Induction Treatise Reading Section 4.2 | Electrodynamics
Motional Induction in Rectangular Geometries
In the study of electromagnetism, few scenarios provide as clear a window into the interplay between mechanics and field theory as a conductor moving through a magnetic field. When a rectangular loop enters a region of uniform magnetic flux density \( \vec{B} \), a series of predictable yet profound responses occur.
I. The Geometry of the System
Consider a coordinate system where a uniform magnetic field \( \vec{B} \) is directed into the page. A rectangular loop lies in the \( xy \)-plane and moves with a constant velocity \( \vec{v} \) in the \( +x \) direction.
Loop
\( \ell \)
\( w \)
v
Figure 1: Loop geometry approaching a uniform \( B \)-field.
The total magnetic flux \( \Phi_B \) through the loop is: \( \Phi_B = B A_{in} \). As the loop moves, the immersed area \( A_{in} \) changes, inducing an EMF according to Faraday's Law.
Checkpoint 1.1
If the magnetic field were not perpendicular to the loop, how would the flux equation be modified?
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Phase I: Entering the Field
As the leading edge enters the field, the immersed area increases linearly with time: \( A(t) = \ell v t \). Consequently, the flux increases:
\[ \Phi_B(t) = B \ell v t \]
The Induced EMF
Faraday's Law states that the induced EMF \( \varepsilon \) is the negative rate of change of magnetic flux:
\[ \varepsilon = - \frac{d\Phi_B}{dt} = - B \ell v \]
Lenz’s Law and Current
Flux into the page is increasing. The loop opposes this change by creating an outward field, resulting in a counter-clockwise (CCW) current: \( I = B\ell v / R \).
The Magnetic Drag Force
Using the Lorentz force law \( \vec{F} = I\vec{\ell} \times \vec{B} \), we find the force on the leading edge points to the left , opposing the loop's velocity.
Conclusion
Entering the field creates a magnetic drag force that resists the physical motion of the loop.
Reflect
Explain why the forces on the top and bottom segments of the loop do not contribute to the net horizontal force.
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Phase II: Fully Immersed
When the loop is entirely within the field, the area inside the field is constant: \( A = \ell \cdot w \).
The Vanishing EMF
Because the magnetic field is uniform and the area is not changing, the flux is constant. The derivative of a constant is zero:
\[ \frac{d\Phi_B}{dt} = 0 \implies \varepsilon = 0 \]
With no induced EMF, no current flows (\( I = 0 \)), and there is no net magnetic force.
Theoretical Insight
Both vertical segments generate individual motional EMFs. However, because they are connected in series but pointing in "opposite" directions relative to the loop's circulation, they cancel each other out perfectly.
Analyze
If the magnetic field were not uniform, would the EMF still be zero in this phase? Justify your answer.
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Phase III: Exiting the Field
As the loop exits, the flux into the page decreases. Faraday’s Law yields a positive EMF:
\[ \varepsilon = - \frac{d\Phi_B}{dt} = + B \ell v \]
Lenz’s Law Response
To replace the lost flux, the loop induces a clockwise (CW) current. This current creates its own magnetic field directed into the page.
The Braking Effect
Only the trailing edge remains in the field. In this segment, current flows upward, and the magnetic field points in. The resulting force \( \vec{F}_B \) is directed to the left .
Conservation Principle
The magnetic force always opposes the change—in this case, it resists the loop leaving the field. This is the mechanical manifestation of Lenz's Law.
Verify
Using the RHR, confirm the direction of force for the trailing edge during the exit phase.
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IV. Summary of Findings
Property Entering Inside Exiting Flux Increasing Constant Decreasing Induced EMF Negative (-) Zero (0) Positive (+) Current Direction CCW None CW Magnetic Force LEFT Zero LEFT
Power Dissipation
The mechanical power required to maintain motion equals the electrical heat dissipated:
\[ P = \frac{B^2 \ell^2 v^2}{R} \]
Synthesis Challenge
Calculate the total energy dissipated in the resistor from the moment the loop starts to enter until it has completely left the field.
