Download Energy Storage in Inductors and Magnetic Fields - Prof. Phillip Duxbury and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 29 Chapter 9 of PS, Chapters 6,7 of Griffiths A. Energy stored in inductors and in magnetic fields An external voltage source must be used to set up a magnetic field in an inductor. The rate at which work is done by the external source is, P = dW dt = EextI (1) We have shown that the voltage across an inductor is −LdI/dt, so we have, dW dt = −(−LI dI dt ) (2) The total energy stored in an inductor is the integral of the work so that, U = ∫ dW dt dt = ∫ i 0 LI dI dt dt = 1 2 Li2 (3) This energy is stored in the magnetic field of the inductor. By considering a solenoid we can find the energy density in the magnetic field, through U = 1 2 Li2 = 1 2 µ0n 2AlI2 = 1 2µ0 B2Al (4) Using Al = volume, the energy density in the magnetic field is then to be, u = 1 2µ0 B2 (5) We have introduced the concept of self-inductance which characterizes the ability of a wire configuration to store energy in the form of magnetic fields. This energy storage can be large and has been proposed as an alternative to batteries, provided superconducting wires are used in order to minimize the resistive losses. As you will see in the assignment very large amounts of energy are stored in the earth’s magnetic field and in the magnetic fields of galaxies. B. Mutual inductance Self-inductance is defined through Nφ = Li = λ, where λ is the flux linkage. Self inductance is a measure of the ability of a conducting system to react to a current flow in the system. However, from consideration of Faraday’s law, it is clear that the magnetic field 1 of one current carring wire configuraiton may influence a second one that is not physically connected to it. This coupling is described by the mutual inductance. We write, N2φ2 = M12i1 so that V2 = −N2 dφ2 dt = −M12 di1 dt (6) This coupling is the basis of antennas, transformers and a variety of other essential elements of technology. Calculations of mutual inductance are similar to that of self-inductance. We consider two examples. A rectangular loop near a wire carrying current i1, and N2 = N1 = 1 We take the loop to have its long sides, of length l2, parallel to a long straight wire carrying current i1. The side which is closest to the wire is distance a from the wire. The short sides have length b1, and the normal to the loop is in the φ̂ direction with respect to the axis of the wire. In that case, we need to find φ2 = M12i1, so we have a by now quite familiar calculation, φ2 = ∫ l 0 dy ∫ a+b a µ0i 2πr = µ0il 2π ln((a+ b)/a), so that M12 = µ0l 2π ln((a+ b)/a) (7) Two solenoids - one inside the other The two solenoids are characterized by cross-sectional areas A1, A2 turns N1, N2 and lengths l1, l2, with the former having A1 > A2. We also take the lengths of both solenoids to be much larger than their radii, so the formula B = niµ0 applies, with n = N/l. If a current i1 flows in the larger solenoid, the flux in the smaller solenoid is then φ2 = N1i1A2µ0/l1 and the so using, N2φ2 = M12i1 implies that M12 = N1N2µ0A2/l1 (8) Now lets consider the reverse situation where a current i2 flows in the smaller solenoid. In that case there is no magnetic field outside the smaller solenoid, so the flux in the larger solenoid is, φ1 = N2i2µ0/l2, then using N ′1φ1 = M12i2 implies that M21 = N ′ 1N2µ0A2/l2 (9) In this expression we use A2 because we assume that there is no field outside the smaller solenoid. N ′1 is the number of loops in length l2, so N ′ 1 = l2n1 = l2N1/l1. With this substitution it is evident that M21 = M12, so the mutual inductance is symmetric. Now 2
