AP Physics 2- 12.4 Electromagnetic Induction and Faraday’s Law- Study Notes- New Syllabus
AP Physics 2- 12.4 Electromagnetic Induction and Faraday’s Law – Study Notes
AP Physics 2- 12.4 Electromagnetic Induction and Faraday’s Law – Study Notes – per latest Syllabus.
Key Concepts:
- Magnetic Flux
- Faraday’s Law of Electromagnetic Induction
- Lenz’s law
- Electromagnetic Induction: Conducting Rod on Rails
Magnetic Flux
Magnetic flux (\( \Phi_\mathrm{B} \)) is a measure of the amount of magnetic field passing through a given surface area. It quantifies the total magnetic field lines penetrating a surface.
Formula:
\( \Phi_\mathrm{B} = \mathrm{B} \cdot \mathrm{A} \cdot \cos\theta \)
- \(\mathrm{B}\): Magnetic field strength (in teslas, T)
- \(\mathrm{A}\): Area of the surface (in \( \mathrm{m^2} \))
- \(\theta\): Angle between the magnetic field vector and the normal (perpendicular) to the surface
Key Concepts:
- If the magnetic field is perpendicular to the surface (\(\theta = 0^\circ\)): \(\Phi_\mathrm{B} = \mathrm{BA}\) (maximum flux)
- If the magnetic field is parallel to the surface (\(\theta = 90^\circ\)): \(\Phi_\mathrm{B} = 0\) (no flux passes through)
- Magnetic flux depends on both the strength of the field and the orientation of the surface.
Units:
- The SI unit of magnetic flux is the weber (Wb).
- 1 Wb = 1 T·m².
Relation to Faraday’s Law:
- A change in magnetic flux through a loop induces an electromotive force (emf) in the loop (basis of electromagnetic induction).
Example :
A uniform magnetic field of \(0.20 \, \mathrm{T}\) passes perpendicularly through a circular loop of radius \(0.10 \, \mathrm{m}\). Find the magnetic flux through the loop.
▶️ Answer/Explanation
Step 1: Area of loop: \(\mathrm{A} = \pi r^2 = \pi (0.10)^2 = 0.0314 \, \mathrm{m^2}\).
Step 2: Since the field is perpendicular, \(\cos\theta = 1\).
Step 3: Flux: \(\Phi_\mathrm{B} = \mathrm{BA}\cos\theta = (0.20)(0.0314)(1) = 0.00628 \, \mathrm{Wb}\).
Answer: \(\Phi_\mathrm{B} \approx 6.3 \times 10^{-3} \, \mathrm{Wb}\).
Faraday’s Law of Electromagnetic Induction
Faraday’s law states that a changing magnetic flux through a circuit induces an electromotive force (emf) in the circuit. The induced emf is proportional to the rate of change of magnetic flux through the circuit.
Mathematical Expression:
\( \mathcal{E} = -N \dfrac{d\Phi_\mathrm{B}}{dt} \)
- \(\mathcal{E}\): Induced emf (in volts, V)
- \(N\): Number of turns of the coil
- \(\Phi_\mathrm{B}\): Magnetic flux through a single loop
- \(-\) sign: Represents Lenz’s Law (the induced emf produces a current whose magnetic field opposes the change in flux).
Key Concepts:
- A changing magnetic field (in strength, direction, or orientation) induces emf.
- The faster the change in flux, the larger the induced emf.
- Constant magnetic flux (no change) → no emf induced.
Units:
- Induced emf (\(\mathcal{E}\)) is measured in volts (V).
Lenz’s Law (Direction of Induced emf):
- The direction of induced current always opposes the change in magnetic flux that produced it.
- This is a consequence of conservation of energy.
Example :
A circular loop of radius \(0.05 \, \mathrm{m}\) with \(N = 20\) turns is placed in a magnetic field that increases uniformly from \(0.2 \, \mathrm{T}\) to \(0.6 \, \mathrm{T}\) in \(0.5 \, \mathrm{s}\). Find the induced emf.
▶️ Answer/Explanation
Step 1: Area of one loop: \(\mathrm{A} = \pi r^2 = \pi (0.05)^2 = 7.85 \times 10^{-3} \, \mathrm{m^2}\).
Step 2: Change in flux per loop: \(\Delta \Phi_\mathrm{B} = \Delta \mathrm{B} \cdot \mathrm{A} = (0.6 – 0.2)(7.85 \times 10^{-3}) = 3.14 \times 10^{-3} \, \mathrm{Wb}\).
