Lec 15 Notes: Electric Generators, Inductance, Magnetic Energy, Transformers

Lec 15 Notes: Electric Generators, Inductance, Energy in Magnetic Fields, Transformers

• Topic overview: Units and concepts related to electric generators, inductance, magnetic energy storage, and transformers.

Faraday’s Law of Induction

• Faraday’s law (emf induced when magnetic flux changes in time):

  • Emf: $\varepsilon = -\frac{N\,\Delta\Phi}{\Delta t} = -N\frac{\Phi<em>f - \Phi</em>i}{t<em>f - t</em>i}$

  • Magnetic flux through a loop: $\Phi = \mathbf{B} \cdot d\mathbf{A} = B A \cos \theta$

• Key q. ualitative idea: An emf appears only if the magnetic flux through a loop changes with time.

• Magnetic flux is a measure of how many magnetic field lines cross a given area.

Motional Emf (emf due to motion in a magnetic field)

• Flux change caused by a moving conductor in a magnetic field is due to the changing area of the loop:$\Delta\Phi = B\,\Delta A = B\, (\ell\,\Delta x)$ where ℓ is the length of the rod and Δx is its displacement.

• Quantitative motional emf (in a rod of length ℓ moving with speed v in a uniform B):

  • $\varepsilon = -N \frac{\Delta\Phi}{\Delta t} = -N\frac{B\,\ell\,\Delta x}{\Delta t} = -N B \ell v$

• Current in the circuit: $I = \frac{\varepsilon}{R} = \frac{-N B \ell v}{R}$

• Magnetic force on the rod (Lenz’s law): $F_m = I\,\ell\,B\sin\theta$ (for typical perpendicular orientation, sinθ = 1)

• Gravitational force on the rod: $F_g = mg$

• When the rod reaches constant speed, net force is zero: $F<em>m = F</em>g$, giving constant velocity motion.

• Energy accounting (not free energy): gravitational potential energy is converted into internal forms (Uheat, Ulight) and kinetic energy as appropriate: $E<em>{total} = K</em>i + U<em>{total} = ext{constant}$ with components like $Ug$, $U{heat}$, $U{light}$.

• Change in flux and emf when external force is used to move the rod: moving the rod changes the loop area, producing a motional emf:

  • Change in flux: $\Delta\Phi = B\,\Delta A = B\, v\, \ell \Delta t$

  • Induced emf: $\varepsilon = N \frac{\Delta\Phi}{\Delta t} = N B v \ell$

• Sign convention and interpretation: the induced emf drives a current that opposes the change in flux (Lenz’s law).

Induced current and external work in a moving rod system

• Current magnitude:

  • $I = \frac{\varepsilon}{R} = \frac{B v \ell}{R}$ (for a single loop with resistance R and rod length ℓ)

• Magnetic force on the rod when current flows: $F_m = I\ell B = \frac{B v \ell}{R} \cdot \ell \cdot B = \frac{B^2 v \ell^2}{R}$

• To maintain constant speed, an external force must balance the magnetic force:

  • $F<em>{ext} = F</em>m$ when the rod is moving at constant velocity.

• Mechanical vs electrical power (ideal energy conversion):

  • Mechanical power input by external agent: $P<em>{mech} = F</em>{ext} \cdot v = F_m\cdot v = \frac{B^2 v^2 \ell^2}{R}$

  • Electrical power dissipated in the circuit (bulb or resistor): $P_{elec} = I^2 R = \left(\frac{B v \ell}{R}\right)^2 R = \frac{B^2 v^2 \ell^2}{R}$

  • In an ideal setup, $P<em>{mech} = P</em>{elec}$, demonstrating energy conservation.

• Important correction to a common misprint in the notes: the correct mechanical power includes velocity, i.e., $P<em>{mech} = F</em>m v = \frac{B^2 v^2 \ell^2}{R}$, not proportional to just a single power of v.

• Summary: External work supplies energy that is converted into electrical energy in the circuit; no free energy is produced.

Change in magnetic flux, generator principle, and practical considerations

• Ways to change magnetic flux through a loop (generator principle):

  • Change the magnitude of the magnetic field B.

  • Change the area A of the loop within the field.

  • Change the angle θ between B and the loop so that cosθ changes.

• Generator equation recap: $\varepsilon = -N \frac{d\Phi}{dt}$, with $\Phi = B A \cos\theta$.

Energy storage in magnetic fields (Inductors)

• Inductance as a proportionality constant: $L = \frac{N\,\Delta\Phi}{\Delta I}$

• For a long solenoid (ideal case):

  • Magnetic flux through one turn: $\Phi = B A = (\mu_0 n I) A$ with turns per length $n = \frac{N}{\ell}$.

  • Inductance of a solenoid: $L = \frac{\mu_0 N^2 A}{\ell}$

  • Alternatively, in terms of n: $L = \mu<em>0 n^2 A \ell$ (note: standard form is $L = \dfrac{\mu</em>0 N^2 A}{\ell}$; the dependence on geometry is what matters).

• Inductance and geometry recap: increasing N or A or decreasing length increases L.

