Lecture 18, Oct. 18, 2000

Recall from last lecture:

Generators: emf = N B A w sin(wt)

20.7 Self-Inductance

Consider a circuit consisting of a solenoid, a resistor, a battery, and a switch (sketch circuit). If initially the switch is open, then no current flows in the circuit and the magnetic field (and flux) in the solenoid is zero. When the switch is closed, current begins to flow through the circuit. The current in the solenoid produces magnetic flux in the solenoid. According to Lenz's law, this increase in flux will be opposed by an induced emf (and corresponding current) in the windings of the solenoid. This opposing emf is such as to limit the current to less than its nominal value of i₀ = V/R. A short instant later, the induced emf will abate, and the current through the solenoid can increase. The increase in current produces another increase in flux, and a corresponding opposing emf. This process continues until the current reaches the nominal value. This phenomenon is known as self-induction because the changing flux through a circuit comes from the circuit itself.

Now let us look at the situation more quantitatively. The induced emf is given by Faraday's law:

emf = -N DF / Dt.

Let's assume that we have an ideal solenoid, in which case the magnetic field inside is B = m₀nI, and the flux through the solenoid is F = BA = m₀nIA, where A is the cross-sectional area of the solenoid. Any change in flux must be produced by a change in the current, DI, since m₀, n, and A remain constant. Therefore, DF = m₀n A DI. Going back to the expression for the induced emf, we have:

emf = -N m₀ n A DI / Dt = -L DI / Dt

where L is a proportionality constant called the inductance of the device. This result says that the induced emf is proportional to the time rate of change of the current.

The inductance is the proportionality constant between the time rate of change of the current and the induced emf. Note the minus sign in the expression for self-induced emf. This means that if the current is increasing, the emf is negative, so as to oppose the increase, and if the current is decreasing, the emf is positive, opposing the decrease.

The unit of inductance is called the henry, abbreviated H. From the expression for self-induced emf we see that 1H = 1Vs/A. In the special case of an ideal solenoid,

L = m₀ N n A.

Recall that n = N/L, so we can write the inductance of an ideal solenoid as

L = m₀ N² A / L.

This is the analog of the expression for the capacitance of a parallel plate capacitor, C = e₀A/d.

Example: P20.36

A solenoid of radius 2.5cm has 400 turns and a length of 20 cm. Find (a) its inductance and (b) the rate at which current must change through it to produce an emf of 75mV.

(a) Using the expression for the inductance of a solenoid:

L = m₀ N² A / L = 4p×10^-7 (400)² p (0.025m)² / 0.20m = 2.0×10^-3H = 2.0mH.

To induce an emf of 75mV, the current must change at the rate (ignore the minus sign for this calculation, it only effects whether the current increases or decreases, but the value of the rate remains the same)

DI/Dt = emf/L = (75mV)/(2.0mH) = 38 A/s.

20.8 RL Circuits

There are many parallels between inductance and capacitance. We've already seen the parallel between the expressions for the capacitance of a parallel plate capacitor and the inductance of a solenoid, and though they may seem unrelated, the relations between voltage and charge for a capacitor, Q = CV, and between induced emf and rate of change of current for an inductor, emf = -L DI / Dt.

Now let's consider the parallel of the capacitor, the inductor. An inductor is a circuit element having inductance, L, and is represented in circuit schematics with a symbol that looks like a coil of wire (sketch).

A circuit containing a resistor and inductor is called an RL circuit. Let's look at a circuit containing a resistor, R, an inductor, L, a battery, V, and switch, and ask what happens when the switch is closed. While the switch is open, no current can flow through the circuit, and there is no magnetic field in the inductor. When the switch is closed, current begins to flow, magnetic flux is created in the inductor, and an emf is setup opposing the flux. Due to the resistance, the maximum current that can flow is I₀ = V/R. The actual current flow is initially less, due to the presence of the inductor:

I = (V/R)(1 - e^-Rt/L) = I₀ (1 - e^-t/t)

where t = L/R is the time constant for the circuit, similar to the time constant for RC circuits.

20.9 Energy Stored in a Magnetic Field

There is yet another similarity between capacitors and inductors, they both store energy. Recall that the energy stored in a capacitor with voltage V between its plates, holding charge Q is:

PE_C = ½QV = ½CV² = Q²/2C

We find it useful to think of this energy as being stored in the electric field created between the plates of the capacitor. Inductors don't have electric fields, but there is a magnetic field, and it takes energy to create. An inductor L, carrying current I, holds energy equal to:

PE_L = ½LI²

Example: P20.45

A 24V battery is connected in series with a resistor and an inductor, where R = 8.0W and L = 4.0H. Find the energy stored in the inductor (a) when the current reaches its maximum value and (b) one time constant after the switch is closed.

(a) The maximum value of the current is I₀ = V/R = 24V / 8.0W = 3.0A. The energy stored in the inductor is then:

PE_L = ½LI² = (4.0H)(3.0A)²/2 = 18J

(b) One time constant after the switch is closed (that is, after current begins flowing in the circuit), I = I₀(1 - e^-1) = 0.37 I₀ = 1.11A. Then energy stored at this current is:

PE_L = ½LI² = (4.0H)(1.11A)²/2 = 2.5J.

Summary of the properties of circuit elements.
	Resistor	Capacitor	Inductor
units	ohm, W = V / A	farad, F = C / V	henry, H = V s / A
symbol	R	C	L
relation	V = I R	Q = C V	emf = -L (DI / Dt)
power dissipated	P = I V = I² R = V² / R	0	0
energy stored	0	PE_C = C V² / 2	PE_L = L I² / 2