AP Physics C E&M- 13.4 Inductance - Study Notes- New Syllabus
AP Physics C E&M- 13.4 Inductance – Study Notes
AP Physics C E&M- 13.4 Inductance – Study Notes – per latest Syllabus.
Key Concepts:
- Physical and Electrical Properties of an Inductor
- Energy Storage and emf in Inductors
Physical and Electrical Properties of an Inductor
An inductor is a circuit element that stores energy in its magnetic field and resists changes in current. Inductance (\(L\)) is the property of a conductor or coil that quantifies its opposition to changes in current.
General Properties:
- Straight wires are usually modeled as having negligible inductance (\(L \approx 0\)).
- A tightly wound coil (solenoid) has significant inductance and is commonly used as an inductor in circuits.
- The opposition to current change arises because a changing current produces a changing magnetic flux, which induces an emf that opposes the change (Lenz’s Law).
Inductance of a Solenoid:
\( \mathrm{L = \mu \dfrac{N^2 A}{\ell}} \)
- \( \mathrm{\mu} \) = magnetic permeability of the core material
- \( \mathrm{N} \) = total number of turns in the solenoid
- \( \mathrm{A} \) = cross-sectional area of the solenoid
- \( \mathrm{\ell} \) = length of the solenoid
Key Ideas:
- Inductance depends only on the geometry of the coil and the magnetic properties of its core, not on current.
- Inductors resist sudden changes in current, making them important in filters, transformers, and energy storage systems.
Example:
A solenoid of length \(0.40 \, m\), with \(500\) turns, and cross-sectional area \(1.0 \times 10^{-3} \, m^2\), is wound on a core with magnetic permeability \( \mu = 4\pi \times 10^{-7} \, H/m \) (free space). Find its inductance.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{L = \mu \dfrac{N^2 A}{\ell}} \).
Step 2: Substitute values: \( \mathrm{L = (4\pi \times 10^{-7}) \dfrac{(500)^2 (1.0 \times 10^{-3})}{0.40}} \).
Step 3: Simplify: \( \mathrm{L = (4\pi \times 10^{-7}) \dfrac{2.5 \times 10^5 \times 1.0 \times 10^{-3}}{0.40}} \).
\( \mathrm{L = (4\pi \times 10^{-7})(625)} \).
Step 4: \( \mathrm{L \approx 7.85 \times 10^{-4} \, H = 0.785 \, mH} \).
Final Answer: The inductance of the solenoid is approximately \( \mathrm{0.785 \, mH} \).
Energy Storage and emf in Inductors
Inductors store energy in the magnetic field generated by current flowing through them. This stored energy can later be dissipated through resistors as heat or transferred to capacitors in oscillating circuits. The processes obey the principle of conservation of energy.
Energy Stored in an Inductor:
\( \mathrm{U = \tfrac{1}{2} L I^2} \)
- \( \mathrm{L} \) = inductance of the inductor
- \( \mathrm{I} \) = current through the inductor
- The energy is stored in the magnetic field created by the current.
Induced emf in an Inductor:
\( \mathrm{\mathcal{E}_L = -L \dfrac{dI}{dt}} \)
- A time-varying current induces an emf that opposes the change in current (Lenz’s Law).
- This emf arises directly from Faraday’s law applied to the changing magnetic flux of the inductor.
Energy Transfer:
The energy stored in the magnetic field of an inductor can be transferred to:
- A resistor, where it is dissipated as thermal energy.
- A capacitor, where it is stored as electric potential energy in an LC circuit.
This transfer follows conservation of energy:
\( \mathrm{ \Delta U_{inductor} + \Delta U_{resistor} + \Delta U_{capacitor} = 0 } \)
Key Idea: Inductors resist changes in current and store energy in magnetic fields. The energy stored (\( \tfrac{1}{2} L I^2 \)) can be released into other parts of the circuit, always obeying conservation laws.
Example:
An inductor of \(0.20 \, H\) carries a current of \(2.5 \, A\). If the current decreases uniformly to zero in \(0.10 \, s\), calculate
(a) the energy stored initially, and
(b) the magnitude of the induced emf.
▶️ Answer/Explanation
Part (a): Energy stored
Formula: \( \mathrm{U = \tfrac{1}{2} L I^2} \).
\( \mathrm{U = \tfrac{1}{2}(0.20)(2.5^2) = 0.5 \times 0.20 \times 6.25 = 0.625 \, J} \).
Part (b): Induced emf
Formula: \( \mathrm{\mathcal{E}_L = -L \dfrac{dI}{dt}} \).
Rate of change: \( \mathrm{\dfrac{dI}{dt} = \dfrac{0 – 2.5}{0.10} = -25 \, A/s} \).
\( \mathrm{\mathcal{E}_L = -(0.20)(-25) = 5.0 \, V} \).
Final Answer:
- Initial energy stored: \( \mathrm{0.625 \, J} \)
- Induced emf: \( \mathrm{5.0 \, V} \)