Physics Lab

Unit 11: EMI, AC & EM Waves

EMI + AC + EM Waves carries 8–10 % of JEE-Main and is a stronghold of Advanced. The topic is unified by Faraday's law: a changing magnetic flux induces an EMF. From this single statement you derive motional EMF, self-induction, mutual induction, LC oscillations, AC circuits, transformer action, and (with Maxwell's correction) the entire theory of light.

Expect:

  • JEE-Main: One motional-EMF problem (rod, disc, rotating ring), one self-inductance / mutual inductance / LR transient, one AC LCR analysis (impedance, phase, power, resonance), one EM-waves conceptual or numerical.
  • JEE-Advanced: Combined problems — e.g. a rod sliding on rails forms an LR-EMF circuit; or a capacitor with changing field has a displacement current giving rise to a magnetic field, asked numerically.

The cleanest mental model:

  • ε=dΦ/dt\varepsilon = -d\Phi/dt (Faraday).
  • AC steady state is best handled with impedances ZL=jωL,ZC=1/(jωC),ZR=RZ_L = j\omega L, Z_C = 1/(j\omega C), Z_R = R, then Ohm's law in phasor form.
  • EM waves in vacuum: E,B\vec E, \vec B perpendicular to each other and to k\vec k; c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0}.

Concept Map

EMI
│
├── Flux Φ = ∫B·dA
├── Faraday:  ε = -dΦ/dt
├── Lenz: induced current opposes the change
├── Motional EMF:  ε = ∫(v×B)·dℓ
│     • rod sliding: ε = BvL
│     • rotating rod: ε = ½BωL²
│     • rotating disc: ε = ½BωR²
├── Self-inductance L:  ε_L = -L dI/dt
│     • Solenoid: L = μ₀n²V
├── Mutual inductance M:  ε₂ = -M dI₁/dt
│     • Reciprocity: M₁₂ = M₂₁
├── LR transient (growth/decay), τ = L/R
├── Energy in L: ½LI², density B²/(2μ₀)
└── LC oscillations:  ω = 1/√(LC)

AC CIRCUITS
│
├── v(t) = V₀ sin(ωt)
├── RMS:  V_rms = V₀/√2,  I_rms = I₀/√2
│
├── Pure elements
│     ├── R: i in phase with v
│     ├── L: i lags v by π/2
│     └── C: i leads v by π/2
│
├── Series LCR
│     Z = √(R² + (X_L − X_C)²),  X_L = ωL, X_C = 1/ωC
│     tan φ = (X_L − X_C)/R
│     Resonance: ω₀ = 1/√(LC), Z = R
│     Q = ω₀L/R = (1/R)√(L/C)
│
├── Power
│     P_avg = V_rms I_rms cos φ
│     Wattless: φ = π/2
│
└── Transformer
       V₂/V₁ = N₂/N₁ = I₁/I₂ (ideal)
       Efficiency, losses (Cu, hysteresis, eddy)

EM WAVES
│
├── Displacement current J_d = ε₀ dE/dt
├── Maxwell's equations (4)
├── c = 1/√(μ₀ε₀) ≈ 3×10⁸ m/s
├── E ⊥ B ⊥ k, E/B = c
├── Energy density u = ½ε₀E² + B²/(2μ₀); both equal
├── Intensity I = ½ε₀E₀²c = avg Poynting
├── Radiation pressure p = I/c (absorbed), 2I/c (reflected)
└── Spectrum: γ → X → UV → vis → IR → μw → radio

Topic 1: Electromagnetic Induction (Faraday & Lenz)

Sub-topic A: Magnetic Flux

For a flat coil in B\vec B:

Φ=NBAcosθ\Phi = NBA\cos\theta

For non-uniform BB or curved surface:

Φ=NBdA\Phi = N\int \vec B\cdot d\vec A

Unit: weber (Wb) = T·m² = V·s.

Sub-topic B: Faraday's Law

ε=dΦdt\boxed{\varepsilon = -\frac{d\Phi}{dt}}

The induced EMF in a closed loop equals the negative rate of change of magnetic flux through it. The sign embodies Lenz's law.

If the loop has resistance RR, induced current I=ε/RI = \varepsilon/R.

Sub-topic C: Lenz's Law

The induced current flows in such a direction that its own magnetic field opposes the change in flux that produced it. This is the conservation of energy in induction — if it were the other way, you would have run-away currents (perpetual motion).

Practical use: sketch the original flux direction; sketch how it's changing; the induced current opposes that change.

