Chapter 30 INDUCTANCEmorse/P272ASp13-Ind.pdf · Calculating self- inductance and self -induced emf (30.3+30.4.) • Determine the self -inductance of a toroidal solenoid with cross-sectional

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INDUCTANCE Chapter 30


Goals for Chapter 30

• To learn how current in one coil can induce an emf in another unconnected coil

• To relate the induced emf to the rate of change of the current

• To calculate the energy in a magnetic field

• To analyze circuits containing resistors and inductors

• To describe electrical oscillations in circuits and why the oscillations decay


Introduction • How does a coil induce a

current in a neighboring coil.

• A sensor triggers the traffic light to change when a car arrives at an intersection. How does it do this?

• Why does a coil of metal behave very differently from a straight wire of the same metal?

• We’ll learn how circuits can be coupled without being connected together.


Mutual inductance • Mutual inductance: A

changing current in one coil induces a current in a neighboring coil.

"2 = ¡N2d©B2

dt

N2©B2 = M21i1

N2d©B2

dt = M21di1dt

"2 = ¡M21di1dt

M21 = N2©B2

i1

"2 = ¡M di1dt and "1 = ¡M di2

dt

M = N2©B2

i1= N1©B1

i2


Mutual inductance examples 30.1. and 30.2. • In one form of Tesla coil, a long solenoid with length l and cross-

sectional area A is closely wound with N1 turns of wire. A coil with N2 turns surrounds it at its center. Find the mutual inductance M. (N1 = 1000, N2 = 10 turns, l = 0.50 m, A = 10 cm2)

• Suppose the current i2 in the outer coil is given by i2 = 2e6 A/s t.

• At t = 3.0 µs find the average magnetic flux through each turn of the solenoid due to the current in the outer coil? What is the induced emf in the solenoid?


Self-inductance • Self-inductance L: A varying current in a circuit induces an emf in

that same circuit.

L = N©Bi

N d©Bdt = L di

dt

" = ¡L didt


Inductors as circuit elements Purpose: oppose any variations of current through the circuit despite

fluctuations in the applied emf.

Device with a particular value of inductance is called inductor or a choke.

How to include inductor in the circuit analysis? H ¡!

En ¢ d¡!l = ¡d©B

dtH ¡!En ¢ d

¡!l = ¡L di

dt

¡!Ec +

¡!En = 0

Assume negligible resistance.

R ba

¡!En ¢ d

¡!l = ¡L di

dtR ba

¡!E c ¢ d

¡!l = L di

dt

Vab = Va ¡ Vb = L didt Potential difference across inductor.


Potential across an inductor

• The potential across an inductor depends on the rate of change of the current through it.

• Figure at the right compares the behavior of the potential across a resistor and an inductor.

• The self-induced emf does not oppose current, but opposes a change in the current.


A small, circular ring of wire (shown in blue) is inside a

larger loop of wire that carries a current I as shown. The small

ring and the larger loop both lie in the same plane. If I

increases, the current that flows in the small ring

Q30.1

A. is clockwise and caused by self-inductance.

B. is counterclockwise and caused by self-inductance.

C. is clockwise and caused by mutual inductance.

D. is counterclockwise and caused by mutual inductance.

Small ring

I

I

Large loop


A small, circular ring of wire (shown in blue) is inside a

larger loop of wire that carries a current I as shown. The small

ring and the larger loop both lie in the same plane. If I

increases, the current that flows in the small ring

A30.1

A. is clockwise and caused by self-inductance.

B. is counterclockwise and caused by self-inductance.

C. is clockwise and caused by mutual inductance.

D. is counterclockwise and caused by mutual inductance.

Small ring

I

I

Large loop


A current i flows through an inductor L in the direction from point b toward

point a. There is zero resistance in the wires of the inductor. If the current is

decreasing,

Q30.2

A. the potential is greater at point a than at point b.

B. the potential is less at point a than at point b.

C. The answer depends on the magnitude of di/dt compared to the magnitude of i.

D. The answer depends on the value of the inductance L.

E. both C. and D. are correct.


A current i flows through an inductor L in the direction from point b toward

point a. There is zero resistance in the wires of the inductor. If the current is

decreasing,

A30.2

A. the potential is greater at point a than at point b.

B. the potential is less at point a than at point b.

C. The answer depends on the magnitude of di/dt compared to the magnitude of i.

D. The answer depends on the value of the inductance L.

E. both C. and D. are correct.


Calculating self-inductance and self-induced emf (30.3+30.4.)

• Determine the self-inductance of a toroidal solenoid with cross-sectional area A and mean radius r, closely wound with N turns of wire on a nonmagnetic core. Assume that B is uniform across a cross-section (that is, neglect the variation of B with distance from the toroid axis).

• If the current in the toroidal solenoid increases uniformly from 0 to 6 A in 3 µs, find the magnitude and direction of the self-induced emf.


