Inductance of a solenoid

As we know, when a solenoid has a current passing through it, there is an approximately uniform magnetic field passing through its core. In the circuit at right, the battery is a source of emf, and the current is determined only by the internal resistance of the battery and any resistance in the wire (via I=V/R). But what would happen if we had a variable emf, and could increase the current? Lenz’s Law tells us that the solenoid would create a magnetic field trying to oppose the increase in B, and Faraday’s Law tells us that the solenoid itself would develop an emf opposite in direction to the battery (reducing the total emf). μ dΦ d d ⎛ N ⎞ ⎛ μ N 2 A ⎞ dI solen ⎜ 0 ⎟ emf = ΔV = − = −N ()BA = −N ⎜ 0 IA⎟ = −⎜ ⎟ dt dt dt ⎝ l ⎠ ⎝ l ⎠ dt The last quantity in parentheses is the “inductance” of a solenoid, “L”:

2 dI L unit: V ⋅s μ0 N A General: 1 H = 1 Lsolen = V = −L “Henry” l dt A A solenoid is an example of an “inductor” (L)

This is a new circuit element with the following symbol:

What does an inductor “do” in a circuit: (1) Like a resistor or a capacitor, it can have a voltage drop or rise across its terminals. A resistor develops a voltage difference V, when a current flows through it. But an inductor develops a voltage difference V when the current through it changes. (2) A capacitor can store charge, but an inductor stores magnetic field. (3) An inductor opposes rapid changes of the current passing through it.

Inductors can be made out of loops, solenoids, toroids, or any other shape that creates a magnetic field threading its own loops.

The inductance of any such device can be greatly increased by filling its magnetic volume with ferromagnetic material. Comparing behavior of circuit devices: R vs L Energy stored in an inductor dI For all types of inductor, we have the basic equation: V = −L dt As an inductor is charged, the power into the inductor is the rate at which energy is being deposited in the device: dI dU P = −VI = LI = dt dt So during each time interval dt, we have simply: dU = LIdI We can integrate dU to find the total potential energy stored in the inductor as it is energized, from I = 0 to the final current: UI B 1 2 U = dU = LIdI = LI = U B ∫∫002

This is magnetic potential energy stored in the magnetic field inside the inductor when it has reached the final current. Note the similarities with the corresponding capacitor derivation, both in the steps taken, and in the form of the final result. Energy density of a magnetic field

Just as we used a capacitor to find the energy density of an electric field, we can use a solenoidμ to find the energy density of a magnetic field: N 2 A L = 0 = μ n2 Al and B = μ nI l 0 0 μ Putting these into the expression for the stored magnetic energy: 2 1 1 ⎛ μB ⎞ B 2 2 2 ⎜ ⎟ U B = LI = ( 0 n Al)⎜ ⎟ = ()Al 2 2 ⎝ 0 n ⎠ 2μ0 But Al is the solenoid volume. We divide by it to find the energy density in the magnetic field: B 2 uB = 2μ0

This result also applies to magnetic fields in free space. We can combine it with the electric field energy density to find the expression for the total field energy density in any given region: ε 1 ⎛ B 2 ⎞ ⎜ 2 ⎟ utotal = uE + uB = ⎜ 0 E + ⎟ 2 ⎝ μ0 ⎠ Equivalent inductance of inductors in series or parallel

Series: Leq = L1 + L2 + ... 1 1 1 Parallel: = + + ... Leq L1 L2

These equations are identical to those used for finding equivalent resistance. As in that case, in series, the same current goes through each device. In parallel, all devices see the same voltage difference. Increasing the inductance

The inductance of any inductor may be increased by filling its volume—where the magnetic field is stored—with a ferromagnetic material. The very large increase in the magnetic field caused by the magnetization factor of the material increases the induced emf (the voltage across the terminals) by the same factor. The RL circuit: de-energizing an inductor

This circuit has a resistor in series with an inductor which has been energized by t = 0. We want to find the current as a function of time, as the resistor dissipates energy and the inductor de-energizes.

The sum of voltage drops around the loop = 0: dI V = V RI = −L (Minus sign since L is de-energizing.) R L dt Now separate the equation and integrate each side: dI R ⎛ I ⎞ R t = − dt ln(I) − ln(I 0 ) = ln⎜ ⎟ = − t = − I L ⎝ I 0 ⎠ L τ t − Now exponentiate: I = e τ I 0

t R − − t L Giving: τ L τ = I = I 0e = I 0e with R Recall RC circuit! The RL circuit: energizing an inductor

If we start instead with a de-energized inductor,

and close switch S1, the battery will energize the inductor through the resistor, R. The analysis is similar to that on the previous slide.

The sum of voltage drops around the loop = 0: dI V +V +V = V − RI − L = 0 batt R L batt dt

After more mathematical thrashing (see text):

t V ⎛ − ⎞ L I = batt ⎜1− e τ ⎟ with τ = ⎜ ⎟ R R ⎝ ⎠ The LC Oscillator Circuit

We’ve seen that in RC and RL circuits, the current rises or falls exponentially with time. What happens if we connect an inductor and capacitor in series—with nothing else in the circuit—and then energize one of them? dI Q The sum of voltage drops around the loop = 0: − L − = 0 dt C dI d ⎛ dQ ⎞ d 2Q Since current is the flow of charge: = ⎜ ⎟ = dt dt ⎝ dt ⎠ dt 2 Putting this into the loop equation we get: d 2Q Q d 2Q 1 − L − = 0 or + Q = 0 dt 2 C dt 2 LC This equation says that the charge Q is oscillating with time, as we now show: ω ω 2 dQ ω d Q 2 Try: Q = Q0 cos(ωt) = − Q sin(ωt) = − Q0 cos(ωt) dt 0 ω dt 2

Put this Q, and its second derivative, 2 1 1 − Q0 cos( t) + Q0 cos(ωt) = 0 ω = into the loop equation, and cancel! LC LC The LC oscillator: “watching the oscillations” The RLC circuit: oscillator circuit with damping

1. Charge the capacitor with the switch, as shown. 2. Move the switch to position a and the LC circuit will start to oscillate. 3. The resistor will dissipate energy as the current flows, damping the oscillations.