Induction & Flux Tricks (IPhO E&M)

IPhO E&M lesson on Faraday's law, motional emf, flux linkage, inductance, and fast flux tricks with energy consistency checks.

International Physics Olympiad preparation

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Induction problems are a mix of geometry, sign discipline, and energy checks. The fastest IPhO solutions compute the flux through a clever surface, differentiate, then use Lenz’s law to set the direction. If motion is involved, the cleanest consistency check is power: mechanical power in equals electrical dissipation plus the rate of change of stored magnetic energy.

Choose a loop direction and a surface normal (right-hand rule links them).
Compute flux Φ = ∬ vector B · d vector A through a convenient spanning surface.
Faraday: E = -dΦ/dt (include turns: E = -N dΦ/dt).
Lenz: induced current direction opposes the change in flux.
Connect to circuit: I = E/R (or include inductance if relevant).
If there is motion, check power: Fv = I²R + dU_mag/dt.

1. Definitions (Must Know)

Magnetic flux:
Φ = ∬ vector B · d vector A.
Faraday’s law (single loop):
E = -dΦ/dt.
Flux linkage (coil with N turns):
E = -d(NΦ)/dt.
Motional emf form (for a moving conductor in static vector B):
E = ∮ (vector v × vector B) · d vector ℓ.
Self-inductance and back emf:
E_L = -LdI/dt, U_L = (1/2)LI².
Mutual inductance:
E₂ = -MdI₁/dt.

2. Key Ideas (What Earns Marks)

Flux is geometry. Many questions are solved by identifying what part of the loop actually threads magnetic field.
You can choose the surface. Any surface whose boundary is the loop is allowed, and the smartest choice makes Φ easy.
Lenz is a direction rule, not a formula. Compute the sign by asking: what current would reduce the change in flux?
Motional emf is field-line cutting. For a rod of length ℓ moving with speed v perpendicular to a uniform B:
E = Bℓ v.
Power check kills sign mistakes. If a magnetic force opposes motion, you must supply mechanical power, and it should match I²R (plus any stored energy change).

3. Detailed Explanations

A. Sign discipline: loop direction and normal.
Pick a loop traversal direction. The right-hand rule gives you the corresponding positive normal vector n hat. Flux is positive if vector B has a component along n hat.

Then Faraday says E = -dΦ/dt. If Φ is increasing, the emf is negative in your chosen direction, meaning the actual current runs opposite your assumed direction.

B. Motional emf in one line.
In a moving conductor, charges feel q vector v × vector B. In steady motion, electrostatic separation builds until E balances this, giving an emf along the conductor. The loop integral form

E = ∮ (vector v × vector B) · d vector ℓ

is a reliable way to keep geometry and sign consistent.

C. Sliding rod on rails: the standard power identity.
A rod of length ℓ slides on rails in uniform B (perpendicular to the plane), closing a circuit with resistance R. If the rod moves with speed v:

E = Bℓ v, I = (Bℓ v)/R.

The magnetic force on the rod has magnitude

F = Iℓ B = (B²ℓ²/R)v,

opposing the motion (Lenz). Mechanical power required to maintain constant v is

P_mech = Fv = (B²ℓ² v²)/R = I²R,

matching electrical dissipation.

D. Inductors: energy storage and “back emf”.
An inductor resists changes in current:

E_L = -LdI/dt.

The minus sign means the inductor’s emf opposes the change. The stored magnetic energy is U_L = 1/2 LI², which is a powerful check in switching problems.

E. Mutual inductance and transformer-style reasoning.
If changing current I₁(t) in coil 1 produces a changing flux through coil 2, then

E₂ = -MdI₁/dt.

In many olympiad setups, you do not need M explicitly: you compute flux through the secondary due to the primary and differentiate.

4. Common Mistakes

Forgetting the cosine in flux: Φ = BA cos θ.
Using the wrong area: only the part of the loop that spans the field contributes.
Sign errors from changing the loop direction mid-solution.
Treating Lenz’s law as “current opposes flux” rather than “current opposes the change in flux.”
Missing the power check: getting a force that accelerates the rod while also claiming I²R heating with no energy source.

5. Exam Tips

Put a dot or cross on your diagram to show the direction of vector B through the page.
Choose a surface for flux that makes Φ a simple product (uniform B, rectangular area, constant angle).
If the loop moves but B is uniform, the only thing that changes is area or orientation.
When a loop is partly inside a field region, write flux as Φ = B × (overlap area).

6. Worked Examples

Example 1 (sliding rod: current, force, and power).
A conducting rod of length ℓ slides at constant speed v on rails in a uniform magnetic field B perpendicular to the plane of the rails. The circuit resistance is R. Find the induced current, the magnetic force magnitude on the rod, and the mechanical power required to maintain the speed.

Solution sketch

Motional emf:

E = Bℓ v.

Current:

I = E/R = (Bℓ v)/R.

Force magnitude on the rod:

F = Iℓ B = (B²ℓ²/R)v,

opposing the motion (Lenz). Mechanical power:

P = Fv = (B²ℓ² v²)/R = I²R.

Example 2 (loop entering a uniform field region).
A rectangular loop of width w and height h moves to the right at speed v into a region of uniform magnetic field B into the page. The loop has resistance R. Find the induced emf magnitude as a function of time during entry, fully inside, and exit.

Solution sketch

During entry, the overlap area increases at rate dA/dt = h v (the boundary sweeps across at speed v). Flux magnitude is Φ = B A, so

|E| = |dΦ/dt| = BdA/dt = Bhv.

While fully inside the uniform field, the overlap area is constant, so dΦ/dt = 0 and E = 0. During exit, the overlap area decreases at the same rate, so the emf magnitude is again Bhv but the current direction reverses (Lenz).

If needed, current magnitude during entry/exit is I = |E|/R = Bhv/R.

Example 3 (induced emf from a changing current in a solenoid).
A long solenoid with n turns per unit length carries current I(t). A single circular loop of wire (resistance R) tightly encloses the solenoid. The solenoid radius is b. Find the induced emf and induced current in the loop.

Solution sketch

Inside an ideal long solenoid:

B = μ₀ n I(t).

The loop encloses the solenoid’s cross-sectional area A = π b², so flux is

Φ = BA = μ₀ n I(t) π b².

Faraday:

E = -dΦ/dt = -μ₀ n π b²dI/dt.

Current in the loop:

Iₗₒₒₚ = E/R.

Direction is set by Lenz: it produces a magnetic field that opposes the change in the solenoid’s field.

7. Mind Stretchers

Mind stretcher: flux surface choice around an ideal solenoid

A circular loop of radius a surrounds a long solenoid of radius b, with b smaller than a. The solenoid current changes with time. Many students say “the magnetic field outside the solenoid is zero, so the flux through the loop is zero, so the induced emf is zero.” Explain what is wrong with that reasoning and compute the induced emf magnitude.

What is wrong: for a loop that encircles the solenoid, any surface whose boundary is that loop must intersect the solenoid cross-section. You cannot choose a spanning surface that avoids the interior region where vector B is nonzero.

Using the solenoid cross-section as the spanning surface:

Φ = μ₀ n I(t) π b² ⇒ |E| = μ₀ n π b²|dI/dt|.

The direction follows Lenz’s law.

8. Practice

For three different geometries, compute Φ by explicitly writing the surface normal and vector B · d vector A.
Do one motional-emf problem and verify Fv = I²R.
Do one inductance problem where you compute E from Φ(I) and then check energy using U_L = 1/2 LI².

Syllabus and review details

No official syllabus alignment is listed for this lesson.