Electromagnetism - Lecture 7

Inductance & AC Circuits

• Self Inductance

• Mutual Inductance

• Magnetic Energy

• AC Circuits

• Electrical Impedance

• Resonance in LCR Circuits

1

Self Inductance

A current loop produces a dipole magnetic field with m=IAn̂

⇒ Change in I causes change in B

⇒ Change in ΦB through loop causes emf E that opposes change

E = −dΦB

dt= −L

dI

dt

where the self inductance of the loop is:

L =dΦB

dI

Self inductance is a purely geometric property of the loop

The unit of inductance is the Henry H = 1 Vs/A

2

Mutual Inductance

⇒ Change in current in a loop induces emf in a nearby loop

Mutual Inductance is symmetric between the two loops:

E1 = −MdI2dt

E2 = −MdI1dt

M =dΦ2

dI1=

dΦ1

dI2

M is a function of the geometry of the two loops

A general formula can be derived using the magnetic vector

potential:

M =

∮

1

∮

2

µ0

dl1.dl2

4πr12

Can write M in terms of the self inductances of the loops:

M = k√

L1L2

where k is the flux coupling coefficient between the loops

3

Notes:

Diagrams:

4

Inductances of Solenoids

For an infinite solenoid the magnetic field along the axis is:

Bz = µ0nI

and the flux through n loops of radius a per unit length l is:

ΦB =

∫

A

B.dS = µ0nIπa2nl

The self inductance of the solenoid is:

L =dΦB

dI= µ0n

2πa2l

For two coaxial solenoids of radii a and b < a the mutual

inductance is:

M = µ0n1n2πb2l

The flux coupling between the solenoids is k = b/a

5

Energy Stored in an Inductor

Work is done to create a current in a loop against the induced emf∫

dW =

∫

EI

∫

Wdt = −

∫

LIdI = −

∫

IdΦB

This work is stored as magnetic energy

UM = −

∫

Wdt =1

2LI2

For two coils including mutual inductance:

UM =1

2L1I1

2 +1

2L2I2

2 + MI1I2

The energy stored in a solenoid is:

UM =1

2µ0n

2I2πa2l =1

2µ0

|B|2πa2l

6

Magnetic Energy Density

Magnetic energy is stored in an inductor through the creation of

the magnetic field

The energy density of a magnetic field is proportional to the

square of its amplitude:

dUM

dτ=

|B|2

2µ0

The energy density can also be expressed in terms of the magnetic

vector potential and the current density:

ΦB =

∮

L

A.dldUM

dτ=

J.A

2

A full derivation can be found in Grant & Phillips Pp.243-246

7

Notes:

Diagrams:

8

Alternating Currents

A coil rotating in a magnetic field generates an alternating current

through a resistance R by induction:

V = V0 cosωt V = IR I =V0

Rcosωt

If instead of a circuit with resistance R we have a capacitance C:

V =Q

C

dV

dt=

I

CI = −ωCV0 sinωt

Finally a circuit with an inductance L:

V = LdI

dtI =

1

L

∫

V dt I =V0

ωLsinωt

9

Electrical Impedance

Impedance is a generalization of the idea of resistance:

Z =V

I

for an AC circuit containing any combination of L, C and R

The change from cosωt to sinωt for C and L is represented by a

complex impedance:

ZR = R ZL = iωL ZC = −i

ωC

For a general circuit the impedance is a complex number with an

amplitude and a phase between V and I:

Z = Re(Z) + Im(Z) = ZR + i(ZC + ZL) = Z0eiφ

10

Combining Impedances

Sum of impedances in series:

Z =∑

i

Zi

R = R1 + R2 L = L1 + L2

1

C=

1

C1

+1

C2

Sum of impedances in parallel:

1

Z=

∑

i

1

Zi

1

R=

1

R1

+1

R2

1

L=

1

L1

+1

L2

C = C1 + C2

11

Notes:

Diagrams:

12

Impedance in LCR Circuits

The complex impedance of an LCR circuit is:

Z = R + i

(

ωL −1

ωC

)

The magnitude of this impedance is:

Z0 =

√

R2 +

(

ωL −1

ωC

)2

and the current has the general form:

I0 cosφ =V0R

Z2

0

where φ is the phase angle between V and I:

tanφ =ωL − 1/ωC

R

13

Resonance in LCR Circuits

The minimum impedance is at the resonance frequency ω0:

Z0 = R φ = 0 ω0 =√

1/LC

At the resonance V and I are in phase, and I is at a maximum

The power dissipated in an LCR circuit is averaged over one cycle:

< P >=< IV >=1

2V0I0 cosφ

< P >=V 2

0

R

(

1

1 + (1/R)2(ωL − 1/ωC)2

)

At resonance the maximum power is dissipated in the resistance:

< P >=V 2

0

R

14

Q Factor in LCR Circuits

Near the resonance:

∆ω = ω − ω0 |∆ω| � ω0

The power dissipation has the form:

< P >=V 2

0

R

(

1

1 + (4L2/R2)(∆ω)2

)

The power falls to half its maximum value at resonance for:

∆ω1/2 = ±R

2L

The Q factor for the circuit is:

Q =ω0

2|∆ω1/2

|=

ω0L

R

15

Notes:

Diagrams:

16