32N 30 Inductance

30.1 Mutual Inductance Derivation

The flux F2 in coil 2 is proportional to

the current in coil 1 or vice versa

(WHY?) and the constant of

proportionality is the mutual

inductance

=

=

The induced emf in the coils are

=

=

Arlyn D. Macasero-Roque 30 Inductance 1

The SI unit of inductance is Tm2/A,

defined as henry:

Example 1:

In one form of Tesla coil (a high-voltage regulator), a long solenoid

with length l and cross-sectional area A is closely wound with N1turns of wire. A coil with N turns surrounds it at its center. Find the

mutual inductance.

Solution:

Remember that a long solenoid carrying a

current i1 produces a magnetic field B1 that

points along the axis of the solenoid. The

field magnitude B is proportional to I and to

n1 , the number of turns per unit length:

0 1 11 0 1 1

N iB n i

l


The flux through a cross section of the

solenoid equals B1A. Since a very long

solenoid produces no magnetic field

outside of its coil, this is also equal to the

flux trough each turn of the outer,surrounding coil, no matter what the

cross-sectional area of the outer coil. The

mutual inductance is then given as:

22 2 0 1 1 0 1 22 1

1 1 1

BN N N i A AN NN B AM

i i i l l

F


Example 1:

Example 2:

Two coils have mutual inductance M = 0.275 H. The

current in the first coil increases at a uniform rate of

0.0500 A/s.

(a) What is the induced emf in the second coil?

(b) Suppose the current described is in the second coil

rather than in the first. What is the induced emf in the

first coil?

12 (0.275 H)( 0.0500 A/s) 0.0138 V

diM

dt

21 (0.275 H)( 0.0500 A/s) 0.0138 V

diM

dt


Example 3:

Two toroidal solenoids are wound on the same form so

that the magnetic field of one passes through the

windings of the other. Solenoid 1 has 800 turns and

solenoid 2 has 300 turns. When the current in solenoid

1 is 3.55 A, the average flux through each turn of

solenoid 2 is 0.0280 Wb.

(a) What is the mutual inductance of the pair of

solenoids?

(b) When the current in solenoid 2 is 1.60 A, what is the

average flux through each turn of solenoid 1?

2 2

1

(300)(0.0280 Wb)2.37 H

3.55 A

NM

i

F

321

1

(1.60 A)(2.37 H) 4.73 10 Wb

800

iM

N

F


In mutual inductance, if two coils are near each other, a current in the first coil produces a magnetic flux through the second coil. If we

change this flux by changing current, an induced emf appears in the second coil. But this change of current ALSO changes the flux within

the first coil, so an induced emf appears in the first coil as well.

This process is called self-induction, and the emf that appears is

called self-induced emf.

30.2 Self Inductance


The inductance

is then defined

as

BNLi

F

Recall that a capacitor can be used to produce electric field between

conductors and that the parallel-plate arrangement was used as a

basic type of capacitor.

Similarly, an inductor can be used to produce a magnetic field and

that a solenoid can be used as a basic type of inductor.

30.2 Inductors


The inductance of a LONG SOLENOID of cross-section is

=

The inductance per unit length near its center is

=

The inductance of a TOROIDAL SOLENOID is

=

.

Inductance, like capacitance, depends only on the geometry.


30.2 Inductance of Derivation

Energy Stored in an Inductor

Let the current at some instant be i and let its rate of change be di/dt; the current is increasing, so di/dt > 0. The voltage between the

terminals a and b of the inductor at this instant is Vab = Ldi/dt, and the rate P at which energy is being delivered to the inductor (equal to

the instantaneous power supplied by the external source) is

ab

diP V i Li

dt

30.3 Magnetic Field Energy


After the current has reached its final steady value I, di/dt = 0 and no more energy is input to the inductor. When there is no current, the

stored energy U is zero; when the current is I, the energy is LI2. Inductor with current I, energy is stored. Resistor with current I, energy is

dissipitated.

The total energy U supplied while

the current increases from zero to a

final value I is

= 0

=

Self-Inductance of an Ideal Toroid within its coil

=

2

2

The energy stored in the toroidal solenoid with current

=1

2

2

22

The energy density stored in the toroid with volume = 2A

=

=1

222

2 2

=

(in vacuum) =

(in material)

30.3 Magnetic Energy Density of a Toroid


The energy in an inductor is actually stored in the magnetic field within the coil, just as the energy of a capacitor is stored in the electric field

between plates.

From Faradays Law:

The direction of L is found fromLenzs Law: the self-induced emf acts

to oppose the change in the current.

The above circuit diagrams

illustrate that self induction

opposes the change in the

current.

We can define a self-induced potential

difference VL across the inductor. If the

inductor is ideal (zero resistance),

VL = L.

L

diL

dt

+

- +

-


30.2 Self-Induced Potential Difference


30.2 Inductors as Circuit Elements

The potential difference across a resistor depends on the current

while the potential difference across an inductor depends on the

rate of change of the current.

Example 4:

An inductor has an inductance of 0.540 H and carries a

current in the direction as shown and is decreasing at a

steady rate of 0.0300 A/s.

(a) What is the self-induced emf?

(b) Which end of the inductor (A or B) is at a higher

potential?

From Lenzs law we find A to be the higher potential

i

A B

(0.540 H)( 0.0300 A/s) 0.0162 VLdi

Ldt

Arlyn D. Macasero-

Roque

30 Inductance 13

Example 5:

The inductor has inductance 0.260 H and carries a

current in the direction as shown in the figure below.

