PHYS1008 Induction

A magnetic Induction machine c 1830. © 2007, American Artifacts, Taneytown, Maryland.

Magnetic Induction and A.C. circuits Objectives: by the end of this you will be able to Understand what magnetic flux is and find it in simple cases. Relate flux and EMF Calculate inductance for simple cases Find the energy produced by changing fields Understand what happens when you use inductances in a circuit. Understand AC in circuits

Outline:

• moving charge ⇒ mag. field

• so mag. field ⇒ moving charge??

• Faraday expt. Current in (A) produces mag. field in solenoid, so look for current in (B),

• He actually saw no effect for steady current in (A) but a momentary current in (B) when switch is opened or closed.
• i.e. changing mag. field ⇒ electric current.
• A huge amount of technology depends on this simple observation

Move conducting rod through uniform mag. field Magnetic force on electrons FM = q v B but moving electrons will cause electric field to be created (like Hall effect) FE = q E In equilibrium FE = FM so E = vB and hence potential V = E L= vBL

• e.g. train goes south, 30 ms^-1 in Canada, where B ~ 1/2 G. What is voltage across axles if they are 2m long?
• Faraday tried this expt. on Thames in London, v ~ 2ms^-1, L~ 100m, B ~ 1/2 Gauss.
• What V should he have found?
• Note: Work is being done against mag. field : i.e. mech. work ⇒ electrical energy: this is the basis for the generator (later).

Magnetic Flux

• Mag. Flux = Field x area

• but in general we need to worry that the field can be changing as well as being at an angle

• Units are Webers: 1Wb = 1 Tm² (note that this is not 1W (watt)!)

Pot. Difference in Loop (B) X Time = Change of Flux in (A)

Vδt = -δφ

We can get current and hence power out of the arrangement above if we let sliding conductor move on fixed one. In time δt bar will move distance d = v δt

δA = L d = L v δt 

• Change in flux

  δφ = BδA
  

• Hence

  δφ = BLv
  δt
  

• But we just showed

  V = BLv
  

• Hence Faraday's law follows (in this case) from Lorentz force: what we have made is a very crude generator.

• Lenz's Law tells direction of field
• \color{red}{B_A } is applied field.
• Suppose \color{red}{B_A } increases,
• then current will be induced so as to oppose change in field:
• i.e. \color{red}{B_A \to B_A + \delta B_A }
• \color{red}{\delta B_A \Rightarrow I}
• \color{red}{B_A \to B_A + \delta B_A \Rightarrow I \Rightarrow B_I }
• \color{red}{B_I } is in opp. direction to \color{red}{\delta B_A }
• Lenz's law follows from energy conservation: if the induced field was in the same direction as the applied field, it would induce a larger current which would produce a larger field which....

A loop of wire has area 30cm2. The field through it changes by 1000 Gauss. What is the change in flux? 3 Wb 30000 Wb 3x10-4 Wb 30 Wb

e.g. A loop of wire has resistance 5Ω, area 30cm2. The flux through it changes by 3x10-4 Wb in .05 seconds. What is the induced EMF? 6 V 6x10-3V 6000 V .06 V

A loop of wire has resistance 5Ω, area 30cm². The flux through it changes by .0003 Wb in .05 seconds, giving an induced EMF of 6x10-3 V. What is the current? 1.2 A 1.2x10-3 A 1.2x10-6 A 12001 A

Generators

Coil of wire with many turns is spun in mag. field (Yes: this looks like a motor. In fact simple motors and simple generators are the same). A simulation

• For N turns, in any positions: φ= NBAcos (θ)

• If θ = ωt then φ = NBAcos (ωt) so ξ = NBA ωsin(ωt)

• This gives us alternating current (AC): the direction of the current keeps changing

1. .37 T
2. .06 T
3. .37 Wb
4. .06 Wb

Eddy Currents

• Induced currents in large volumes of conductor which produce heating If field out of surface increases, current will flow, producing heat by resistance.

• Partially overcome by laminating metal, so each current must have a smaller E.M.F., so the power loss is much less

• A useful application is Electromagnetic Braking Wheel rotating inside mag. field will have current induced in it, induced mag. field will interact with external field to slow down wheel, so rotational K.E. converted to heat, but no friction to cause wear.