Key Takeaway: Induction creates a mechanical resistance that preserves the symmetry of energy.
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V. Graphical Signatures of Induction
Magnetic Flux (\( \Phi_B \)) vs. Time
Start Entry Full Entry Start Exit Full Exit
\( \Phi_{max} \)
0
Figure 2: The magnetic flux through the loop rises linearly during penetration, remains constant while fully immersed, and falls linearly during the exit.
Induced EMF (\( \varepsilon \)) vs. Time
\( +B\ell v \)
\( -B\ell v \)
Entry Inside Exit
Figure 3: The induced EMF is a derivative of flux. Since flux is linear in transition phases, EMF is a constant non-zero value during those periods.
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Induction Insights Handout Loop Logic
AP Physics C: Induction
PHY-C-4.2
Researcher:
Date:
A rectangular loop (l × w ) with resistance R enters a region of uniform magnetic field B at velocity v .
LOOP l w
v
Phase Flux change Induced I Force Entering Increasing CCW LEFT Inside Steady Zero Zero Exiting Decreasing CW LEFT
I. Derivation of EMF
\[ \Phi_B = \iint \vec{B} \cdot d\vec{A} = B l v t \]
\[ \varepsilon = - d\Phi_B / dt = - B l v \]
Instructional Graphics
Flux Profile
Entry Inside Exit
Max
EMF Profile
+Blv
-Blv
Entry Inside Exit
Vector Analysis Exercise
Using the formula F = I_l_ × B , justify why the magnetic force is leftward during both transition phases. Explain which side of the loop is active.
FBD: Entry
FBD: Exit
Loop Logic Problems Worksheet Loop Logic Practice
Electromagnetic Induction | AP Physics C
ID: IND-004
Researcher:
Date:
The Scenario: A square loop of wire with side length \( s = 0.20 \, \text{m} \) and total resistance \( R = 0.50 \, \Omega \) moves at a constant speed \( v = 4.0 \, \text{m/s} \) through a region containing a uniform magnetic field \( B = 0.80 \, \text{T} \) directed into the page. The field exists between \( x = 0 \) and \( x = 0.50 \, \text{m} \).
v
x=0 x=0.5m
1. PHASE ANALYSIS: ENTRY
As the leading edge of the loop enters the field, calculate the magnitude of the induced EMF (\( \varepsilon \)) and the resulting induced current (\( I \)).
2. DIRECTIONAL LOGIC
Justify the direction of the induced current (CW or CCW) using Lenz's Law. Be specific about the flux change and the system's response.
3. MECHANICAL COUPLING
Calculate the magnitude and direction of the external force required to keep the loop moving at the constant speed \( v \) while it is entering the field.
Set ID: IND-004 | Page 1/2
4. GRAPHICAL SYNTHESIS
Sketch the Magnetic Force (\( F_B \)) acting on the loop as a function of time. Let \( t=0 \) be the moment the loop starts to enter the field.
Force Time Forward (+) Backward (-)
5. ENERGY CONSERVATION
Prove that the total mechanical work done by the external force during the entry phase is exactly equal to the electrical energy dissipated as heat in the resistor.
6. CHALLENGE: NON-UNIFORM FIELD
Suppose the magnetic field varied as \( B(x) = B_0 (1 + \alpha x) \). Determine the induced EMF \( \varepsilon(t) \) as the loop enters the field from \( x=0 \) at \( t=0 \). Note: The loop side length is \( s \).
AP PHYSICS C | UNIT 4
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Flux Flow Slides Loop Logic
Vector & Graphical Analysis
Phase 1: Entering Field
Lenz's Logic
Flux (IN) is increasing
System opposes change
Current: CCW (+)
Force Vector:
Drag acts on leading wire to the LEFT.
\( F_B \)
Phase 2: Fully Inside
Steady State Analysis
Flux is Constant
\[ \varepsilon = - d\Phi_B / dt = 0 \]
"No flux change means zero induced potential. The loop coasts at equilibrium."
