mutual inductance electrical circuit, Exams of Electrical Engineering

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Phys-

Lecture 18 Mutual InductanceSelf-InductanceR-L Circuits

Mutual Inductance^ If we have a constant current i

in coil 1, 1 a constant magnetic field is created andthis produces a constant magnetic flux incoil 2. Since the

Φis constant, there NOB^ induced current in coil 2.If current iis time varying, then the^1

ΦB

flux is varying and this induces an emf

ε^2 in coil 2, the emf isWe introduce a ratio, called

mutual inductance

, of flux in coil 2

divided by the current in coil 1.

dB^2 N 2^ dt

Φ −= ε 2

N Φ^22 B M = 21 i^1

Mutual InductanceThe induced emf,has the following features;

di^2 M −= ε 1^ dt

-^ The induced emf opposes the magnetic flux change (lenz’s Law)• The induced emf increases if the current changes very fast• The induced emf depends on M, which depends only the geometryof the two coils and not the current.• For a few simple cases, we can calculate M, but usually it is justmeasured.

Two coils have mutual inductance of 3.

−^4 × 10 H. The current

in the first coil increases at a uniform rate of 830 A/s .A) What is the magnitude of induced emf in the 2

nd^ coil? Is it

constant?B) suppose that the current is instead in the 2

nd^ coil, what is the

magnitude of the induced emf in the 1

st^ coil?

Problem 30.

A^ Vs diMdt H

(^27). (^0) ) 830 1 4 )(^1025. (^3) ( 2

−= −= ε^ −×−= di^2^ VM^^27.^0 −=−= ε 1 dt

Nicolai Tesla (1856-1943)Born in Croatia, graduated from University ofPrague. Arrived in New York with 4cents and wentto work for Edison. Tesla invented polyphasealternating-current system, induction motor,alternating-current power transmission, Teslacoil transformer, wireless communication, radio,and fluorescent lights. He set up a Tesla coil in

Colorado Springs in 1899, belowis a photo of this lab. He lightedlamps 40Km away. He also claimedto receive messages from anotherplanet!! In honor of his contributionsto electromagnetic phenomena, theMagnetic field intensity was namedin units of “Tesla”

Applications of Mutual Inductance • Transformers^ – Change one AC voltage into another

iron ∼ ε N^ N (^12) (primary)^ (secondary) V^1^ V^2

N^2 V V = 2 1^ N^1

• Airport Metal Detectors^ – Pulsed current

^ pulsed magnetic field
^ Induces emf in metal– Ferromagnetic metals “draw in” more B
^ larger mutual inductance

^ larger emf

• Emf^
  ^ current (how much, how long itlasts, depends on the resistivity of thematerial)– Decaying current produces decayingmagnetic field^
  ^ induces current in receiver coils– Magnitude & duration of signal dependson the composition and geometry of themetal object.

Self Inductance We previously considered induction between 2 coils. Now we considerthe situation where a single isolated coil induces emf on itself. This isCalled “back emf” and if the current changes, there is a self induced emfthat opposes the change in current. We form the same ratio, now calledSelf-Inductance,

L, and we have the back emf,

N Φ B L =^ i^ di^ L −= ε^ dt

Behavior of isolated coil in circuits^ Resistor with current Ihas potential drop, V=iRfrom a to b

Coil witha)^ constant current i has NOVoltage dropb) di/dt>0, potential decreasesfrom a to b, V=Ldi/dtc) di/dt<0, potential increasesfrom a to b, V=-L|di/dt| Remember, emf in coil opposes current change.

Clicker:Two simple pieces of wire A and B are shapedinto almost complete loops. The loose ends ofeach loop are connected to identical batteries.Assume the loops have the same totalresistance, and that they do not interfere witheach other.2) Which loop has the greatest flux through it (assume the loopshave the same current in them)?a)^ A b)^ B c)^ same 3) Which loop has the greatest self inductance?a)^ A b)^ B c)^ same

Self inductance is defined asso if^ I^ is the same,

L >^ L.^ A^ B

Φ B L ≡ I Recall the B-field of current loop (at the center):

B =^ μ I/(2R)^0 The area of the loop is

(^2) A = π R Flux is ⇒^ The flux through the loop increases with

rr •≡Φ SdBB ∫ (^) R.

⇒^ The flux through loop A is bigger than the flux through loop B

.

Clicker problem

-^ Consider the two inductors shown:^ – Inductor 1 has length

l ,^ N^ total turns and has inductance

L.^1

• Inductor 2 has length 2

l , 2 N^ total turns and has inductance

L.^2

• What is the relation between

L and^1 L?^2^ (a)^ L <^ L^2

(b)^ L 1 2 =^ L (c) 1

l r N turns L >^ L 2 1 2 l r r 2 N turns

How to combine inductors

l^2 l^ r r^ r N turns^2 N^ turns

• To calculate

L , we need to calculate the flux.• Since^ B

= B , the flux through any (^12) given turn is the same in eachinductor• There are twice as many turns in inductor 2; therefore the net fluxthrough inductor 2 is twice the flux through inductor 1!

Therefore,

L = 2 L.^2 1^ Inductors in series

add^ (like resistors):

eff^1

L^ L^ L =^ +^2

And inductors in paralleladd like resistors in parallel:

eff^1

1 1 1 =^ + L^ L^ L^2

Self Inductance of toroidal solenoid The magnetic field in a toroid wasand the net mag.flux isHence the self inductance is,

N^ A^ r

NiNA ri

NBL i

μ π μ^ π^22

2 0

Φ^0 ==

Ni μ 0 B =^ r π 2 =

Ni^ A^ r μ^0 BA ==Φ B^ π^2 Example 30.3 N=200, A=5cm2, r=0.1m

H L

μ π^ π

(^40105) 104200 ) (^1). (^0) ( 2

=× ×=

Magnetic field Energy in a toroid Consider a toroid magnet, the B fieldis , B =^ μμμμ NI/2^0

ππππ r (ex.28.11). The energy is, Substituting the B field into theEqn., we have,

(^11 ) ANI r LIU

       μ ==^ π ( )^

(^ )^

2 1 22 1 22 2

IN^ B r IN r U^ rA

 =      =       =  

densityEnergy BU volume U^ rA

==

2 20 2

Energy in Electric Fields and Magnetic Fields In chapter 24.3, we discussed energy in a parallel plate with area Aand separation d, The electric field energy in the capacitor was

(^ )^

(^ )^

(^22) ( ) (^0)

1 2 1 2 (^11 )

EAd AEdd EdC CVU

ε^

=   =  ==

(^20) E 2 ε energy U == volume Ad Now we find the magnetic field energy in the toroid magnet is

12 20 2

B energy U volume rA

π^

== The^

rr^22 fields are proportional to the energy density , EB