text{N}}{-14.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\text{N}}\right)=34\text{°}\text{.}\hfill \end{array}[/latex]

   
   The magnitude of the force can also be calculated using Equation 11.2. The velocity in this question, however, has three components. The z-component of the velocity can be neglected, because it is parallel to the magnetic field and therefore generates no force. The magnitude of the velocity is calculated from the x– and y-components. The angle between the velocity in the xy-plane and the magnetic field in the z-plane is 90 degrees. Therefore, the force is calculated to be:
   

   [latex]\begin{array}{ccc}\hfill |\stackrel{\to }{\textbf{v}}|& =\hfill & \sqrt{{\left(2\right)}^{2}+{\left(-3\right)}^{2}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\frac{\text{m}}{\text{s}}=3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\frac{\text{m}}{\text{s}}\hfill \\ \hfill F& =\hfill & qvB\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =\left(3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-19}\text{C}\right)\left(3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{m/s}\right)\left(1.5\phantom{\rule{0.2em}{0ex}}\text{T}\right)\text{sin}\left(90\text{°}\right)=1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\text{N.}\hfill \end{array}[/latex]

   
   This is the same magnitude of force calculated by unit vectors.

Significance

The cross product in this formula results in a third vector that must be perpendicular to the other two. Other physical quantities, such as angular momentum, also have three vectors that are related by the cross product. Note that typical force values in magnetic force problems are much larger than the gravitational force. Therefore, for an isolated charge, the magnetic force is the dominant force governing the charge’s motion.

Representing Magnetic Fields

The representation of magnetic fields by magnetic field lines is very useful in visualizing the strength and direction of the magnetic field. As shown in Figure 11.6, each of these lines forms a closed loop, even if not shown by the constraints of the space available for the figure. The field lines emerge from the north pole (N), loop around to the south pole (S), and continue through the bar magnet back to the north pole.

Magnetic field lines have several hard-and-fast rules:

1. The direction of the magnetic field is tangent to the field line at any point in space. A small compass will point in the direction of the field line.
2. The strength of the field is proportional to the closeness of the lines. It is exactly proportional to the number of lines per unit area perpendicular to the lines (called the areal density).
3. Magnetic field lines can never cross, meaning that the field is unique at any point in space.
4. Magnetic field lines are continuous, forming closed loops without a beginning or end. They are directed from the north pole to the south pole.

The last property is related to the fact that the north and south poles cannot be separated. It is a distinct difference from electric field lines, which generally begin on positive charges and end on negative charges or at infinity. If isolated magnetic charges (referred to as magnetic monopoles) existed, then magnetic field lines would begin and end on them.