Phase 3: Exiting Field
Lenz's Logic
Flux (IN) is decreasing
Replacement field created
Current: CW (-)
Force Vector:
Drag acts on the trailing wire to the LEFT.
Force
Calculus Signatures
Flux Profile
Entry Inside Exit
MAX
EMF Profile
+Bℓv ZERO -Bℓv
Entry Inside Exit
Directional Logic Guide Directional Logic Guide
Lenz's Law & Force Vectors
Phase 1: Entry Logic
Step A: Flux Analysis
Loop moves right, penetration \( x \) increases. Area inside field \( A = \ell x \) increases. Flux into page (\( \Phi \)) is increasing .
Step B: EMF Direction
System creates OUTWARD field to oppose increase. Right hand rule (thumb out) results in CCW current. EMF is negative (\( -B \ell v \)).
FORCE: CCW current on RIGHT wire segment (pointing UP) feeling B-field (pointing IN). Force is LEFT.
Phase 2: Middle Logic
Step A: Flux Analysis
Loop is fully inside. Area is constant. \( \Phi = B \ell w = \text{const} \).
Step B: EMF Result
Derivative of a constant is zero . No EMF, no current.
FORCE: Zero net magnetic force.
Phase 3: Exit Logic
Step A: Flux Analysis
Loop moves right, penetration \( x \) decreases. Flux into page (\( \Phi \)) is decreasing .
Step B: EMF Direction
System creates INWARD field to replace lost flux. Right hand rule (thumb in) results in CW current. EMF is positive (\( +B \ell v \)).
FORCE: CW current on LEFT wire segment (pointing UP) feeling B-field (pointing IN). Force is LEFT.
The Golden Rule of Induction
Induced currents and the resulting magnetic forces always act to oppose the physical motion causing the flux change. This is the mechanical manifestation of the Law of Conservation of Energy.
Loop Logic Answer Key Teacher Answer Key
Loop Logic Practice | AP Physics C
Internal Use Only
1. Phase Analysis: Entry
Calculations
Variables: \( s = 0.20 \, m, R = 0.50 \, \Omega, v = 4.0 \, m/s, B = 0.80 \, T \)
Induced EMF:
Induced Current:
2. Directional Logic
Lenz's Law Justification
As the loop enters the field, the magnetic flux directed into the page (\( \Phi_B \)) is increasing. To oppose this change, the system induces a magnetic field directed out of the page . According to the Right-Hand Rule (thumb pointing out), the induced current must circulate in a Counter-Clockwise (CCW) direction.
3. Mechanical Coupling
Force Vector Analysis
The magnetic force acts only on the leading vertical segment: \( F_B = IsB \).
\( F_B = (1.28 \, A)(0.20 \, m)(0.80 \, T) = 0.2048 \, N \)
Direction of \( F_B \): Left (opposing velocity).
External Force: To maintain constant velocity, \( F_{ext} = F_B \).Right .
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4. Graphical Synthesis
Magnetic Force (\( F_B \)) Key
Right (+) Left (-) ENTRY Inside Exit
Key Insight: Magnetic force always opposes motion (Left) during both flux changes.
5. Energy Conservation
Energy Balance Proof
Work:
\( W = F \cdot \Delta x = (IsB) \cdot s = (\frac{Bsv}{R} s B) s = \frac{B^2 s^3 v}{R} \)
Heat:
\( E = P \cdot \Delta t = (I^2 R) \cdot (\frac{s}{v}) = (\frac{Bsv}{R})^2 R \cdot \frac{s}{v} = \frac{B^2 s^3 v}{R} \)
Conclusion: \( W = E \). Conservation of energy is satisfied.
6. Challenge Solution
Calculus of Non-Uniform Fields
\( \Phi(x) = \int_{0}^{x} B_0(1+\alpha u) s \, du = B_0 s (x + \frac{\alpha x^2}{2}) \)
Substitute \( x = vt \):
Induced EMF: \( \varepsilon = -d\Phi/dt \)
\( \varepsilon(t) = -B_0 s v (1 + \alpha v t) \)
AP Physics C: Electricity and Magnetism
