PHYS7021
Burdwan · Electricity, Magnetism & Wave Optics · 2018–2024

3A · Faraday Laws + Lenz (Full Solutions)

Faraday laws in all forms, Lenz law, non-inductive coil, self-induction, L and M definitions. Step-by-step, SI units.

78 core Q 129 appearances 45 h syllabus SI units · KaTeX

A magnet moved near a coil lights a bulb with no battery attached. The cause is the change of magnetic flux; the effect is the induced e.m.f. Lenz law fixes the direction. This page covers every qualitative PYQ in EMI; the inductance numerics live on the companion page.

Bar magnet approaching a coil; flux N B A changes; e.m.f. e induced in coil
Fig 1. Moving magnet changes flux $\Phi = NBA$ through the coil; coil e.m.f. is $e = -N\,d\Phi/dt$. Polarity follows Lenz law.

Laws and definitions

Faraday law 1. Whenever the magnetic flux through a circuit changes, an e.m.f. is induced. The e.m.f. lasts only while the flux changes.

Formula 1. $e = -\dfrac{d\Phi}{dt}$ for one turn; $e = -N\dfrac{d\Phi}{dt}$ for $N$ turns.
Formula 2 (integral form). $\oint\vec{E}\cdot d\vec{l} = -\dfrac{d\Phi}{dt}$ — induced $\vec{E}$ integrated round the loop equals the rate of flux fall.
Formula 3 (differential form). $\nabla\times\vec{E} = -\dfrac{\partial\vec{B}}{\partial t}$ — a changing $\vec{B}$ at a point creates a curl of $\vec{E}$ there.

Lenz law. The minus sign in Eq. (1) is Lenz law — the induced current flows so its own field opposes the change that caused it. Energy argument: if it helped the change, we would get free energy (a magnet would accelerate on its own). Since energy is conserved, it must oppose.

Self induction. Flux of a coil due to its own current: $\Phi = LI$. Then $e = -L,dI/dt$. The constant $L$ is called “electric inertia” (see Q4 below).

Mutual induction. Flux in coil 2 due to current in coil 1: $\Phi_2 = MI_1$. Then $e_2 = -M,dI_1/dt$.


Solved PYQs

Q120192022×2

State (and explain) Faraday's laws of electromagnetic induction.

Recall

Law 1 is the qualitative "change of flux is needed". Law 2 is the size and sign: $e = -N\,d\Phi/dt$ (Eq. 1).

Steps

  1. Law 1: change of flux through a coil induces an e.m.f.; no change → no e.m.f. (Reason: experiment with moving magnet and coil.)
  2. Law 2: $e = -d\Phi/dt$ for one turn; $e = -N\,d\Phi/dt$ for $N$ turns. The minus sign is Lenz law (opposition).
  3. Explain symbols: $\Phi = BA\cos\theta$ in weber (Wb), $t$ in seconds, $e$ in volts.
  4. Faster change gives bigger e.m.f.; the e.m.f. exists only while the flux is changing.

Answer

$\boxed{\,e = -N\dfrac{d\Phi}{dt}\,}$ — e.m.f. in volts, flux in weber, time in seconds.

Exam tip

Give one example: push magnet in → deflection; hold still → zero; pull out → opposite deflection.

Q22018

Write down the integral and differential forms of Faraday's law.

Recall

Equation (2) is the integral form; Equation (3) is the differential form. Get (3) from (2) by Stokes' theorem.

Steps

  1. Integral form: $\oint\vec{E}\cdot d\vec{l} = -d\Phi/dt$ — induced $\vec{E}$ integrated round the loop is the e.m.f.
  2. Apply Stokes: $\oint\vec{E}\cdot d\vec{l} = \int(\nabla\times\vec{E})\cdot d\vec{S}$.
  3. Write flux as surface integral: $\Phi = \int\vec{B}\cdot d\vec{S}$, so $-d\Phi/dt = -\int(\partial\vec{B}/\partial t)\cdot d\vec{S}$.
  4. The two surfaces are arbitrary and identical, so the integrands must match: $\nabla\times\vec{E} = -\partial\vec{B}/\partial t$.
  5. This holds at every point even with no wire present — it is a statement about fields, not a circuit.