A steady current flows through an inductor. If the current is doubled while the inductance remains constant, the

amount of energy stored in the inductor

Q30.3

A. increases by a factor of .

B. increases by a factor of 2.

C. increases by a factor of 4.

D. increases by a factor that depends on the geometry of the inductor.

E. none of the above

2


A steady current flows through an inductor. If the current is doubled while the inductance remains constant, the

amount of energy stored in the inductor

A30.3

A. increases by a factor of .

B. increases by a factor of 2.

C. increases by a factor of 4.

D. increases by a factor that depends on the geometry of the inductor.

E. none of the above

2


Magnetic field energy • The energy stored in an inductor U.

• Lets assume that the current is increasing at di/dt rate.

The energy dU supplied to inductor during dt is:

So energy stored in inductor is:

Vab = L didt P = Vabi = Li didt

dU = Li di

U = LR I0

idi = 12LI2


Magnetic energy density

Energy in the inductor is stored in the magnetic field within the coil.

L = ¹0N2A

2¼r

U = 12LI2 = 1

2¹0N

2A2¼r I2

u = U2¼rA = 1

2¹0N2I2

(2¼r)2

B = ¹0NI2¼r

N2I2

(2¼r)2 = B2

¹20

u = B2

2¹0u = B2

2¹


Magnetic field energy • The energy stored in an inductor is U = 1/2 LI2.

• The energy density in a magnetic field is u = B2/2µ0 (in vacuum) and u = B2/2µ (in a material).


Example 30.5: Storing energy in an inductor The electric power industry would like to find efficient ways to

store electric energy generated during low-demand hours to help meet customer requirements during high-demand hours. Could a large inductor be used? What inductance would be needed to store 1 kWh of energy in a coil carrying 200-A current?


The R-L circuit

• An R-L circuit contains a resistor and inductor and possibly an emf source.

• At t = 0, switch S1 closes.

• If we apply Kirhoff’s loop rule:

• At t = 0, i = 0 and vab = 0:

vab = iR vbc = L didt

"¡ iR¡ L didt = 0

didt = "¡iR

L = "L ¡

RL i

( didt )initial = "L

Greater the inductance, the slower the current increase.

As the current approaches final value I, rate of increase drops to zero:

( didt )final = 0 = "L ¡

RL I and I = "

R

I = "R


Current growth in an R-L circuit

• Lets derive current as a function of time in R-L circuit.

• So the current is:

• At t = 0, I = 0 and di/dt = ε/L.

• At t∞, iε/R and di/dt 0

• Time constant measures how quickly current builds up to its final value:

dii¡("=R) = ¡R

Ldt

i = "R (1¡ e¡(R=L)t)

¿ = LR "i = i2R + Li didt


Current decay in an R-L circuit

Current varies according to:

Energy analysis for the current

decay:

It is all very similar to charging

and discharging capacitor in RC circuit.

i = I0e¡(R=L)t

0 = i2R + Li didt


Example 30.6. A sensitive electronic device of resistance R= 175 Ω is to be

connected to a source of emf (of negligible internal resistance) by a switch. The device is designed to operate with a 36 mA current, but to avoid damage to the device, the current can rise to no more than 4.9 mA in the first 58 µs after the switch is closed. An inductor is therefore connected in series with the device: the switch in question is S1.

a) What is the required emf source ε1?

b) What is the required inductance L?

c) What is the R-L time constant τ?


Example 30.7: Energy in R-L circuit When the current in an R-L circuit is decaying, what fraction of

the original energy stored in the inductor has been dissipated after 2.3 time constants?


The L-C circuit

• An L-C circuit contains an inductor and a capacitor and is an oscillating circuit.


Electrical oscillations in an L-C circuit

¡L didt ¡

qC = 0

Kirhoff’s loop rule:

i = dq=dt and di=dt = d2q=dt2

d2qdt2 + 1

LC q = 0

q = Qcos(!t + Á) ! =q

1LC

i = ¡!Qsin(!t + Á)


Electrical and mechanical oscillations

• Table on the right summarizes the analogies between SHM and L-C circuit oscillations.


Example 30.8: An oscillating circuit A 300-V dc power supply is used to charge a 25-µF capacitor.

After the capacitor is fully charged, it is disconnected from the power supply and connected across a 10-mH inductor. The resistance in the circuit is negligible.

a) Find the frequency and period of oscillation of the circuit.

b) Find the capacitor charge and the circuit current 1.2-ms after the inductor and capacitor are connected.


The L-R-C series circuit • An L-R-C circuit exhibits

damped harmonic motion if the resistance is not too large. ¡iR¡ L di

dt ¡qC = 0

d2qdt2 + R

Ldqdt + 1

LC q = 0

q = Ae¡(R=2L)tcos(q

1LC ¡

R2

4L2 t + Á)

!0 =q

1LC ¡

R2

4L2


Example 30.10: An underdamped L-R-C circuit What resistance R is required (in terms of L and C) to give an

L-R-C series circuit a frequency that is one-half the undamped frequency?

Documents

Chapter 30 INDUCTANCEmorse/P272ASp13-Ind.pdf · Calculating self- inductance and self -induced emf (30.3+30.4.) • Determine the self -inductance of a toroidal solenoid with cross-sectional