Chapter 32 Inductance and Magnetic Materials The appearance of an induced emf in one circuit due to changes in the magnetic field produced by a nearby circuit is called mutual induction. The response of the circuit is characterized by their mutual inductance. Can we find an induced emf due to its own magnetic field changes? Yes! The appearance of an induced emf in a circuit associated with changes in its own magnet field is called self-induction. The corresponding property is called self-inductance. A circuit element, such as a coil, that is designed specifically to have self-inductance is called an inductor. 1 32.1 Inductance Close: As the flux through the coil changes, there is an induced emf that opposites this change. The self induced emf try to prevent the rise in the current. As a result, the current does not reach its final value instantly, but instead rises gradually as in right figure. Open: When the switch is opened, the flux rapidly decreases. This time the self- induced emf tries to maintain the flux. When the current in the windings of an electromagnet is shut off, the self-induced emf can be large enough to produce a spark across the switch contacts. 2 32.1 Inductance (II) Since the self-induction and mutual induction occur simultaneously, both contribute to the flux and to the induced emf in each coil. The flux through coil 1 is the sum of two terms: N1Φ1 = N1(Φ11 + Φ12 ) The net emf induced in coil 1 due to changes in I1 and I2 is d V = −N (Φ + Φ ) emf 1 dt 11 12 3 Self-Inductance It is convenient to express the induced emf in terms of a current rather than the magnetic flux through it. The magnetic flux is directly proportional to the current flowing through it. N1Φ11 = L1I1 where L1 is a constant of proportionality called the self- inductance of coil 1. The SI unit of self-inductance is the henry (H). The self-inductance of a circuit depends on its size and its shape. The self-induced emf in coil 1 due to changes in I1 takes the form dI1 Vemf_11 = −L1 dt 4 Mutual Inductance The magnetic flux produced by coil 2 is proportional to I2. Thus the flux contributed from coil 2 may be written as N1Φ12 = MI2 where the constant of proportionality is M is called the mutual inductance of the two coils. The SI unit of mutual inductance is the henry (H). The mutual inductance of the two coils depends on its separation and orientation. The emf induced in coil 1 due to changes in I2 takes the form dI V = −M 2 emf_12 dt Are the mutual inductances M12 and M21 the same? Why? 5 Example 32.1 A long solenoid of length L and cross-sectional area A has N turns. Find its self-inductance. Assume that the field is uniform throughout the solenoid. Solution: N Φ = BA = µ IA 0 L NΦ = LI 2 L = µ0n AL Notice that the self-inductance is proportional to the square of the number density. This result may be compared to the capacitance of two parallel plates (C=ε0A/d), which depends on the geometry of the capacitor. 6 Example 32.2 A coaxial cable consists of an inner wire of radius a that carries a current I upward, and an outer cylindrical conductor of radius b that carries the same current downward. Find the self-inductance of a coaxial cable of length L. Ignore the magnetic flux within the inner wire. Solution: µ0 I µ0 I B = , dΦ = BdA = ldx 2πx 2πx b µ0 I µ0 Il b Φ = ldx = ln = LI ∫a 2πx 2π a µ b L = 0l ln 2π a Hint1: The direction of the magnetic field. Hint2: What happens when considers the inner flux? 7 Example 32.3 A circular coil with a cross-sectional area of 4 cm2 has 10 turns. It is placed at the center of a long solenoid that has 15 turns/cm and a cross-sectional area of 10 cm2, as shown below. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance? Solution: Φ12 = B2 A1 = µ0n2 I2 A1 N1Φ12 M = = µ0n2 N1 A1 I2 = (4π ×10−7 )(1500)(10)(0.0004) = 7.54 uF Notice that although M12=M21, it would have been much difficult to find Φ21 because the field due to the coil is quite nonuniform. 