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Related post:
At this point, let us consider one example of the right hand rule in nature. The magnetic force \(\vec{F}_m\) experienced by a moving charge \(q\) with velocity \(\vec{v}\) in a region with magnetic field \(\vec{B}\) is given by:
\[\begin{equation}\label{eqn:fm} q \vec{v} \times \vec{B} = \vec{F}_m\end{equation}\]
Just as in the previous post, let us use color codes for the vectors involved in the cross product:
The first vector \(q\vec{v}\) is red.
The second vector \(\vec{B}\) is green.
The third vector (the result of the cross product) \(\vec{F}_m\) is blue.
Before we proceed, notice that I specifically included the \(q\) (which is a scalar quantity) in the first vector. This has an important implication. The direction of the first vector is not simply the direction of the velocity; instead, it should include the effect of the sign of the charge \(q\).
Test you understanding: An electron is moving to the left. What is the correct unit vector for the direction of the first vector in Equation (\(\ref{eqn:fm}\))? ANSWER: \(+\hat{i}\). [Can you explain why?]
At this point, we need to use the discussion about the right hand rule (RHR) and cross products from the previous post to solve for representative examples.
Example 1. An electron \(q = -e = -1.60 \times 10^{-19} \textrm{ C}\) is moving to the left with speed \(v = 3.00 \times 10^{3} \textrm{ m/s}\) before encountering a region of uniform magnetic field \(B = 1.00 \times 10^{-5} \textrm{ T}\) directed into the page. What will be the force on this charge at the instant it encounters the magnetic field?
SOLUTION. Let us first identify our vectors:
\(q\vec{v} = \left[-1.60 \times 10^{-19} \textrm{ C}\right] \left[3.00 \times 10^{3} \textrm{ m/s} (-\hat{i})\right] = \left[4.80 \times 10^{-16} \textrm{ Cm/s} \right] \hat{i}\) [negative sign cancels out]
\(\vec{B} = \left[1.00 \times 10^{-5} \textrm{ T}\right] (-\hat{k})\) [negative because into the page]
\(\vec{F}_m\) [to be solved using Equation (\(\ref{eqn:fm}\))]
Now, just like what we always do in class for handling vector quantities, always solve for the magnitude and direction of the vector quantity separately.
MAGNITUDE: Recall that the magnitude of the cross product is given by the product of the individual vector magnitudes multiplied by the sign of the angle between them. Also, note that magnitudes are always positive.
\(\left|\vec{F}_m\right| = F_m = qvB \sin 90^\circ = qvB = 4.8 \times 10^{-21} \textrm{ N}\) [\(90^\circ\) because perpendicular]
DIRECTION: Using the RHR, you can verify that the cross product of the unit vectors given above yields:
\(\hat{i} \times (-\hat{k}) = \hat{j}\)
[How to do that again? STEP 1: Put your four fingers (right hand) towards \(\hat{i}\) (right); STEP 2: Your palm should open towards \(-\hat{k}\) (into the page or away from you); STEP 3: Curling the four fingers yield a thumb that points \(\hat{j}\) (upward).]
FINAL ANSWER: \(\boxed{\vec{F}_m = \left[4.8 \times 10^{-21} \textrm{ N}\right] \hat{j}}\).
Exercise 1. This time, use the same values as Example 1, but only with a positron instead of an electron. How does your answer compare with the answer given above?
The problem given in Example 1 is illustrated in Figure 1 below. Notice how the directions of the specified vectors with their corresponding colors are shown here. The green arrows that point into the page is the region of uniform magnetic field \(\vec{B}\). Again, it is worth emphasizing that \(q\vec{v}\) has a different direction from the velocity vector \(\vec{v}\) (here shown in black) because of the sign of the charge of the electron. You can also check using the RHR that we got the correct direction for \(\vec{F}_m\).
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| Figure 1. Illustrating Example 1. |
At this point, let us check what will happen to the electron upon entering this region. I want you to focus on \(\vec{v}\) (black vector) and \(\vec{F}_m\) (blue vector). Notice that the force is perpendicular to the velocity. From kinematics, can you remember what happens when this is the case?
Recall that when the force is in the same direction as the velocity, then it tends to increase the velocity vector: the object speeds up. On the other hand, when the force is in the exact opposite direction as the velocity, it tends to decrease the velocity vector: the object slows down. Here, neither of these cases is true; the force has no component that is parallel to the velocity.
But according to Newton's Second Law, the force will produce an acceleration for the massive object. So how will this force produce an acceleration when it can't change the speed of the object? Well, in this case, it will change the direction of the velocity vector. In this case, without changing how fast the electron moves, the magnetic force tends to change the direction of the velocity. The next instant of the electron's motion is schematically illustrated in Figure 2 below.
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| Figure 2. Next instant after Figure 1, greatly exaggerated. Same vector magnitudes, but different directions. |
The force changed the trajectory of the electron's motion without changing the magnitudes of the velocity, and, thus, the magnetic force. But then again, one of the things that we will realize here is that the magnetic force is still perpendicular to the velocity! So at the next instant, the electron will again change direction due to the perpendicular magnetic force, but it will maintain its speed (the magnitude of its velocity), and so on. Figure 3 shows the eventual motion of the electron:
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| Figure 3. Eventual motion of the electron. Magnitudes of the velocity and magnetic force are not changing, just their directions. The velocity is always perpendicular to the magnetic field. |
Wait, have you encountered something like this before? Yes. The case when the velocity and force magnitudes are constant, but are always perpendicular, results in UNIFORM CIRCULAR MOTION. In Example 1, therefore, the electron will trace a circular path as long as it is inside the region with magnetic field. Note that when the electron leaves the region with magnetic field, there will be no more magnetic force (and thus, no acceleration), so the velocity will be unchanged (both magnitude and direction are constant, resulting in uniform linear motion).
Exercise 2. This time, use the same values as Example 1, but only with a positron instead of an electron. How will the positron move? How similar/different is the motion from that of an electron?
The motion of charged particles in regions with uniform magnetic fields is called cyclotron motion, and the simple uniform circular motion discussed above is one of the possible cases of cyclotron motion. Other examples are discussed in our book.
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