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Addendum – Chapter 21. Mutual inductance –. Circulation of currents in one coil can generate a field in the coil that will extend to a second, close by device. Flux Changes. Suppose i 1 CHANGES. Current (emf) is induced in 2 nd coil. Mutual Inductance.

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mutual inductance
Mutual inductance –
  • Circulation of currents in one coil can generate a field in the coil that will extend to a second, close by device.

Flux Changes

Suppose i1 CHANGES

Current (emf) is

induced in 2nd

coil.

mutual inductance1
Mutual Inductance
  • i1 creates a field that (partially) passes through the second coil.
  • As i1 changes, the flux through coil 2 changes and an emf (and current i2) are created.
  • The two coils are mutually linked by what we call an “inductance”

i2

Induction

watch out
Watch Out!
  • Exam #2 one week from today.
    • Chapters 20 & 21
    • Same format but possibly one set of multiple choice questions that you hate.
    • You should already be studying.
  • QUIZ on Friday – Chapter #21
  • Today we continue with the chapter. We should finish it on Friday. Maybe.
  • No study session on Monday next week
  • We will have a study session on Tuesday morning like last time. Details to follow.

Induction

slide5

This schedule is now in effect:

If I am not there … find me!

Induction

mutual inductance2
Mutual Inductance

i2

mutual Inductance

Induction

note the form
Note the form:

UNIT: henry

Think of this when we define INDUCTANCE (L) of

a small coil in the next section.

Induction

the two coils
The two coils

Remember – the magnetic

field outside of the solenoid

is pretty much zero.

Two fluxes (fluxi?) are the same!

Induction

slide9

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid

Length = 0.5 meters

N=6750 coils

n=6750/.5=1.35E04 turn/meter

Magnetic field INSIDE the smaller coil

is the same as in the larger coil and

is given by:

Check My Arithmetic Please!

Induction

slide10

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid

Induction

slide11

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 cm in diameter and contains 15 coils. The current in the outer solenoid is changing at 37.5 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the inner solenoid

Check My Arithmetic Please!

Induction

self inductance
Self-inductance –
  • Any circuit which carries a varying current self-induced from it’s own magnetic field is said to have INDUCTANCE (L).
inductance defined
Inductance Defined

If the FLUX changes a bit during a short time Dt, then the current will change by a small amount Di.

Faraday says this is the emf!

This is actually a

calculus equation

Induction

slide17
So …

There should be

a (-) sign but we

use Lenz’s Law

instead!

E=

The UNIT of “Inductance – L” of a coil is the henry.

SYMBOL:

Induction

consider ac voltage
Consider “AC” voltage

Minimum Change@Dt

V1

Maximum Change@Dt

Induction

the transformer
The transformer

FLUX is the same through

both coils (windings).

Induction

remember that a capacitor stored energy u 1 2 cv 2
Remember that a Capacitor stored ENERGY? U=(1/2)CV2

U=Area=(1/2)LI2

i

Li

LI

Li

DU

i

Di

I

Induction

Induction

slide24
SO …

Energy Stored in a capacitor

The energy stored in a capacitor with capacitance C and a voltage V is

U=(1/2)LI2

Induction

the energy stored is in the magnetic field
The Energy stored is in the Magnetic Field

Consider a solenoid with N turns that is very long. We assume that the field is

uniform throughout its length, ignoring any “end effects”. For a long enough solenoid, we can get away with it for the following argument. Maybe.

Induction

rl or lr series circuit
RL or LR Series Circuit
  • Switch is open .. no current flows for obvious reasons.
  • Switch closed for a long time:
    • Steady current, voltage across the inductor is zero. All voltage (E) is across the resistor.
    • i=E/R

Induction

rl or lr series circuit1
RL or LR Series Circuit

When the switch opens, current change is high and back emf from L is maximum.

i

E/R

t

As the current increases, more voltage is across R, the rate of change of I decreases

and as the current increases, it increases more slowly.

Induction

rl circuit
RL Circuit
  • When L=0, the current rises very rapidly (almost instantly)
  • As L increases, it takes longer for the current to get to its maximum.

Induction

the graphic result current growth
The Graphic Result – Current Growth

}

63% of

maximum

e= 2.71828…

Induction

decay short out the battery
Decay – Short out the battery
  • Magnetic field begins to collapse, sending its energy into driving the current.
  • The energy is dissipated in the resistor.
  • i begins at maximum (E/R) and decays.

Induction

solution
Solution

Induction