Sub-topic D: Motional EMF — Three Standard Cases

Case 1: Straight rod sliding on rails.

Rod of length LL moves with velocity vv perpendicular to its length, in field B\vec B perpendicular to the plane.

Force on a free charge in the rod: qv×Bq\vec v\times\vec B, which separates positive charge to one end (creates an electric field). At equilibrium qE=qvBqE = qvB, so the EMF (PD across the rod) is:

ε=vBL\varepsilon = vBL

Equivalently, by Faraday: Φ=BLx\Phi = BL x, dΦ/dt=BLv=εd\Phi/dt = BLv = \varepsilon.

Case 2: Rotating rod.

Rod of length LL rotates with angular velocity ω\omega about one end, in field BB perpendicular to the plane of rotation.

Element dξd\xi at distance ξ\xi from the pivot moves with speed ωξ\omega\xi. EMF in element: dε=B(ωξ)dξd\varepsilon = B(\omega\xi)d\xi. Integrate:

ε=12BωL2\boxed{\varepsilon = \tfrac{1}{2}B\omega L^2}

Case 3: Rotating disc (Faraday disc).

A conducting disc of radius RR rotates at ω\omega in field BB perpendicular to the disc. EMF between centre and rim is the integral of dε=B(ωr)drd\varepsilon = B(\omega r)dr from 00 to RR:

ε=12BωR2\varepsilon = \tfrac{1}{2}B\omega R^2

Sub-topic E: Induced Electric Field

A changing magnetic field induces a non-conservative electric field even in the absence of charges. From Faraday in integral form:

Ed=dΦBdt\oint \vec E\cdot d\vec\ell = -\frac{d\Phi_B}{dt}

Unlike the static electric field, this E\vec E has non-zero curl and its line integral around a closed path is non-zero.

For an axially symmetric region of uniform B(t)B(t) in a circle of radius RR:

E(r)={r2dB/dt(r<R)R22rdB/dt(r>R)E(r) = \begin{cases} -\tfrac{r}{2}dB/dt & (r < R) \\ -\tfrac{R^2}{2r}dB/dt & (r > R) \end{cases}

Worked Problem 1

A horizontal conducting rod of length 11 m falls freely under gravity in a horizontal magnetic field of 0.50.5 T, with the rod perpendicular to B\vec B. Find the EMF after 22 s.

Solution. v=gt=20v = gt = 20 m/s. ε=BvL=0.5201=10\varepsilon = BvL = 0.5\cdot 20\cdot 1 = 10 V.

Worked Problem 2 (JEE-Advanced)

A square loop of side aa and resistance RR moves with constant velocity vv into a region of uniform BB perpendicular to the loop's plane. The leading edge enters the field at t=0t = 0. Find: (a) induced current while entering, (b) force needed to keep it moving, (c) power dissipated.

Solution. (a) While entering, only the leading edge is in the field; ε=Bav\varepsilon = Bav, I=Bav/RI = Bav/R. (b) Force on the leading edge Fmag=BIa=B2a2v/RF_{\text{mag}} = BIa = B^2a^2v/R (opposing motion). External force needed: Fext=B2a2v/RF_{\text{ext}} = B^2a^2v/R. (c) Power dissipated: P=I2R=(Bav)2/RP = I^2 R = (Bav)^2/R. Equals FextvF_{\text{ext}}\cdot v — energy balance.

Once fully inside (or fully outside) Φ\Phi no longer changes, I=0I = 0, no force needed.


Topic 2: Inductance

Sub-topic A: Self-Inductance

A current II through a coil produces a flux Φ=LI\Phi = LI, where LL is the self-inductance (henry: H = Wb/A = V·s/A).

When II changes:

εL=LdIdt\varepsilon_L = -L\frac{dI}{dt}

This "back EMF" opposes the change in II.

Sub-topic B: Self-Inductance of a Long Solenoid

nn turns per unit length, area AA, length \ell. Flux per turn Φ1=μ0nIA\Phi_1 = \mu_0 n I\cdot A. Total flux linkage NΦ1=(n)μ0nIA=μ0n2AIN\Phi_1 = (n\ell)\mu_0 n I A = \mu_0 n^2 \ell A\cdot I. So

L=μ0n2A=μ0n2VL = \mu_0 n^2 \ell A = \mu_0 n^2 V

where VV is the volume.

For a solenoid filled with a material of relative permeability μr\mu_r: Lμ0μrn2VL \to \mu_0\mu_r n^2 V.

