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{\Phif - \Phii}{tf - ti}
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: Fm = Fg, 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{total} = Ki + U{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{ext} = Fm when the rod is moving at constant velocity.
Mechanical vs electrical power (ideal energy conversion):
Mechanical power input by external agent: P{mech} = F{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{mech} = P{elec}, demonstrating energy conservation.
Important correction to a common misprint in the notes: the correct mechanical power includes velocity, i.e., P{mech} = Fm 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 = \mu0 n^2 A \ell (note: standard form is L = \dfrac{\mu0 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: uB = \frac{B^2}{2\mu0} (valid for any magnetic field region).
Total energy in a solenoid from energy density perspective: U = uB \cdot V = \frac{B^2}{2\mu0} \; (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 = \mu0 n I and L = \frac{\mu0 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: Np, Secondary turns: Ns
Primary voltage: Vp, Secondary voltage: Vs
Transformer equation (ideal, Faraday’s law applied to both coils):
\frac{Vs}{Vp} = \frac{Ns}{Np}
Equivalently: \frac{Vp}{Vs} = \frac{Np}{Ns}
Power conservation (ideal transformer, no losses):
P{in} = P{out} \Rightarrow Vp Ip = Vs Is
Hence, \frac{Is}{Ip} = \frac{Vp}{Vs} = \frac{Np}{Ns}
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: uB = \frac{B^2}{2\mu0}, 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: Fm = Fg = m g
Mechanical vs electrical power (ideal):
P{mech} = F{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: uB = \frac{B^2}{2\mu0}
Transformer relations (ideal):
M \frac{Vs}{Vp} = \frac{Ns}{Np}
Vp Ip = Vs Is, hence \frac{Is}{Ip} = \frac{Np}{Ns}
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.