The current is changing at a constant rate. (a) The

potential between points and is Vab = 1.04 V, withpoint at higher potential. Is the current increasing ordecreasing? (b) If the current at = . is 12.0 A,what is the current at = . ?

a) From Lenz Law, i should be decreasing in order to

induced an emf in that direction.

b) (di/dt) is negative and hence i2 = 4 A

Arlyn D. Macasero-

Roque

30 Inductance 14

When the switch S1 is closed on a,

the current in the resistor starts to

rise. If the inductor were not in the

circuit, the current would rapidly rise

from zero to a steady value of /R butbecause the inductor is present, the

current would gradually rise from

zero.

In this circuit, the inductor acts to

oppose changes in the current

through it. Some time later, it acts like

an ordinary connecting wire when the

current approaches /R.


30.4 R-L Circuits

Loop rule gives us

+ = 0

The initial rate of change of current is

=

The final rate of change of current is

= 0

And the steady-state current is

=


30.4 Current Growth in R-L Circuits

This differential equation has a solution given by

Defining the time constant as

We can rewrite the solution as

63%


30.4 Current Growth in R-L Circuits

Reversing the switch to point b, removes the

battery from the circuit and the current through

the resistor begins to decrease. However, it

cannot drop immediately to zero but must decay

to zero over time.

Without the battery, the loop rule equation now becomes

The solution to this equation is

37%

Arlyn D. Macasero-

Roque

30 Inductance 18

30.4 Current Decay in R-L Circuits

Example 6:

A sensitive electronic device of 175-W resistance is to be

connected to an emf source by a switch. The device is

designed to operate at a 36-mA current, but to avoid damage

the current must not rise above 4.9 mA in the first 58 s after

the switch is closed. So to protect the device, it is connected

in series with an inductor.

(a) Assuming negligible internal resistance, what emf must be

used?

(b) What inductance is required?

(c) What is the time constant?

(0.036 A)(175 ) 6.3 VIR W

( / ) (175 / )58 s6.3 V( ) (1 ) 4.9 mA (1 ) 69 mH175

R L t Li t e e LR

69 mH390 s

175 L

L

R

W


(a) What is the initial current in the resistor just after closing the switch at b?

(b) What is the current in the resistor after 0.200 ms?

(c) What is the potential difference across the inductor after 0.200 ms?

(d) How long does it take the current to decrease to half its initial value?

Example 7:

( / ) 0120 V( ) (0) 0.300 A400

R L ti t e i eR

W

( / ) (400 /0.200 H)0.200 ms

0( ) (0.3 A) 0.201 AR L ti t I e e

(0.201 A)(400 ) 80.4 VLV iR W

Suppose = 120V, R = 400 W, and L =0.200 H. The switch is closed at a until

a constant current is established. The

switch is then closed at b.

( / ) 00

0 0

0.5( ) 0.200 H( ) ln ln 0.347 ms

400

R L t IL i ti t I e tR I I

W


Example 8:

(a) Just after closing the switch at a, what are the voltages across R and

L?

(b) After several time constants with the switch still at a, what are the

voltages across R and L?

(c) What are the voltages across R and L during the time when i = 0.0500

A?

Suppose = 80.0V, R = 400 W, and

L = 0.200 H. Initially, there is no

current in the circuit. The switch is

closed at a.



31.6 Transformers

Composed of 2 coils or windings, electrically insulated from each other

but wound on the same core.

All magnetic field lines are confined to the iron core (possible with

cores of large permeability)

TRANSFORMERS can easily step-up/step-down AC voltages (more

difficult with DC voltages).

Neglecting resistance in the windings,

the induced emfs in the primary and

secondary coils are:

=

and =

The magnetic flux is the same for

each coil, hence

=

and since 1 and 2 both oscillate at the same frequency as the acsource, then

1 11 1 2 2

2 2

V N

V I V IV N

(from energy considerations)

Step-up: N2 > N1 Step-down: N2 < N1

31.6 Transformers


When the switch is closed, the potential difference across R

is

1) V(N2/N1)

2) V(N1/N2)

3) V

4) 0

Check Your Understanding!


The primary coil of a transformer is connected to a battery, a

resistor and a switch. The secondary coil is connected to an

ammeter. When the switch s closed, the ammeter reads

1) Zero current

2) A nonzero current for a short instant

3) A steady current

Check Your Understanding!



In the figure, the current i1 in coil 1 sets

up a magnetic field B, some of which

pass through coil 2. The flux F2 in coil

2 is proportional to the current in coil 1

(why?)

22 = 211

=

where M21 is a proportionality constant.



When F2 and i1 change with time, we

can take time derivatives on both

sides:

22 =

211

2

2 = 21

1

=

Similarly,

=


Consider a long solenoid of cross-section A. What is the

inductance per unit length near its center?

First we calculate the flux linkage NFB set up by the current in the

solenoid. The flux linkage can be expressed as

=

where n is the number of turns per unit length and B is the

magnetic field within the solenoid:

=


30.2 Inductance of a Long Solenoid Derivation

The inductance is then

==

=

Thus, the inductance per unit length for a long solenoid near the

center is

=


30.2 Inductance of a Long Solenoid

Figure (a) shows a toroidal solenoid, which we may describe as a

solenoid bent into a torus (doughnut) with rectangular cross section.

From the symmetry we see the field forms concentric circles inside the

toroid as shown in figure (b).

path of integration


30.2 Inductance of a Toroid

According to Amperes law, the

magnetic field within the toroid is:

=

The flux can be obtained by

integration:

= =

=

2

=

So the inductance of a toroidal

solenoid is expressed as:

=


30.2 Inductance of a Toroid Derivation

32N 30 Inductance

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