11.2 Magnetic Fields and Lines

An Alpha-Particle Moving in a Magnetic Field

An alpha-particle $\left(q=3.2\times {10}^{\mathrm{-19}}\text{C}\right)$ 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 $5.0\times {10}^4\text{m/s?}$ (b) in the negative y-direction with a speed of $5.0\times {10}^4\text{m/s?}$ (c) in the positive z-direction with a speed of $5.0\times {10}^4\text{m/s?}$ (d) with a velocity $\overrightarrow{v}=\left(2.0\widehat{i}-3.0\widehat{j}+1.0\widehat{k}\right)\times {10}^4\text{m/s?}$

Strategy

We are given the charge, its velocity, and the magnetic field strength and direction. We can thus use the equation $\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ or $F=qvB\text{sin}\theta$ to calculate the force. The direction of the force is determined by RHR-1.

Solution

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:
   $$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}=\left(3.2\times {10}^{\mathrm{-19}}\text{C}\right)\left(5.0\times {10}^4\text{m/s}\widehat{i}\right)\times \left(1.5\text{T}\widehat{k}\right)=\mathrm{-2.4}\times {10}^{\mathrm{-14}}\text{N}\widehat{j}.$$
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:
   $$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}=\left(3.2\times {10}^{\mathrm{-19}}\text{C}\right)\left(\mathrm{-5.0}\times {10}^4\text{m/s}\widehat{j}\right)\times \left(1.5\text{T}\widehat{k}\right)=\mathrm{-2.4}\times {10}^{\mathrm{-14}}\text{N}\widehat{i}.$$
   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:
   $$F=qvB\text{sin}\theta =(3.2\times {10}^{\mathrm{-19}}\text{C})(5.0\times {10}^4\text{m}\text{/}\text{s})(1.5\text{T})\text{sin}(90\text{°})=2.4\times {10}^{\mathrm{-14}}\text{N.}$$
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:
   $$\begin{array}{cc}\hfill \overrightarrow{F}& =q\overrightarrow{v}\times \overrightarrow{B}=\left(3.2\times {10}^{\mathrm{-19}}\text{C}\right)\left(\left(2.0\widehat{i}-3.0\widehat{j}+1.0\widehat{k}\right)\times {10}^4\text{m/s}\right)\times \left(1.5\text{T}\widehat{k}\right)\hfill \\ & =\left(\mathrm{-14.4}\widehat{i}-9.6\widehat{j}\right)\times {10}^{\mathrm{-15}}\text{N.}\hfill \end{array}$$
   This solution can be rewritten in terms of a magnitude and angle in the xy-plane:
   $$\begin{array}{ccc}\hfill \left|\overrightarrow{F}\right|& =\hfill & \sqrt{F_x^2+F_y^2}=\sqrt{{(\mathrm{-14.4})}^2+{(\mathrm{-9.6})}^2}\times {10}^{\mathrm{-15}}\text{N}=1.7\times {10}^{\mathrm{-14}}\text{N}\hfill \\ \hfill \theta & =\hfill & {\text{tan}}^{\mathrm{-1}}\left(\frac{F_y}{F_x}\right)={\text{tan}}^{\mathrm{-1}}\left(\frac{\mathrm{-9.6}\times {10}^{\mathrm{-15}}\text{N}}{\mathrm{-14.4}\times {10}^{\mathrm{-15}}\text{N}}\right)=34\text{°}\text{.}\hfill \end{array}$$
   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:
   $$\begin{array}{ccc}\hfill \left|\overrightarrow{v}\right|& =\hfill & \sqrt{{(2)}^2+{(\mathrm{-3})}^2}\times {10}^4\frac{\text{m}}{\text{s}}=3.6\times {10}^4\frac{\text{m}}{\text{s}}\hfill \\ \hfill F& =\hfill & qvB\text{sin}\theta =(3.2\times {10}^{\mathrm{-19}}\text{C})(3.6\times {10}^4\text{m/s})(1.5\text{T})\text{sin}(90\text{°})=1.7\times {10}^{\mathrm{-14}}\text{N.}\hfill \end{array}$$
   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.