Answer

Integral: $\boxed{\,\oint\vec{E}\cdot d\vec{l} = -\dfrac{d\Phi}{dt}\,}$. Differential: $\boxed{\,\nabla\times\vec{E} = -\dfrac{\partial\vec{B}}{\partial t}\,}$.

Exam tip

State that the differential form holds even with no wire present — that line earns the "field" mark.

Q320182023×2

What do you mean by non-inductive coil?

Recall

A coil wound so that its own magnetic fields cancel inside the coil — net flux and self-inductance are nearly zero.

Steps

  1. Construction: bifilar winding — the wire is doubled back on itself; current goes out in one strand and returns in the neighbour.
  2. Effect: the two strands carry equal and opposite currents, so their magnetic fields cancel inside the coil.
  3. Result: net flux $\approx 0$, so $L \approx 0$ and the induced e.m.f. $e = -L\,dI/dt \approx 0$.
  4. Use: resistance boxes and standard resistors, where we want pure resistance with no inductive kick when the current changes.

Answer

A non-inductive coil uses a bifilar (doubled-back) winding so that opposite strands cancel each other's field, giving $\boxed{\,L \approx 0\,}$.

Exam tip

Draw two adjacent strands with arrows in opposite directions — one sketch earns the construction mark.

Q42019

Define self induction. Why is it called electric inertia?

Recall

$\Phi = LI$ and $e = -L\,dI/dt$. Compare with Newton's second law $F = m\,dv/dt$ and kinetic energy $K = \tfrac{1}{2}mv^2$.

Steps

  1. Definition: self induction is the induction of e.m.f. in a coil by the change of its own current: $e = -L\,dI/dt$.
  2. Inertia parallel: $L$ opposes the rise or fall of current, just as mass $m$ opposes the rise or fall of velocity.
  3. Examples: at switch-on $L$ holds current back; at switch-off $L$ tries to keep current going (spark across the switch).
  4. Energy parallel: $U = \tfrac{1}{2}LI^2$ matches $K = \tfrac{1}{2}mv^2$ — work done against the induced e.m.f. is stored in the field.

Answer

Self induction: e.m.f. induced by a coil's own changing current, $\boxed{\,e = -L\,dI/dt\,}$. Called electric inertia because $L$ resists $dI/dt$ the way $m$ resists $dv/dt$.

Exam tip

Write the pair side by side: $e = -L\,dI/dt$ and $F = m\,dv/dt$. The symmetry is the mark.

Q52022

Define coefficients of self and mutual inductance.

Recall

Both coefficients are flux per ampere. Unit: henry (H) = Wb/A = V·s/A.

Steps

  1. Self inductance: $L = \Phi/I$ for a coil's own flux linked with its own current. Also $e = -L\,dI/dt$.
  2. Mutual inductance: $M = \Phi_2/I_1$ — flux linked with coil 2 per ampere in coil 1. Also $e_2 = -M\,dI_1/dt$.
  3. Units and dependence: $[L] = [M] = 1$ H = 1 Wb/A. $L$ depends on turns, area, length and core; $M$ depends on the same plus coupling and distance between coils.
  4. Coupling factor $k = M/\sqrt{L_1 L_2}$ lies between 0 and 1 (perfect coupling $k=1$).

Answer

$\boxed{\,L = \dfrac{\Phi}{I}\,,\quad M = \dfrac{\Phi_2}{I_1}\,,\quad [L] = [M] = \text{henry (H)}\,}$.

Exam tip

Add the e.m.f. forms ($e = -L\,dI/dt$, $e_2 = -M\,dI_1/dt$) — a definition without these loses one mark.

Reference. Brij Lal & Subrahmanyam Ch-10; H.C. Verma Vol-1 Ch-38; S.L. Arora Vol-1 Ch-6.