Energy stored in a magnetic field (solenoids and general magnetic fields)

• Energy stored in an inductor: $U = \frac{1}{2} L I^2$

• Energy density of a magnetic field: $u<em>B = \frac{B^2}{2\mu</em>0}$ (valid for any magnetic field region).

• Total energy in a solenoid from energy density perspective: $U = u<em>B \cdot V = \frac{B^2}{2\mu</em>0} \; (A\ell)$ where V = Aℓ is the solenoid volume. This is consistent with $U = \frac{1}{2} L I^2$ when using the solenoid relation $B = \mu<em>0 n I$ and $L = \frac{\mu</em>0 N^2 A}{\ell}$.

• Note: In many slides the algebra is presented with small notational inconsistencies; the standard, consistent forms above should be used for problem solving.

Transformers: basics and ideal transformer equations

• What a transformer does: changes voltage in an alternating current (AC) circuit by inductively coupling two coils.

• Definitions:

  • Primary turns: $N<em>p$, Secondary turns: $N</em>s$

  • Primary voltage: $V<em>p$, Secondary voltage: $V</em>s$

• Transformer equation (ideal, Faraday’s law applied to both coils):

  • $\frac{V<em>s}{V</em>p} = \frac{N<em>s}{N</em>p}$

  • Equivalently: $\frac{V<em>p}{V</em>s} = \frac{N<em>p}{N</em>s}$

• Power conservation (ideal transformer, no losses):

  • $P<em>{in} = P</em>{out} \Rightarrow V<em>p I</em>p = V<em>s I</em>s$

  • Hence, $\frac{I<em>s}{I</em>p} = \frac{V<em>p}{V</em>s} = \frac{N<em>p}{N</em>s}$

• Practical implication: If the output voltage is lower, the current in the secondary is higher (power is conserved).

• Practical notes: Real transformers have losses (core, copper, stray losses), but the ideal relationships illustrate the basic trade-offs between voltage and current.

Energy storage and induction recap

• Inductors store energy in magnetic fields; energy scales with the square of current: $U = \frac{1}{2} L I^2$.

• Magnetic energy density relates to the magnetic field strength: $u<em>B = \frac{B^2}{2\mu</em>0}$, with total energy in a magnetic region equal to energy density times the volume.

Containment structure (context snippet)

• Containment/production chain snippet from the notes (educational context):

  • Containment Structure -> Reactor -> Vessel -> Turbine -> Control Rods -> Generator -> Condenser

  • Power Production framework: Higher reservoir -> Electricity -> Transformer -> Generator -> Turbine

• Practical takeaway: This outlines a generalized flow from energy production to electrical output in large-scale power plants, contextualizing the role of generators, transformers, and turbines.

Quick practice questions (from slides)

• Question: If you stretch the wire of a solenoid so the diameter doubles but the number of turns and the length remain unchanged, by what factor does the inductance increase?

  • Answer: 4 (Factor of 4). Reason: Inductance scales as $L = \dfrac{\mu_0 N^2 A}{\ell}$; doubling diameter increases cross-sectional area A by a factor of 4; thus L increases by factor 4.

• Question: A conducting rod slides on a conducting track in a constant B field directed into the page. What is the direction of the induced current?

  • Answer approach: Use motional emf $\varepsilon = B v \ell$ and Lenz’s law to determine polarity; current direction depends on the direction of rod motion; without a clear diagram the direction cannot be uniquely determined here. (Use clockwise vs. counterclockwise convention around the loop and apply the right-hand rule.)

Summary of key formulas to remember

• Faraday’s law: $\varepsilon = -\frac{N\,\Delta\Phi}{\Delta t}$, with $\Phi = B A \cos \theta$

• Flux change for a moving rod: $\Delta\Phi = B\, \Delta A = B\, (\ell \Delta x)$, hence $\varepsilon = -N B \ell v$

• Current: $I = \frac{\varepsilon}{R}$

• Magnetic force on a current-carrying rod: $F_m = I \ell B$

• Constant-velocity condition: $F<em>m = F</em>g = m g$

• Mechanical vs electrical power (ideal):

  • $P<em>{mech} = F</em>{ext} v = F_m v = \frac{B^2 v^2 \ell^2}{R}$

  • $P_{elec} = I^2 R = \frac{B^2 v^2 \ell^2}{R}$

• Inductance: $L = \frac{N\,\Delta\Phi}{\Delta I}$, and for a solenoid, $L = \frac{\mu_0 N^2 A}{\ell}$

• Energy in an inductor: $U = \frac{1}{2} L I^2$

• Energy density of a magnetic field: $u<em>B = \frac{B^2}{2\mu</em>0}$

• Transformer relations (ideal):

  • M $\frac{V<em>s}{V</em>p} = \frac{N<em>s}{N</em>p}$

  • $V<em>p I</em>p = V<em>s I</em>s$, hence $\frac{I<em>s}{I</em>p} = \frac{N<em>p}{N</em>s}$

Note: The transcript contains a few typographical inconsistencies (e.g., some missing v factors in mechanical power, and a few inconsistent dimensional forms for L). The standard, widely accepted formulas above should be used for exams and problem solving.