Step 3: Rate of change of flux: \(\dfrac{\Delta \Phi_\mathrm{B}}{\Delta t} = \dfrac{3.14 \times 10^{-3}}{0.5} = 6.28 \times 10^{-3} \, \mathrm{Wb/s}\).
Step 4: Induced emf: \(\mathcal{E} = N \dfrac{\Delta \Phi_\mathrm{B}}{\Delta t} = (20)(6.28 \times 10^{-3}) = 0.126 \, \mathrm{V}\).
Answer: \(\mathcal{E} \approx 0.13 \, \mathrm{V}\).
Lenz’s Law
Lenz’s law states that the direction of the induced emf (and hence the induced current) is always such that it opposes the change \((\delta)\) in magnetic flux that produced it. This law ensures the conservation of energy in electromagnetic induction.
Mathematical Connection:
\( \mathcal{E} = -N \dfrac{\delta \Phi_\mathrm{B}}{\delta t} \)
- The negative sign explicitly represents Lenz’s law: opposition to change.
- \(\Phi_\mathrm{B} = \mathrm{BA}\cos \theta\) is the magnetic flux through the loop.
Key Concepts:
- If the magnetic flux through a loop is increasing, the induced current will create a magnetic field that opposes the increase.
- If the magnetic flux is decreasing, the induced current will create a magnetic field that opposes the decrease (i.e., it tries to maintain the original flux).
- Lenz’s law prevents violation of the law of conservation of energy. Without it, induced emf would reinforce the change, creating energy from nothing.
Example :
A conducting loop is placed near a magnet. The magnet is pushed toward the loop. What is the direction of the induced current?
▶️ Answer/Explanation
Step 1: The magnet’s north pole approaches the loop, so the magnetic flux through the loop increases into the page.
Step 2: According to Lenz’s law, the induced current will produce a magnetic field that opposes this increase. So, the induced magnetic field must point out of the page.
Step 3: Using the right-hand rule, a field out of the page requires a counterclockwise current in the loop.
Answer: The induced current is counterclockwise.
Electromagnetic Induction: Conducting Rod on Rails
Setup:
- A conducting rod of length \(\mathrm{L}\) slides on frictionless conducting rails.
- The system is placed in a region of uniform magnetic field \(\vec{B}\), directed into the page.
- The rod is pulled with velocity \(\vec{v}\) perpendicular to its length.
Magnetic Flux:
\( \Phi_\mathrm{B} = \mathrm{BA}\cos \theta = \mathrm{B(Lx)} \quad \text{(with } \theta = 0^\circ \text{, since } \vec{B} \perp \text{area}) \)
- As the rod moves, the area \(A = \mathrm{Lx}\) increases.
- Hence, the magnetic flux through the loop changes with time.
Induced emf (Faraday’s Law):
\( \mathcal{E} = – \dfrac{\delta \Phi_\mathrm{B}}{\delta t} = – \dfrac{\delta (\mathrm{BLx})}{\delta t} = – \mathrm{BL} \dfrac{\delta x}{\delta t} \)
\( \mathcal{E} = – \mathrm{BLv} \)
- Magnitude of induced emf: \(\mathcal{E} = \mathrm{BLv}\).
- Direction is determined using Lenz’s law: the induced current opposes the change in flux.
Motional Force on the Rod:
The current-carrying rod in the magnetic field experiences a magnetic force:
\( F = I \mathrm{L} B = \dfrac{\mathrm{BLv}}{R} \cdot \mathrm{LB} = \dfrac{\mathrm{B^2 L^2 v}}{R} \)
- This force opposes the applied pulling force (consistent with Lenz’s law).
Example :
A conducting rod of length \(0.5 \, \mathrm{m}\) slides with speed \(2 \, \mathrm{m/s}\) on rails in a uniform magnetic field of \(0.4 \, \mathrm{T}\). The rails are connected to a \(4 \, \Omega\) resistor.
Find the induced emf, current, and power dissipated in the resistor.
▶️ Answer/Explanation
Step 1: Induced emf: \(\mathcal{E} = \mathrm{BLv} = (0.4)(0.5)(2) = 0.4 \, \mathrm{V}\).
Step 2: Current: \( I = \dfrac{\mathcal{E}}{R} = \dfrac{0.4}{4} = 0.1 \, \mathrm{A} \).
Step 3: Power dissipated: \( P = I^2R = (0.1)^2 (4) = 0.04 \, \mathrm{W} \).
Answer: \(\mathcal{E} = 0.4 \, \mathrm{V}, \, I = 0.1 \, \mathrm{A}, \, P = 0.04 \, \mathrm{W}\).