Mutual Inductance Mutual inductance: a changing current in one coil will induce a current in a second coil: And vice versa; note that the constant M, known as the mutual inductance, is the same:

Mutual Inductance Unit of inductance: the henry, H: 1 H = 1 V·s/A = 1 Ω·s. A transformer is an example of mutual inductance. Figure 30-1. A changing current in one coil will induce a current in the second coil.

Mutual Inductance Example A long thin solenoid of length l and cross-sectional area A contains N1 closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance. Solution: Assuming all the flux from the solenoid stays within the secondary coil, the flux is BA = μ0 (N1/l) I1A, and M = μ0 (N1N2/l) A.

Self-Inductance A changing current in a coil will also induce an emf in itself: Here, L is called the self-inductance:

Example Solenoid inductance. (a) Determine a formula for the self-inductance L of a tightly wrapped and long solenoid containing N turns of wire in its length l and whose cross-sectional area is A. (b) Calculate the value of L if N = 100, l = 5.0 cm, A = 0.30 cm2, and the solenoid is air filled. Solution: a. B = μ0nI and is constant; the flux is μ0NIA/l and the self-inductance is μ0N2A/l. b. Plugging in the numbers gives L = 7.5 μH.

Self-Inductance Conceptual Example: Direction of emf in inductor. Current passes through a coil from left to right as shown. (a) If the current is increasing with time, in which direction is the induced emf? (b) If the current is decreasing in time, what then is the direction of the induced emf? Figure 30-4. Example 30–4. The + and - signs refer to the induced emf due to the changing current, as if points A and B were the terminals of a battery (and the coiled loops were the inside of the battery). Solution: a. The induced emf opposes the change that caused it, so the induced emf acts to oppose the current. b. Now the induced emf acts to reinforce the current (as it is decreasing).

Self-Inductance Example: Coaxial cable inductance. Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r1 and the outer conductor has a radius r2. Assume the conductors are thin hollow tubes so there is no magnetic field within the inner conductor, and the magnetic field inside both thin conductors can be ignored. The conductors carry equal currents I in opposite directions. Solution: The flux through a rectangle of width dr and length l is dΦB = (μ0I/2πr) l dr. Integrating over r to find the total flux gives ΦB = (μ0Il/2π) ln (r2/r1). Therefore, L = (μ0l/2π) ln (r2/r1).

Energy Stored in a Magnetic Field Just as we saw that energy can be stored in an electric field, energy can be stored in a magnetic field as well, in an inductor, for example. Analysis shows that the energy density of the field is given by

LR Circuits A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor. Figure 30-6. (a) LR circuit; (b) growth of current when connected to battery.

LR Circuits The sum of potential differences around the loop gives Integrating gives the current as a function of time: . The time constant of an LR circuit is . .

LR Circuits If the circuit is then shorted across the battery, the current will gradually decay away: . Figure 30-7. The switch is flipped quickly so the battery is removed but we still have a circuit. The current at this moment (call it t = 0) is I0. Figure 30-8. Decay of the current in Fig. 30–7 in time after the battery is switched out of the circuit.

LR Circuits Example 30-6: An LR circuit. At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor’s magnetic field? Solution: a. At t = 0 the current is zero. b. The time constant is L/R = 7.3 ms. c. The maximum current is V0/R = 0.40 A. d. When I = ½ Imax, e-t/τ = ½ then t = 5.0 ms. e. P = IV = 2.4 W. f. U = ½ LI2, so dU/dt = LI dI/dt = I(V0-RI) = 1.2 W. The rest of the power is dissipated in the resistor.

LC Circuits and Electromagnetic Oscillations An LC circuit is a charged capacitor shorted through an inductor. Figure 30-10. An LC circuit.

Circuits and Electromagnetic Oscillations Summing the potential drops around the circuit gives a differential equation for Q: This is the equation for simple harmonic motion, and has solutions . .

Circuits and Electromagnetic Oscillations Substituting shows that the equation can only be true for all times if the frequency is given by The current is sinusoidal as well:

LC Circuits and Electromagnetic Oscillations The charge and current are both sinusoidal, but with different phases. Figure 30-11. Charge Q and current I in an LC circuit. The period T = 1/f = 2π/ω = 2π(L/C)1/2.

LC Circuits and Electromagnetic Oscillations The total energy in the circuit is constant; it oscillates between the capacitor and the inductor: Figure 30-12. Energy UE (red line) and UB (blue line) stored in the capacitor and the inductor as a function of time. Note how the energy oscillates between electric and magnetic. The dashed line at the top is the (constant) total energy U = UE + UB.

LC Circuits and Electromagnetic Oscillations Example: LC circuit. A 1200-pF capacitor is fully charged by a 500-V dc power supply. It is disconnected from the power supply and is connected, at t = 0, to a 75-mH inductor. Determine: (a) the initial charge on the capacitor; (b) the maximum current; (c) the frequency f and period T of oscillation; and (d) the total energy oscillating in the system. Solution: a. Q0 = CV = 6.0 x 10-7 C. b. Imax = ωQ0 = 63 mA. c. f = ω/2π = 17 kHz; T = 1/f = 6.0 x 10-5 s. d. U = Q02/2C = 1.5 x 10-4 J.

LC Oscillations with Resistance (LRC Circuit) Any real (nonsuperconducting) circuit will have resistance. Figure 30-13. An LRC circuit.

LC Oscillations with Resistance (LRC Circuit) Now the voltage drops around the circuit give The solutions to this equation are damped harmonic oscillations. The system will be underdamped for R2 < 4L/C, and overdamped for R2 > 4L/C. Critical damping will occur when R2 = 4L/C.

LC Oscillations with Resistance (LRC Circuit) This figure shows the three cases of underdamping, overdamping, and critical damping. Figure 30-14. Charge Q on the capacitor in an LRC circuit as a function of time: curve A is for underdamped oscillation (R2 < 4L/C), curve B is for critically damped (R2 = 4L/C), and curve C is for overdamped (R2 > 4L/C).

LC Oscillations with Resistance (LRC Circuit) The angular frequency for underdamped oscillations is given by . The charge in the circuit as a function of time is .

Oscillations with Resistance (LRC Circuit) Example: Damped oscillations. At t = 0, a 40-mH inductor is placed in series with a resistance R = 3.0 Ω and a charged capacitor C = 4.8 μF. (a) Show that this circuit will oscillate. (b) Determine the frequency. (c) What is the time required for the charge amplitude to drop to half its starting value? (d) What value of R will make the circuit nonoscillating? Solution: a. R2 is less than 4L/C, so this circuit will oscillate. b. f = ω/2π = 360 Hz. c. t = 2L/R ln 2 = 18 ms. d. For R2 ≥ 4L/C, R must be ≥ 180 Ω.