Sub-topic C: Self-Inductance of a Toroid

Toroid of NN turns, mean radius rr, cross-section AA:

L=μ0N2A2πrL = \frac{\mu_0 N^2 A}{2\pi r}

Sub-topic D: Mutual Inductance

If coil 1 has I1I_1 and produces flux Φ21\Phi_{21} through coil 2:

Φ21=M21I1\Phi_{21} = M_{21} I_1

MM is the mutual inductance; symmetric: M12=M21=MM_{12} = M_{21} = M (Reciprocity Theorem).

Two coaxial solenoids: small one (length \ell, area AA, n1n_1 turns/m) inside a larger one (n2n_2 turns/m).

When current I2I_2 flows in the outer solenoid, field inside is B=μ0n2I2B = \mu_0 n_2 I_2. Flux through the inner solenoid per turn: BA=μ0n2I2ABA = \mu_0 n_2 I_2 A. Total flux linkage with the inner solenoid: n1μ0n2I2An_1\ell \cdot\mu_0 n_2 I_2 A. So

M=μ0n1n2AM = \mu_0 n_1 n_2 \ell A

Also M=kL1L2M = k\sqrt{L_1 L_2} where 0k10 \le k \le 1 is the coupling coefficient. k=1k = 1 for perfect (no leakage) flux linkage.

Sub-topic E: LR Circuit — Growth and Decay

Growth (switch closed at t=0t=0, battery ε\varepsilon, resistor RR, inductor LL):

ε=IR+LdIdt\varepsilon = IR + L\frac{dI}{dt}

Solving:

I(t)=εR(1et/τ),τ=L/RI(t) = \frac{\varepsilon}{R}(1 - e^{-t/\tau}),\quad \tau = L/R

Decay (battery removed at t=0t=0, replaced by short):

I(t)=I0et/τI(t) = I_0 e^{-t/\tau}

Sub-topic F: Energy in an Inductor

Work done by EMF against back-EMF while current builds from 0 to II:

W=0IεIdt=0ILIdIdtdt=0ILIdI=12LI2W = \int_0^I \varepsilon \, I'\, dt = \int_0^I L I' \frac{dI'}{dt}dt = \int_0^I L I'\, dI' = \tfrac{1}{2}LI^2

Energy density in the magnetic field:

uB=B22μ0u_B = \frac{B^2}{2\mu_0}

This is the magnetic analog of uE=12ε0E2u_E = \tfrac{1}{2}\varepsilon_0 E^2.

Sub-topic G: LC Oscillations — Derivation

Charged capacitor CC connected to inductor LL. KVL:

qC+Ldidt=0,i=dq/dt\frac{q}{C} + L\frac{di}{dt} = 0,\quad i = dq/dt Lq¨+q/C=0q¨=1LCq\Rightarrow L\ddot q + q/C = 0 \Rightarrow \ddot q = -\frac{1}{LC}q

SHM in charge! Angular frequency:

ω=1LC\omega = \frac{1}{\sqrt{LC}}

Total energy oscillates between capacitor (q2/2Cq^2/2C) and inductor (12Li2\tfrac{1}{2}Li^2), with sum constant:

E=q22C+12Li2=q022C=12LI02E = \frac{q^2}{2C}+\tfrac{1}{2}Li^2 = \frac{q_0^2}{2C} = \tfrac{1}{2}LI_0^2

Worked Problem 3

A 55 mH inductor and a 20μ20\,\muF capacitor are connected. Find the frequency of LC oscillations.

Solution. ω=1/LC=1/5×10320×106=1/107=103.5=3162\omega = 1/\sqrt{LC} = 1/\sqrt{5\times 10^{-3}\cdot 20\times 10^{-6}} = 1/\sqrt{10^{-7}} = 10^{3.5} = 3162 rad/s. f=503f = 503 Hz.


Topic 3: Alternating Current — Basics

Sub-topic A: Sinusoidal Source

v(t)=V0sin(ωt)v(t) = V_0 \sin(\omega t)

Average over a cycle: zero (positive and negative halves cancel).

RMS (root mean square):

Vrms=v2=V0/20.707V0V_{\text{rms}} = \sqrt{\langle v^2\rangle} = V_0/\sqrt 2 \approx 0.707\, V_0

Same for Irms=I0/2I_{\text{rms}} = I_0/\sqrt 2.

Power: with resistive load P=VrmsIrms=V02/(2R)\langle P\rangle = V_{\text{rms}} I_{\text{rms}} = V_0^2/(2R).

