22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field - College Physics 2e | OpenStax (2024)

Learning Objectives

By the end of this section, you will be able to:

• Describe the effects of magnetic fields on moving charges.
• Use the right hand rule 1 to determine the velocity of a charge, the direction of the magnetic field, and the direction of the magnetic force on a moving charge.
• Calculate the magnetic force on a moving charge.

Right Hand Rule 1

The 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 magnetic force $F$ on a charge $q$ moving at a speed $v$ in a magnetic field of strength $B$ is given by

where $\theta$ is the angle between the directions of $\mathbf{\text{v}}$ and $\mathbf{\text{B}}.$ This force is often called the Lorentz force. In fact, this is how we define the magnetic field strength $B$—in terms of the force on a charged particle moving in a 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). To determine how the tesla relates to other SI units, we solve $F=\text{qvB}\text{sin}\theta$ for $B$.

Because $\text{sin}\theta$ is unitless, the tesla is

(note that C/s = A).

Another smaller unit, called the gauss (G), where $\mathrm{1\; G}={\text{10}}^{-4}\mathrm{T}$, is sometimes used. The strongest permanent magnets have fields near 2 T; superconducting electromagnets may attain 10 T or more. The Earth’s magnetic field on its surface is only about $5\times {\text{10}}^{-5}\mathrm{T}$, or 0.5 G.

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 right hand rule 1 (or RHR-1), which is illustrated in Figure 22.16. RHR-1 states that, to determine the direction of the magnetic force on a positive moving charge, you point the thumb of the right hand in the direction of $\mathbf{\text{v}}$, the fingers in the direction of $\mathbf{\text{B}}$, and a perpendicular to the palm points in the direction of $\mathbf{\text{F}}$. One way to remember this is that there is one velocity, and so the thumb represents it. There are many field lines, and so the fingers represent them. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Figure 22.16 Magnetic fields exert forces on moving charges. This force is one of the most basic known. The direction of the magnetic force on a moving charge is perpendicular to the plane formed by $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$ and follows right hand rule–1 (RHR-1) as shown. The magnitude of the force is proportional to $q$, $v$, $B$, and the sine of the angle between $\mathbf{\text{v}}$ and $\mathbf{\text{B}}$.

Example 22.1

Calculating Magnetic Force: Earth’s Magnetic Field on a Charged Glass Rod

With 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 20-nC 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 22.17.)

Figure 22.17 A positively charged object moving due west in a region where the Earth’s magnetic field is due north experiences a force that is straight down as shown. A negative charge moving in the same direction would feel a force straight up.

Strategy

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation $F=\text{qvB}\text{sin}\theta$ to find the force.

Solution

The magnetic force is

$$F=\text{qvB}\text{sin}\theta .$$

22.4

We see that $\text{sin}\theta =1$, since the angle between the velocity and the direction of the field is $\text{90º}$. Entering the other given quantities yields

$$\begin{array}{lll}F& =& \left(\text{20}\times {\text{10}}^{\mathrm{–9}}\mathrm{C}\right)\left(\text{10 m/s}\right)\left(5\times {\text{10}}^{\mathrm{–5}}\mathrm{T}\right)\\ & =& 1\times {\text{10}}^{\text{–11}}\left(\mathrm{C}\cdot \text{m/s}\right)\left(\frac{\mathrm{N}}{\mathrm{C}\cdot \text{m/s}}\right)=1\times {\text{10}}^{\text{–11}}\mathrm{N.}\end{array}$$

22.5

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.