8 32.2 LR Circuits How does the current rise and fall as a function of time in a circuit containing an inductor and a resistor in series? dI Rise Vemf − IR − L = 0 dt dI Let I = I e−αt + β ⇒ = −αI e−αt 0 dt 0 R e−αt : α = L Vemf 0 : Vemf − Rβ = 0 ⇒ β = R Vemf t = 0 : I0 = −β = − R R V − t ∴I = emf (1− e L ) R The quantity τ=L/R is called the time constant. 9 32.2 LR Circuits Decay dI − IR − L = 0 dt dI Let I = I e−αt ⇒ = −αI e−αt 0 dt 0 R e−αt : α = L V t = 0 : I = emf 0 R R t V − t V − ∴I = emf e L = emf e τ R R The quantity τ=L/R is called the time constant. 10 31.3 Energy Stored in an Inductor The battery that establishes the current in an inductor has to do work against the opposing induced emf. The energy supplied by the battery is stored in the inductor. In Kirchhoff’s loop rule, we obtain di V = iR + L emf dt di iV = i2 R + Li emf dt dU 1 iV = i2 R + L , where U = LI 2 emf dt L 2 power supplied power dissipated energy change rate by the battery in the resistor in the inductor 11 Energy Density of the Magnetic Field We have expressed the total energy stored in the inductor in terms of the current and we know the magnetic field is proportional to the current. Can we express the total magnetic energy in terms of the B-field? Yes. Let’s consider the case of solenoid. 2 L = µ0n Al 2 1 2 1 2 B U L = LI = (µ0nI) Al = Al 2 2µ0 2µ0 B2 uB = (The energy density of a magnetic field in free space) 2µ0 Although this relation has been obtained from a special case, the expression is valid for any magnetic field. 12 Example 31.4 A 50-mH inductor is in series with a 10-Ω resistor and a battery with an emf of 25 V. At t=0 the switch is closed. Find: (a) the time constant of the circuit; (b) how long it takes the current to rise to 90% of its final value; (c) the rate at which energy is stored in the inductor; (d) the power dissipated in the resistor. Solution: (a) The time constant is τ = L/R = 5 ms −t /τ (b) 0.9I0 = I0 (1− e ) ⇒ t = −τ ln 0.1 =11.5 ms dU dI V (c) L = LI , I = emf (1− e−t /τ ) dt dt R 2 dU V L = R emf (e−t /τ − e−2t /τ ) dt R 2 2 Vemf −t /τ −2t /τ (d) PR = I R = R (1− 2e + e ) 13 R Example 31.5 The breakdown electric field strength of air is 3x106 V/m. A very large magnetic field strength is 20 T. compare the energy densities of the field. 1 Solution: U = ε E 2 = (0.5)(8.85×10-12 )(3×106 )2 E 2 0 = 40 J/m3 2 1 2 20 U B = B = -7 2µ0 2× 4π ×10 = 3.2×108 J/m3 Magnetic fields are an effective means of storing energy without breakdown of the air. However, it is difficult to produce such large fields over large regions. 14 Example 31.6 Use the expression for the energy density of the magnetic field to calculate the self-inductance of a toroid with a rectangular cross section. Solution: µ NI B = 0 2πr 2 2 2 B B µ0h(NI) dU B = dV = h(2πrdr) = dr 2µ0 2µ0 4πr 2 2 2 b µ0h(NI) µ0hN I b 1 2 U B = dr = ln( ) = LI ∫a 4πr 4π a 2 µ N 2h b L = 0 ln( ) 2π a Can we use the concept of magnetic flux to derive the self- inductance? 15 31.4 LC Oscillations The ability of an inductor and a capacitor to store energy leads to the important phenomena of electrical oscillations. 16 31.4 LC Oscillations (II) Consider the situation shown in Fig. 32.11. A circuit consists of a inductor and a capacitor in series. According to Kirchhoff’s loop rule, Q dI − L = 0 C dt dQ I = − (note there is a minus sign) dt Q d 2Q + L = 0 C dt 2 Q = Q0 sin(ω0t +φ) 1 where ω = (natural angular frequency), 0 LC Q0 and φ can be detremined from initial condition. 17 31.4 LC Oscillations (III) The current and charge relation Q = Q0 sin(ω0t +φ) I = −I0ω0 cos(ω0t +φ) The energy stored in the system is 1 1 U = Q2 = Q2 sin 2 (ω t +φ) E 2C 2C 0 0 1 1 U = LI 2 = LQ2ω 2 cos2 (ω t +φ) B 2 2 0 0 0 1 = Q2 cos2 (ω t +φ) 2C 0 0 1 2 U = U E +U B = Q0 (constant) 2C 18 The Analogies Between LC Oscillations and the Mechanical Oscillations LC Osc.