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Magnetic Forces Chapter 13: Magnetic Forces Chapter Learning Objectives: After completing this chapter the student will be able to: Calculate the force on a charged particle moving at a uniform speed through a magnetic field. Calculate the force between two current-carrying conductors. Calculate the torque on a loop of current in a constant magnetic field. You can watch the video associated with this chapter at the following link: Historical Perspective: Henrik Antoon Lorentz (1853-1928) was a Dutch theoretical physicist who shared the 1902 Nobel Prize in Physics. His work focused on special relativity and quantum mechanics, but he also did research in the area of electromagnetic fields. The Lorentz force is named after him. Photo credit: https://upload.wikimedia.org/wikipedia/commons/3/33/Hendrik_Antoon_Lorentz.jpg, [Public domain], via Wikimedia Commons. 1 13.1 Magnetic Force on a Charged Particle We have now seen how we can use Ampere’s Law, magnetic vector potential, and the Biot- Savart Law to calculate the magnetic flux density, B, at any point in the vicinity of current density J. But there is still one more piece to the puzzle: What does this magnetic flux density do, exactly? It turns out that moving charges create magnetic flux density, and magnetic flux density creates forces on other moving charges, as shown in Equation 13.1. (Equation 13.1) We can calculate the magnitude of the force by the calculating the product qvB, while the direction can be found using the right-hand rule for the cross-product, reversing the direction of the force if the charge is negative. z F q<0 v y q Fq>0 B x Figure 13.1. Determining the direction of magnetic force Example 13.1: What is the force on an a charge of -0.1C with a velocity of 50ay m/s passing through a magnetic field of 1.5az T? 2 Hopefully you will recall that the electrical force on a charge is the charge multiplied by the electric field: (Equation 13.2) Often, a charged particle will be subject to both electric fields and magnetic fields at the same time. In this case, the force on the particle will be the sum of the two forces, and we call this combination the Lorentz Force, as described in Equation 13.3. (Equation 13.3) The electric force will always be in the same direction as the electric field (for a positive charge), and the magnetic force will always be perpendicular to both the velocity and the magnetic flux density. Example 13.2: What is the Lorentz force on the particle from Example 13.1 if there is also an electric field of 100ax V/m present? 13.2 Gyroradius If there is no electric field, then the Lorentz force reduces back down to Equation 13.1, the force due to the magnetic field. This is a very interesting equation, because it shows that the force is always perpendicular to the direction of motion. Any system that demonstrates force perpendicular to velocity will result in circular motion, as shown in Figure 13.2. v B F B r B Figure 13.2. Magnetic fields cause circular motion. 3 The radius of this rotation, called the gyroradius or the Larmor radius, can be calculated from fundamental physics. We know that the magnitude of the centripetal acceleration necessary for a particle of mass m to rotate with a velocity v and at a radius of ris: (Equation 13.4) Combining Equations 13.1 and 13.4 and considering only the magnitudes, we find: (Equation 13.5) Solving this equation for r, we find the gyroradius to be: (Equation 13.6) Example 13.3: An electron is accelerated through a voltage of 0.1V, and it is then subject to a magnetic field of 0.01T. Calculate the gyroradius. Example 13.4: Repeat example 13.3 where the particle is a singly ionized potassium atom (19 protons, 20 neutrons, and 18 electrons). 4 One final consequence of the fact that the force is always perpendicular to the velocity. Since work is the integral of the dot product between force and direction of motion, and since the dot product of perpendicular vectors is zero, magnetic fields alone do not do any work to a particle: (Equation 13.7) 13.3 Force on a Current-Carrying Conductor So far in this chapter, we have been studying the force on a free charged particle that just happens to be traveling through a magnetic field. But most of the time, charged particles are carried along the length of a conductor. It turns that if the charges within a conductor are subject to a force due to a magnetic field, then they will exert this force on the conductor itself. Our task is to determine the magnitude and direction of the force on this conductor. Beginning with Equation 13.1, and considering a differential charge dQ, we obtain the following equation: (Equation 13.8) If we then replace dQ with rvdv, we find: (Equation 13.9) Charge density rv multiplied by velocity v is equal to current density J, and we can break the differential volume into a differential surface area multiplied by a differential length: (Equation 13.10) Current density multiplied by surface area gives current, and the direction of the current density can combine with the dl to give a vector differential term: (Equation 13.11) Reversing the direction of the cross product (which introduces a negative sign) and taking the integral, we obtain the final version of this equation: (Equation 13.12) 5 Example 13.5: Calculate the force on a wire carrying current I2 near a second parallel wire carrying current I1 in the same direction. Do parallel wires with current flowing in the same direction attract or repel each other? y x z I1 I2 Example 13.6: Calculate the force on a wire carrying current I2 near a second parallel wire carrying current I1 in the opposite direction. Do parallel wires with current flowing in the opposite direction attract or repel each other? y x z I1 I2 13.4 Ampere’s Force Law We can generalize the calculations from Examples 13.5 and 13.6 to handle any current-carrying conductor interacting with another current-carrying conductor. Consider the force on wire #2 caused by the magnetic fields created by wire #1: 6 (Equation 13.13) From the Biot-Savart Law, the magnetic field at the location of wire #2 caused by the current flowing in wire #1 is: (Equation 13.14) Substituting Equation 13.14 into Equation 13.13, we obtain Ampere’s Force Law: (Equation 13.15) Notice that, from this equation, we can calculate the force on wire #2 from the magnetic field of wire #1, given only the shapes of the two wires and the currents flowing through them. Ampere’s Force Law is not often used analytically, but it can be useful to perform computer calculations of force. It has many similarities to Coulomb’s Law, including a dependence on 1/R2. The integrals are shown as being closed-loop integrals since in practice it is impossible to have isolated current elements that are not part of a loop. 13.5 Force and Torque on a Loop of Current In your daily life, you don’t likely encounter many examples of a straight piece of wire being subject to a force caused by a linear magnetic flux density. However, electric motors operate on the principle of loops of current-carrying wire in the presence of a magnetic field, so they such a loop of current merits special attention. Consider Figure 13.3, which shows a rectangular loop of current in the presence of a magnetic field passing through its center. We will assume that the current is being created externally to the loop, and that there is a very small (negligible) break in the loop that allows the current to be inserted into the loop. 7 FTOP I Dx y FLEFT FRIGHT B Dy I x I z I F BOTTOM Figure 13.3. Net force on a loop of current in a magnetic field (side view). Applying Equation 13.1 and the right-hand rule to each of the four sides of this rectangle, we obtain the force on each side. For example, for the top conductor, we could apply Equation 13.11 (simplified for a constant current and a straight wire) to obtain: (Equation 13.16) Similarly, we will find that the force on each of the four sides is pointing outward, and the net force on the loop is zero. If the current flows in the opposite direction, all four forces point inward, and the net force is still zero. But there has to be more to the story, since this is supposedly how electric motors work. There is a second part to the story, and it is the torque on the loop. Consider Figure 13.4, which shows a top view of the current loop when it is not perpendicular to the field. F1 m I B q I F3 Figure 13.4. Torque on a loop of current in a magnetic field (top view). 8 Whereas linear force can be thought of as a push or pull, torque can be thought of as a twist or turn. In Figure 13.4, F1 and F3 will combine to rotate the coil of wire counter-clockwise, bringing the magnetic dipole moment m (described in section 12.4) to line up with the externally applied magnetic flux density. The torque can be calculated as follows: (Equation 13.17) Here, F is the force causing the torque, r is the “moment arm,” which is the distance between the origin and the point where the force is being applied, and q is the angle between the force and the moment arm.
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