Prerequisites

Students should be familiar with the contents of the parallel Lesson and with the concept of the vector product. A review of the properties of the latter is available in an Appendix on The Vector Product.

Learning Outcomes

Students will develop a more general understanding of the force (magnitude and direction) exerted by a magnetic field on a charged particle and will learn to describe this force in a mathematically efficient way.

Instructions

Students should know how the applet functions, as described in Help and ShowMe.

The applet should be open. The step-by-step instructions in the following text are to be done in the applet. You may need to toggle back and forth between instructions and applet if your screen space is limited.

Vector Equation for the Magnetic Force

Appendix

The Vector Product

In the accompanying Lesson, it is shown that the direction and magnitude of the force exerted by a magnetic field on a particle carrying electric charge q are as follows.

1. Direction of the Force. The direction of the force exerted on a particle moving in a magnetic field can be visualized by means of the right-hand rule illustrated in Figure 1 below.

   

   Figure 1

   The force acting on the particle is normal (perpendicular) to the plane spanned by the particle's velocity vector and the magnetic field vector at that point.

   To determine along which normal to this plane the force points, hold your right hand flat, with outstretched fingers in the direction of the velocity , with the thumb pointing off to one side in a direction perpendicular to the plane spanned by and , and with the palm facing in the direction of . When you curl the fingers of your right hand, they should curl towards . The thumb will then be pointing in the direction of the magnetic force .

2. Magnitude of the Force. If the velocity is at right angles to the magnetic field, the magnitude F of the magnetic force exerted on the particle is given by the expression

   F = |q| v B if . (1)

   If the velocity is parallel to the magnetic field, the force is 0,

   F = 0 if || . (2)

It would be nice to be able to give a single vector equation that combines the right-hand rule and Equations (1) and (2) and generalizes these Equations to the case when is neither perpendicular to nor parallel with .

This is indeed possible, but it requires the use of the vector product. A brief review of the vector product is provided in an Appendix on The Vector Product.

In terms of the vector product, the magnetic force in all situations can be written

= q X. (3)

You need to memorize the order of the vector factors in Equation (3): comes first, last. The direction of is reversed if you reverse the order of these factors. The order of the factors is also important in memorizing the right-hand rule. The outstretched fingers of the right hand are in the direction of the first factor, . In subsequently curling the fingers of the right hand, they are curled towards the second factor, .

Exercise 1. Test if Equation (3) is correct by testing with the applet whether the following predictions of the equation are true:

• reverses when the sign of q is changed;
• reverses when is reversed
; (the applet does not let you change the direction of the initial velocity, but you can observe the direction of the force at a point when the velocity has reversed during the motion);
• reverses when is reversed;
• F is equal to |q| v B if is perpendicular to ;
• F = 0 if is parallel to .

Exercise 2. Write down the Cartesian (x,y,z)-components of the magnetic force by transcribing the general vector product equation (A6) in the Appendix for the special case of force expression (3).

Answer. Substituting the symbols from Expression (3) into Equation (A6) gives

= (F_x,F_y,F_z) = q (v_yB_z - v_zB_y, v_zB_x - v_xB_z, v_xB_y - v_yB_x). (4)

Let us choose the x and y axes to be in the orbital plane (plane of the screen), the x-axis horizontal and to the right and the y-axis vertical and upward. Then the z-axis will be perpendicular to this plane and towards the viewer (out of the screen), to make this a right-handed coordinate system.

Suppose is towards the viewer (out of screen). Then

= (B_x,B_y,B_z) = (0, 0, B) . (5)

Substituting this into Expression (4) gives

= (F_x,F_y,F_z) = q (v_yB - 0, 0 - v_xB, 0 - 0) = q B (v_y, -v_x, 0). (6)

For a velocity that is directed in the positive y-direction,

= (v_x,v_y,v_z) = (0, v, 0) , (7)

Expression (6) reduces to

= q B v(1, 0, 0) . (8)

This is a force pointing in the positive x-direction if q > 0 and in the negative x direction if q < 0. Expression (8) gives |q| v B for the magnitude of the force, which agrees with Expression (1).

Exercise 3. Verify Prediction (8) for the direction of the force by running a motion in the applet and pausing it when the velocity is straight up (in the y-direction). From what you know about the direction of the force in uniform circular motion, is the direction of the force as predicted?

Exercise 4. Work out the components of the force at a moment when the velocity is pointing in the north-west direction. Does Expression (6) yield a force vector that has the right direction and magnitude? Hint: What are the components of the velocity vector in this case, expressed in terms of the magnitude v of the velocity?