Sub-topic B: Pure Elements with AC

Pure resistor:

i=V0/Rsinωti = V_0/R \sin\omega t

ii in phase with vv.

Pure inductor:

v=Ldi/dt=V0sinωti=V0/(ωL)cosωt=(V0/ωL)sin(ωtπ/2)v = L\, di/dt = V_0\sin\omega t \Rightarrow i = -V_0/(\omega L)\cos\omega t = (V_0/\omega L)\sin(\omega t - \pi/2)

Current lags voltage by π/2\pi/2. Inductive reactance XL=ωLX_L = \omega L.

Pure capacitor:

q=Cv=CV0sinωti=CV0ωcosωt=(V0/(1/ωC))sin(ωt+π/2)q = Cv = CV_0\sin\omega t \Rightarrow i = CV_0\omega\cos\omega t = (V_0/(1/\omega C))\sin(\omega t + \pi/2)

Current leads voltage by π/2\pi/2. Capacitive reactance XC=1/(ωC)X_C = 1/(\omega C).

Sub-topic C: Phasor Representation

Visualize AC quantities as rotating vectors (phasors) — projection on a chosen axis gives the instantaneous value. Adding two AC quantities of the same frequency reduces to phasor addition.

In phasor diagrams (with current as reference):

  • Resistor voltage: along I^\hat I.
  • Inductor voltage: 90° ahead of I^\hat I.
  • Capacitor voltage: 90° behind I^\hat I.

Topic 4: Series LCR Circuit

Sub-topic A: Impedance and Phase

Series LCR with sinusoidal source. KVL (phasor):

V=VR+VL+VCV = V_R + V_L + V_C

Magnitudes: VR=IRV_R = IR, VL=IXLV_L = IX_L, VC=IXCV_C = IX_C. The VLV_L and VCV_C are antiparallel; the net reactive voltage is VLVCV_L - V_C (if VL>VCV_L > V_C).

Magnitude of total voltage:

V=IR2+(XLXC)2=IZV = I\sqrt{R^2 + (X_L - X_C)^2} = IZ

where

Z=R2+(XLXC)2\boxed{Z = \sqrt{R^2 + (X_L - X_C)^2}}

Phase angle of voltage w.r.t. current:

tanϕ=XLXCR\tan\phi = \frac{X_L - X_C}{R}

If ϕ>0\phi > 0, voltage leads current (inductive); if ϕ<0\phi < 0, voltage lags (capacitive).

Sub-topic B: Resonance

Resonance at XL=XCω0=1/LCX_L = X_C \Rightarrow \omega_0 = 1/\sqrt{LC}, the same frequency as LC oscillations.

At resonance:

  • Z=RZ = R (minimum).
  • II is maximum: I0=V0/RI_0 = V_0/R.
  • ϕ=0\phi = 0: current in phase with applied voltage.
  • VLV_L and VCV_C individually can be much larger than the source voltage but cancel each other out.

Bandwidth Δω=R/L\Delta\omega = R/L. Quality factor:

Q=ω0Δω=ω0LR=1RLCQ = \frac{\omega_0}{\Delta\omega}=\frac{\omega_0 L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}

Higher QQ = sharper resonance.

Sub-topic C: Power and Power Factor

Instantaneous power p(t)=v(t)i(t)p(t) = v(t)\cdot i(t). Average over a cycle:

P=VrmsIrmscosϕ\langle P\rangle = V_{\text{rms}}I_{\text{rms}}\cos\phi

cosϕ\cos\phi is the power factor. Pure inductor or capacitor has ϕ=±π/2\phi = \pm\pi/2, cosϕ=0\cos\phi = 0: no average power dissipated — "wattless current".

In LCR: only RR dissipates power, and that's P=Irms2R\langle P\rangle = I_{\text{rms}}^2 R.

Worked Problem 4

A series LCR has R=100ΩR = 100\,\Omega, L=0.5L = 0.5 H, C=10μC = 10\,\muF. Source V0=100V_0 = 100 V, frequency 5050 Hz. Find XL,XC,Z,I0X_L, X_C, Z, I_0, phase, average power.

Solution. ω=100π314\omega = 100\pi \approx 314 rad/s.

XL=ωL=157ΩX_L = \omega L = 157\,\Omega.

XC=1/(ωC)=1/(314105)=318ΩX_C = 1/(\omega C) = 1/(314\cdot 10^{-5}) = 318\,\Omega.

XLXC=161ΩX_L - X_C = -161\,\Omega (capacitive).