Magnetic Levitation

Current in primary (Ip) produces field in primary produces current in secondary (Faraday's Law) produces opposing mag. field (Lenz's Law) Opposing dipoles repel. Only works with A.C. (Why?)

Inductances

Faraday expt. shows that changing current (and hence flux) in one circuit will produce a changing E.M.F. in second.

• If a circuit can affect a second one, it can affect itself:
• quantifying this gives coef. of self-induction or inductance of a coil.
• Coefficient of self inductance L defined by

  φ = LI
  Ɛ  = - δφ = -LδI
         δt     δt
  

e.g. For a long solenoid, length l, x-sect A: the flux through any one loop due to itself and all the others is δφ = BA = μ₀ N I A l Total flux is just φ = Nδφ so L = μ₀ N²A l Note: in most practical cases we cannot calculate L but it can always be measured. Unit of L is "Henry" L ~ μ₀(length) Proportional to the square of the no. of turns:

A solenoid consists of 1000 turns, 50 cm, x-sect 40 cm²: what is the self-inductance?

1. .01 H
2. 1 H
3. 100 µH
4. 1 mH

Energy stored in an Inductance:

Energy must be supplied to increase the current: if there is no resistance, an inductance can be used to store energy

U = ½ L I²

A solenoid consists of 1000 turns, 50 cm, x-sect 40 cm², so it has a self inductance of 10 mH. If there is a current of 30 A in it, how much energy is stored

1. 4.5 J
2. 4.5x10³ J
3. 4.5x10^-2 J
4. 4.5x10^-6 J

• For a solenoid, \color{red}{B = \mu _0 \frac{{NI}}{\ell }} and \color{red}{L = \mu _0 \frac{{N^2 A}}{\ell }}
• so we can write
  \color{red}{ U = \frac{{B^2 }}{{2\mu _0 }}\ell A}
• or energy/unit volume as
  \color{red}{ u = \frac{U}{V} = \frac{{B^2 }}{{2\mu _0 }}}

1. 160 G?
2. 160 T?
3. 2.5x10⁴ T
4. 2.5x10⁴ G

Transformers

Power loss in transmission is ~ RI²: hence can reduce current by increasing voltage.

• e.g suppose we want to transmit power P₀ = 1 MW over 500 km with a wire of resistance of 10^-8Ω/m,
• what is power loss P_L at 110 V
• what is power loss P_L at 1 MV

However, voltage must be decreased locally: only works for A.C. Important for power transmission

assume no load: then voltage across prim. V₁ = N₁ dφ , V₂ = -N₂ dφ = - V₁ N₂ dt dt N₁ Also flux through, prim. and sec. must be same: N₁I₁ = N₂I₂ Hence Power I₁V₁ = I₂V2 = V₂²/R₂ if resistance R₂ is connected across secondary. This is "step-down" V₂ < V₁ and so I₂ > I₁.

Inductances in a circuit:

What happens if the switch is closed? A current will start to flow, but this will produce a flux in the solenoid, which will produce an EMF in the opposite direction, which will reduce the current... Kirchoff's loop law gives EMF from Battery + EMF from Inductance - RI = 0 or - L dI + V - I R = 0 dt which is very like the equation we had for a capacitance note we usually pretend that an inductance has no resistance

This is a current which increases from zero to the Ohm's law value I = V/R. e.g if our circuit has a battery with an EMF of 12 V, R = 10 Ω and L = 2mH,

1. What will the final current be?
2. How long will the current take to increase to 90% of its final value?

A. C. in Circuits

• Usually talk about "root mean square" (r.m.s.) value of voltage.
  \color{red}{ V_{rms} = \left\langle {\left( {V_m \sin \left( {\omega t} \right)} \right)^2 } \right\rangle ^{1/2} = \frac{{V_m }}{{\sqrt 2 }}}

• Power in AC: as long as we have a purely resistive circuit, V and I are in phase