Exercise 5. Choose both and to be in the x-direction. Work out the force in this case using Expression (6). Does the applet confirm your result?

Exercise 6. Suppose points in the z-direction and the velocity has a non-zero z-component v_z.

• Using the general Expression (4) for the force, what is the force equal to in this case?
• How does the force compare with Expression (6)?
• Let denote the velocity component vector in the (x,y)-plane. Use the expression for the force you have obtained in the first bullet, to write down an expression for the magnitude of the force in terms of the magnitude v of . Hint: What are the components of ?
• What can you conclude about the way v_z is changing in time?
• What will the motion look like when the initial velocity has a non-zero z-component and the magnetic field is in the z-direction?

>>>>> Appendix <<<<<

Figure A1

The vector product, illustrated in Figure A1, is a rule that assigns to two vectors and a third vector . The vector product is written with an "X" for the multiplication sign,

= X . (A1)

The vector product is therefore also called the cross product.

The vector product is defined only in three, not two or one dimensions, and it does not have all of the properties of multiplication that we are used to from the multiplication of numbers.

To see how the product vector can be obtained from the factors and , let us discuss in turn the direction and magnitude of and the calculation of the components of in terms of those of and .

1. Direction of . Two vectors and span a plane. This plane is indicated in Figure A1 by the dark grey surface. Only in the special case when the two vectors are collinear do they not span a plane. In this case, their vector product is equal to zero.

   The vector = X is normal (perpendicular) to the plane spanned by and .

   The only question is: "Along which normal is it?" There is one normal sticking out of one side of a plane and another normal sticking out on the other side. The following right-hand rule, illustrated in Figure A2 below, lets you decide which normal it is.

   

   Figure A2

   Hold your right hand flat with outstretched fingers in the direction of vector and with the thumb pointing off to one side in a direction perpendicular to the plane spanned by and . There are two such directions in which the thumb could point. Which is the right one?

   The right direction for the thumb is the one in which your palm faces in the direction of vector . When you curl the fingers of your right hand, they should curl towards . The thumb will then be pointing in the direction of = X .

   If your palm faces in the wrong direction, rotate your hand by 180^o around the axis that points along the outstretched fingers.

2. Magnitude of . The magnitude of is equal to the area of the parallelogram spanned by and . This parallelogram is shaded light grey in Figure A1 above.

   When and are collinear, the area of the parallelogram and the vector product = X are zero.

   When and are perpendicular to each other, the parallelogram is a rectangle with area equal to ab. Thus,

   c = || = ab if (A2)

   When and are not perpendicular to each other, the area of the parallelogram can be calculated as illustrated by the diagram in Figure A3 below. The trick is to replace the parallelogram (yellow) by a rectangle (black) of the same area.

   

   Figure A3

   The parallelogram has the sides a and b. The rectangle of the same area has the sides a and b. The area of the rectangle is equal to ab.

   Therefore Equation (A2) can be generalized to

   c = || = ab (A3)

   where the proviso is now no longer necessary.

   One can make a right triangle containing sides a and a and the angle q where the angle q is the angle between the vectors and . Basic trigonometry applied to this triangle implies a = a sin q. Substituting this expression into Equation (A3) gives

   c = || = a (sin q) b (A4)

3. Components of . Suppose the Cartesian (x,y,z)-components are given for and :

   = (a_x,a_y,a_z) (A5a)

   = (b_x,b_y,b_z). (A5b)

   Then the Cartesian components of can be shown to be equal to

   = (c_x,c_y,c_z) = (a_yb_z - a_zb_y, a_zb_x - a_xb_z, a_xb_y - a_yb_x). (A6)

   Notice the "cyclic permutation" of the subscripts x,y,z in this equation.

The vector product behaves like a product of numbers in the following sense. If c and d are numbers, then for any two vectors and it is true that

(c ) X (d ) = (cd) ( X ). (A7)

The vector product also obeys the distributive law. If , , and are three arbitrary vectors, then

( + ) X = X + X . (A8)

However, in a number of other ways the vector product does not behave like a product of numbers.

1. The vector product does not obey the commutative law. Instead, for any two vectors and ,

   X = - X . (A9)

2. The vector product does not obey the distributive law. For a given triplet of vectors , , and , in general ( X ) X is not equal to X ( X ).
3. There exists no unit vector such that its vector product with another vector is equal to .
4. For a given vector there exists no "inverse vector", ^-1, such that X ^-1 is equal to a unit vector like the one imagined in the previous bullet, because such an does not exist.

   This means that one can not divide by vectors.