Z=1002+1612=10000+25921189.5ΩZ = \sqrt{100^2 + 161^2} = \sqrt{10000+25921}\approx 189.5\,\Omega.

I0=100/189.50.528I_0 = 100/189.5 \approx 0.528 A.

tanϕ=161/100=1.61ϕ58.2\tan\phi = -161/100 = -1.61 \Rightarrow \phi \approx -58.2^\circ (voltage lags current).

P=VrmsIrmscosϕ=(100/2)(0.528/2)(0.527)13.9\langle P\rangle = V_{\text{rms}}I_{\text{rms}}\cos\phi = (100/\sqrt 2)(0.528/\sqrt 2)(0.527) \approx 13.9 W.

Worked Problem 5 (Resonance)

For the same LCR, find the resonance frequency.

Solution. ω0=1/LC=1/0.5105=1/5×106447\omega_0 = 1/\sqrt{LC} = 1/\sqrt{0.5\cdot 10^{-5}} = 1/\sqrt{5\times 10^{-6}}\approx 447 rad/s, f071.2f_0 \approx 71.2 Hz.


Topic 5: Transformers

Sub-topic A: Principle

A transformer has two coils wound on the same iron core. AC current in the primary creates a changing flux; the secondary picks up the flux through mutual induction. With N1,N2N_1, N_2 turns, for an ideal transformer (no losses):

V2V1=N2N1\frac{V_2}{V_1} = \frac{N_2}{N_1}

Power conservation (ideal): V1I1=V2I2V_1 I_1 = V_2 I_2

I1I2=N2N1\frac{I_1}{I_2} = \frac{N_2}{N_1}

Step-up: N2>N1N_2 > N_1V2>V1V_2 > V_1 (and I2<I1I_2 < I_1). Step-down: N2<N1N_2 < N_1V2<V1V_2 < V_1 (and I2>I1I_2 > I_1).

Sub-topic B: Losses

  1. Copper loss (I2RI^2R in the windings).
  2. Iron / hysteresis loss (energy in magnetisation cycle — minimised by soft iron).
  3. Eddy current loss (induced currents in the core — minimised by laminated cores).
  4. Flux leakage (not all flux from primary reaches secondary).

Real transformer efficiency η=Pout/Pin95\eta = P_{\text{out}}/P_{\text{in}}\approx 9599%99\%.

Transformers don't work on DC (no dΦ/dtd\Phi/dt).


Topic 6: Maxwell's Equations & Electromagnetic Waves

Sub-topic A: Displacement Current

Maxwell noticed: Ampère's law fails for a capacitor being charged. The "current" enclosed depends on which surface you take spanning a given loop! He fixed it by introducing the displacement current:

Id=ε0dΦEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt}

Modified Ampère–Maxwell law:

Bd=μ0(Ic+Id)=μ0Ic+μ0ε0dΦEdt\oint \vec B\cdot d\vec\ell = \mu_0 (I_c + I_d) = \mu_0 I_c + \mu_0\varepsilon_0\frac{d\Phi_E}{dt}

Sub-topic B: Maxwell's Equations (Integral Form)

#NameEquation
1Gauss (electric)EdA=Qenc/ε0\oint \vec E\cdot d\vec A = Q_{\text{enc}}/\varepsilon_0
2Gauss (magnetic)BdA=0\oint \vec B\cdot d\vec A = 0
3FaradayEd=dΦB/dt\oint \vec E\cdot d\vec\ell = -d\Phi_B/dt
4Ampère–MaxwellBd=μ0Ic+μ0ε0dΦE/dt\oint \vec B\cdot d\vec\ell = \mu_0 I_c + \mu_0\varepsilon_0\, d\Phi_E/dt

Equation #4's displacement current is what closes the loop and predicts EM waves.

Sub-topic C: EM Waves in Vacuum

Combining Maxwell's equations in free space gives:

2E=μ0ε02Et2,2B=μ0ε02Bt2\nabla^2 \vec E = \mu_0\varepsilon_0 \frac{\partial^2 \vec E}{\partial t^2}, \qquad \nabla^2 \vec B = \mu_0\varepsilon_0\frac{\partial^2\vec B}{\partial t^2}

These are wave equations with speed

c=1μ0ε0c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}

Plugging numbers: c=1/4π×1078.854×10123×108c = 1/\sqrt{4\pi\times 10^{-7}\cdot 8.854\times 10^{-12}} \approx 3\times 10^8 m/s — exactly the speed of light. Light is an EM wave.