Power in AC

• \color{red}{I = V/R}
• \color{red}{I = \frac{V}{R} = \frac{{V_m \sin \left( {\omega t} \right)}}{R}}
• \color{red}{I = \frac{V}{R} = \frac{{V_m \sin \left( {\omega t} \right)}}{R} = I_m \sin \left( {\omega t} \right)}
• so that the power is \color{red}{P = IV = V_m \sin \left( {\omega t} \right)I_m \sin \left( {\omega t} \right)}
• \color{red}{ = \frac{{V_m ^2 }}{R}\sin ^2 \left( {\omega t} \right)}
• so average \color{red}{P_{rms} = \frac{{V_m ^2 }}{{2R}}\left\langle {1 - \cos \left( {2\omega t} \right)} \right\rangle = \frac{{V_m ^2 }}{{2R}}}

1. What is V_m?
2. What is I_m?
3. What is R?

Capacitors and Inductors and AC

What happens if we have a capacitor and inductance in a circuit? When the switch is thrown, current will start flowing, but cannot flow through cap. so it just gets charged up

• Charge Q starts on C
• Current I starts flowing, but L chokes it back
• When Q = 0, I cannot drop to zero (L keeps it flowing)
• Q reverses sign, current drops to zero
• Go to step 1!
• LC gives the frequency: specifically
  \color{red}{ 2\pi f = \omega _0 = \frac{1}{{\left( {LC} \right)^{1/2} }}}

0. 158 kHz
1. 25 GHz
2. 15.8 Hz
3. .25 Hz

If the battery is replaced by an AC-supply, we get the LRC circuit. If the current is \color{red}{ I = I_0 \sin \left( {\omega t} \right)} EMF produced by the inductance is \color{red}{ V_L = L\frac{{dI}}{{dt}} = LI_0 \frac{{d\sin \left( {\omega t} \right)}}{{dt}}} \color{red}{ V_L = L\frac{{dI}}{{dt}} = LI_0 \frac{{d\sin \left( {\omega t} \right)}}{{dt}} = LI_0 \omega \cos \left( {\omega t} \right)} This can be written as an average (rms) value \color{red}{ V_L ^{rms} = I_L ^{rms} X_L ,X_L = \omega L} (note this looks just like Ohm's law) XL is the "impedance" of the inductor. Note that it increases as ω increases: this means that an inductor behaves like a small resistance at low frequency and a large one at high frequency.

• If the charge is \color{red}{Q = Q_0 \cos \left( {\omega t} \right)} and voltage is \color{red}{V_C = \frac{Q}{C}}
• then \color{red}{I = \frac{{dQ}}{{dt}} = Q_0 \frac{{d\cos \left( {\omega t} \right)}}{{dt}}}
• gives us the current \color{red}{I = \frac{{dQ}}{{dt}} = Q_0 \frac{{d\cos \left( {\omega t} \right)}}{{dt}} = - Q_0 \omega \sin \left( {\omega t} \right)}
• then
  \color{red}{ V_C ^{rms} = I_C ^{rms} X_C ,X_C = \frac{1}{{\omega C}}}
• Again this looks just like Ohm's law where
• so the impedance of the capacitor decreases as ω increases: this means that it behaves like a large resistance at low frequency and a small one at high frequency.

• \color{red}{ X_L = \omega L,X_C = \frac{1}{{\omega C}},X_R = R}
• \color{red}{ X_L = \omega L,X_C = \frac{1}{{\omega C}},X_R = R \Rightarrow X_{tot} = \sqrt {R^2 + \left( {\omega L - \frac{1}{{\omega C}}} \right)^2 } }
• Pictorally:
  

Resonant Circuits<

• \color{red}{ \omega L = \frac{1}{{\omega C}}}
• (Quite complicated to prove). Current in this circuit looks like
  \color{red}{ V^{rms} = I^{rms} X_{tot} \Rightarrow I^{rms} = \frac{{V^{rms} }}{{\sqrt {R^2 + \left( {\omega L - \frac{1}{{\omega C}}} \right)^2 } }}}

• which gives a resonant circuit, which picks out just one frequency \color{red}{ \omega _0 = \frac{1}{{\left( {LC} \right)^{1/2} }}}

• What is the resonant frequency f₀ = ω₀/2π of the circuit?
• What will the current be at resonance if V = 20V?
• What will it be at f = 1/2 f₀?

• Importance: if you are trying to pick a signal of one frequency out of a whole lot of similar ones, you can do it with a tuned circuit of this kind.