Properties of EM waves in vacuum:

  1. Transverse: E,B\vec E, \vec B both perpendicular to direction of propagation k^\hat k.
  2. E\vec E and B\vec B are perpendicular to each other.
  3. In phase (peak together, zero together).
  4. E0=cB0E/B=cE_0 = c B_0 \Leftrightarrow E/B = c.
  5. Direction of propagation k^=E^×B^\hat k = \hat E\times\hat B.

Sub-topic D: Energy and Momentum of EM Waves

Energy densities (in vacuum, instantaneous):

uE=12ε0E2,uB=B2/(2μ0)u_E = \tfrac{1}{2}\varepsilon_0 E^2,\quad u_B = B^2/(2\mu_0)

Using E=cBE = cB and c2=1/(μ0ε0)c^2 = 1/(\mu_0\varepsilon_0):

uB=E2/(c2μ02)=ε0E2/2=uEu_B = E^2/(c^2\mu_0\cdot 2)= \varepsilon_0 E^2/2 = u_E

Equal! Total energy density:

u=uE+uB=ε0E2u = u_E + u_B = \varepsilon_0 E^2

Time-averaged for sinusoidal wave (E2=E02/2\langle E^2\rangle = E_0^2/2):

u=12ε0E02\langle u\rangle = \tfrac{1}{2}\varepsilon_0 E_0^2

Intensity II (avg energy crossing unit area per unit time):

I=uc=12ε0E02c\boxed{I = \langle u\rangle c = \tfrac{1}{2}\varepsilon_0 E_0^2 c}

Poynting vector: S=(1/μ0)E×B\vec S = (1/\mu_0)\vec E\times\vec B (instantaneous energy flux).

Radiation pressure on a perfectly absorbing surface:

Prad=I/cP_{\text{rad}} = I/c

For a perfectly reflecting surface (momentum reversal):

Prad=2I/cP_{\text{rad}} = 2I/c

Momentum density in field: u/c=ε0E2/cu/c = \varepsilon_0 E^2/c.

Sub-topic E: EM Spectrum

RegionWavelength rangeSource / Use
γ-rays<0.01< 0.01 nmNuclear transitions
X-rays0.010.011010 nmInner-shell electron transitions
Ultraviolet1010400400 nmSun, arcs; sterilisation
Visible400400700700 nmSun, lamps
Infrared700700 nm–11 mmThermal, remote control
Microwaves11 mm–11 mRadar, mobile, ovens
Radio>1> 1 mCommunication

All travel at cc in vacuum; ordering is by frequency (γ highest, radio lowest).

Worked Problem 6

A 100 W lamp radiates uniformly in all directions. At distance 11 m, find E0E_0, B0B_0, intensity, and radiation pressure on a perfect absorber.

Solution. I=P/(4πr2)=100/(4π)7.96I = P/(4\pi r^2) = 100/(4\pi) \approx 7.96 W/m².

E0=2I/(ε0c)=27.96/(8.85×10123×108)=599477.4E_0 = \sqrt{2I/(\varepsilon_0 c)} = \sqrt{2\cdot 7.96/(8.85\times 10^{-12}\cdot 3\times 10^8)} = \sqrt{5994}\approx 77.4 V/m.

B0=E0/c=2.58×107B_0 = E_0/c = 2.58\times 10^{-7} T.

Radiation pressure P=I/c=7.96/(3×108)=2.65×108P = I/c = 7.96/(3\times 10^8) = 2.65\times 10^{-8} Pa. Tiny — but used by solar sails.


Problem-Solving Heuristics

  1. For motional EMF, use ε=(v×B)d\varepsilon = \int(\vec v\times\vec B)\cdot d\vec\ell rather than trying to find dΦ/dtd\Phi/dt for rotating systems.
  2. For an LR transient, τ=L/R\tau = L/R; for RC, τ=RC\tau = RC. After 5τ5\tau, you're at steady state.
  3. At steady state in DC, inductors act as wires (VL=LdI/dt=0V_L = L\,dI/dt = 0) and capacitors as open circuits (IC=0I_C = 0).
  4. At t=0+t = 0^+ (just after a switch), inductors prevent sudden change in current (act as open), capacitors prevent sudden change in voltage (act as short).
  5. Phasors: for an LCR problem, always draw the phasor diagram with I\vec I as reference, then VRV_R along it, VLV_L up, VCV_C down, and total VV as the resultant.
  6. At resonance XL=XCX_L = X_C, Z=RZ = R (smallest possible), II maximum.
  7. Power factor: cosϕ=R/Z\cos\phi = R/Z. Wattless if ϕ=±π/2\phi = \pm\pi/2 (pure LL or CC).
  8. For transformer problems, conserve power if ideal; for non-ideal, η=V2I2cosϕ2/(V1I1cosϕ1)\eta = V_2I_2\cos\phi_2/(V_1I_1\cos\phi_1).
  9. In EM waves, kE×B\vec k\propto \vec E\times\vec B (right-hand rule). E0=cB0E_0 = cB_0.
  10. Energy density u=ε0E2u = \varepsilon_0 E^2 total in vacuum (electric and magnetic equal).
  11. For reflected light, radiation pressure is twice the absorber case (2I/c2I/c).
  12. Identify the loop / surface for Faraday consistently — choose normals; sign conventions matter.

Common Traps & Mistakes

  • Lenz's law sign. The induced EMF opposes the change in flux, not the flux itself.
  • Self-inductance is always positive, but the EMF is opposite to dI/dtdI/dt.
  • In an LCR at resonance, VLV_L and VCV_C can each exceed source VV, but they cancel — so VV across L+CL+C = 0 only when summed as phasors.
  • Reactance XLX_L and XCX_C depend on ω\omega — don't use static values.
  • Average power across a pure LL or CC is zero, not just for one quarter cycle.
  • Transformers don't increase or decrease power (ideal), only redistribute V and I.
  • Maxwell's correction (displacement current) isn't real charge motion — it's ε0dE/dt\varepsilon_0\,dE/dt.
  • EM waves carry momentum: p=U/cp = U/c. Most students forget this.
  • For EM waves in a medium, v=c/nv = c/n, λn=λ0/n\lambda_n = \lambda_0/n, but frequency is unchanged.
  • A current in DC doesn't see inductance — only when switching. At steady DC, inductors are wires.
  • Energy in an inductor is 12LI2\tfrac{1}{2}LI^2, not LILI.

Quick Revision Card

  • Flux: Φ=NBAcosθ\Phi = NBA\cos\theta.
  • Faraday: ε=dΦ/dt\varepsilon = -d\Phi/dt; Lenz says induced opposes the change.
  • Motional EMF: rod ε=BvL\varepsilon = BvL; rotating rod 12BωL2\tfrac{1}{2}B\omega L^2; disc 12BωR2\tfrac{1}{2}B\omega R^2.
  • Self-inductance: L=μ0n2VL = \mu_0 n^2 V (solenoid).
  • Mutual inductance: M=μ0n1n2AM = \mu_0 n_1 n_2\ell A, M=kL1L2M = k\sqrt{L_1L_2}.
  • LR: I=(ε/R)(1et/τ)I = (\varepsilon/R)(1 - e^{-t/\tau}), τ=L/R\tau = L/R. Energy 12LI2\tfrac{1}{2}LI^2, density B2/(2μ0)B^2/(2\mu_0).
  • LC: ω=1/LC\omega = 1/\sqrt{LC}.
  • AC: Vrms=V0/2V_{rms} = V_0/\sqrt 2.
  • Pure LL: ii lags vv by π/2\pi/2, XL=ωLX_L = \omega L.
  • Pure CC: ii leads vv by π/2\pi/2, XC=1/(ωC)X_C = 1/(\omega C).
  • Series LCR: Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2}, tanϕ=(XLXC)/R\tan\phi = (X_L-X_C)/R.
  • Resonance: ω0=1/LC\omega_0 = 1/\sqrt{LC}, Z=RZ = R, Q=(1/R)L/CQ = (1/R)\sqrt{L/C}.
  • Power: P=VrmsIrmscosϕ\langle P\rangle = V_{rms}I_{rms}\cos\phi.
  • Transformer: V2/V1=N2/N1=I1/I2V_2/V_1 = N_2/N_1 = I_1/I_2.
  • Displacement current: Id=ε0dΦE/dtI_d = \varepsilon_0\,d\Phi_E/dt.
  • EM wave: c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0}; E=cBE = cB; u=ε0E2u = \varepsilon_0 E^2; I=12ε0E02cI = \tfrac{1}{2}\varepsilon_0 E_0^2 c.
  • Radiation pressure: I/cI/c (absorbed), 2I/c2I/c (reflected).

Formula Sheet

ConceptFormula
Magnetic fluxΦ=NBAcosθ\Phi = NBA\cos\theta
Faraday's lawε=dΦ/dt\varepsilon = -d\Phi/dt
Motional EMF (rod)ε=BvL\varepsilon = BvL
Rotating rod EMFε=12BωL2\varepsilon = \tfrac{1}{2}B\omega L^2
Rotating disc EMFε=12BωR2\varepsilon = \tfrac{1}{2}B\omega R^2
Self-inductance (solenoid)L=μ0n2AL = \mu_0 n^2 \ell A
ToroidL=μ0N2A/(2πr)L = \mu_0 N^2 A/(2\pi r)
Mutual inductanceM=μ0n1n2AM = \mu_0 n_1 n_2\ell A, M=kL1L2M = k\sqrt{L_1 L_2}
LR growthI=(ε/R)(1et/(L/R))I = (\varepsilon/R)(1 - e^{-t/(L/R)})
Inductor energyUL=12LI2U_L = \tfrac{1}{2}LI^2
Magnetic energy densityuB=B2/(2μ0)u_B = B^2/(2\mu_0)
LC frequencyω=1/LC\omega = 1/\sqrt{LC}
RMSVrms=V0/2V_{rms} = V_0/\sqrt 2
Resistor (AC)ii in phase with vv
Inductor (AC)XL=ωLX_L = \omega L, ii lags vv by π/2
Capacitor (AC)XC=1/(ωC)X_C = 1/(\omega C), ii leads vv by π/2
LCR impedanceZ=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2}
LCR phasetanϕ=(XLXC)/R\tan\phi = (X_L - X_C)/R
Resonanceω0=1/LC\omega_0 = 1/\sqrt{LC}, Z=RZ = R
Quality factorQ=ω0L/R=(1/R)L/CQ = \omega_0 L/R = (1/R)\sqrt{L/C}
Average powerP=VrmsIrmscosϕP = V_{rms}I_{rms}\cos\phi
Transformer (ideal)V2/V1=N2/N1=I1/I2V_2/V_1 = N_2/N_1 = I_1/I_2
Displacement currentId=ε0dΦE/dtI_d = \varepsilon_0\,d\Phi_E/dt
Speed of lightc=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0}
E–B relationE=cBE = cB
EM energy densityu=ε0E2u = \varepsilon_0 E^2
EM intensityI=12ε0E02cI = \tfrac{1}{2}\varepsilon_0 E_0^2 c
Radiation pressureP=I/cP = I/c (absorbed), 2I/c2I/c (reflected)

Sub-topics

6 pages

Practice quiz

Quiz
Unit 11: EMI, AC & EM Waves — JEE Quiz
15 questions · pick the best answer
Q1

A rod of length 1 m rotates at 100 rad/s about one end in a perpendicular magnetic field of 0.5 T. The EMF developed between centre and the free end is:

Q2

A square loop of side 10 cm and resistance 1 Ω enters a region of B = 0.5 T (perpendicular to loop) at velocity 2 m/s. The current induced while entering:

Q3

A solenoid of length 0.5 m, area 5×10⁻⁴ m², 1000 turns has self-inductance:

Q4

An LR circuit has L = 1 H, R = 10 Ω, battery EMF = 12 V. Current after 0.1 s (starting from zero):

Q5

A capacitor 20 μF and an inductor 5 mH form an LC oscillator. The frequency of oscillation:

Q6

In a series LCR circuit R = 100 Ω, L = 0.5 H, C = 10 μF, source 100 V at 50 Hz. The power factor is approximately:

Q7

In an LCR series circuit at resonance:

Q8

A transformer has 200 primary turns and 1000 secondary turns. Primary is connected to 110 V AC. Secondary voltage and step-up ratio:

Q9

An EM wave has E₀ = 60 V/m. The corresponding B₀ is:

Q10

A 100 W lamp radiates isotropically. The amplitude of E at 5 m distance:

Q11

Radiation pressure exerted by a 1 kW laser beam of cross-section 1 mm² on a perfect reflector (in Pa):

Q12

The displacement current density in a region where the electric field is changing at dE/dt = 10⁶ V/(m·s) is:

Q13

Two coils of self-inductances L₁ = 16 mH and L₂ = 9 mH are placed close together with k = 0.5. Their mutual inductance is:

Q14

Current in an LR circuit (steady-state) is I₀. The battery is suddenly shorted while keeping L and R in the closed loop. Heat dissipated until current falls to zero:

Q15

An EM wave is travelling along +ẑ in vacuum. At some instant E is along +x̂, E₀ = 6 V/m. The direction and magnitude of B at that instant: