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CalibratePhysics

Chapter 12

Electromagnetic Induction

Maharashtra HSC · 13 subtopics · textbook pages 265–287

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants, units and conventions

  • Permeability of free space μ0=4π×10−7 T m A−1=4π×10−7 H m−1\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1} = 4\pi\times10^{-7}\ \text{H m}^{-1}. The core of every coil in this chapter is air unless the question says otherwise.
  • Magnetic flux ϕ\phi is in weber (Wb), 1 Wb=1 T m21\ \text{Wb} = 1\ \text{T m}^2. Induced emf ee is in volt, and dϕdt\dfrac{d\phi}{dt} is in weber per second.
  • Inductance LL and MM are in henry (H): 1 H=1 WbA=1 voltA/s=1 ohm⋅second=1 T m2A1\ \text{H} = 1\ \dfrac{\text{Wb}}{\text{A}} = 1\ \dfrac{\text{volt}}{\text{A/s}} = 1\ \text{ohm}\cdot\text{second} = 1\ \dfrac{\text{T m}^2}{\text{A}}.
  • Angular frequency ω=2πf=2πT\omega = 2\pi f = \dfrac{2\pi}{T} in rad s−1\text{rad s}^{-1}. A speed given in r.p.m. must be divided by 60 to get ff in revolutions per second (Hz).
  • Flux linkage of a coil of NN turns is NϕN\phi, also in weber. For a closely wound coil every turn links the same flux.
  • Every formula below is written for magnitudes; the minus sign in Faraday's law is Lenz's law and fixes only the direction.

Faraday's laws of electromagnetic induction

FormulaSymbols, units and conditions
e=−dϕdte = -\dfrac{d\phi}{dt}
  • Induced emf in a single-turn circuit; ϕ\phi is the magnetic flux linked with the circuit at time tt. The negative sign is Lenz's law.
e=−Ndϕdt=−d(Nϕ)dte = -N\dfrac{d\phi}{dt} = -\dfrac{d(N\phi)}{dt}
  • Coil of NN turns, each turn linking the same flux ϕ\phi. This is the "flux rule": emf equals the rate at which flux through the circuit changes.
∣e∣=NΔϕΔt=N(ϕfinal−ϕinitial)Δt=NABfinal−BinitialΔt\lvert e\rvert = N\dfrac{\Delta\phi}{\Delta t} = \dfrac{N(\phi_{\text{final}}-\phi_{\text{initial}})}{\Delta t} = NA\dfrac{B_{\text{final}}-B_{\text{initial}}}{\Delta t}
  • Average induced emf when the field changes uniformly (linearly) in a time Δt\Delta t through a fixed area AA held normal to B⃗\vec{B}. If the field is merely reversed, ΔB=2B\Delta B = 2B, not BB.
I=∣e∣R,q=I Δt=N ΔϕRI = \dfrac{\lvert e\rvert}{R},\qquad q = I\,\Delta t = \dfrac{N\,\Delta\phi}{R}
  • Induced current in a closed circuit of total resistance RR, and the charge that flows in time Δt\Delta t.

Flux of the field

FormulaSymbols, units and conditions
dϕ=B⃗⋅da⃗=B dacos⁡θd\phi = \vec{B}\cdot\vec{da} = B\,da\cos\theta
  • Flux of B⃗\vec{B} through a small element of area da⃗\vec{da}; θ\theta is the angle between B⃗\vec{B} and the direction assigned to the area (the outward normal, fixed by the right-handed screw rule).
ϕ=∫SB⃗⋅da⃗\phi = \int_S \vec{B}\cdot\vec{da}
  • Flux through a finite surface SS. B⃗\vec{B} can be taken out of the integral only if it is the same everywhere on SS, in which case ϕ=BAcos⁡θ\phi = BA\cos\theta.
ϕ=BAcos⁡θ,ϕmax⁡=BA when θ=0∘,ϕ=0 when θ=90∘\phi = BA\cos\theta,\qquad \phi_{\max} = BA \text{ when } \theta = 0^\circ,\qquad \phi = 0 \text{ when } \theta = 90^\circ
  • Uniform field through a flat area AA. "Plane of the coil perpendicular to B⃗\vec{B}" means the area vector is parallel to B⃗\vec{B}, so θ=0\theta = 0 and ϕ=BA\phi = BA.
e=−ddt∫SB⃗⋅da⃗e = -\dfrac{d}{dt}\int_S \vec{B}\cdot\vec{da}
  • General statement. Flux can change because BB changes, because the area SS changes, or because the orientation changes — or any combination.
∮E⃗⋅dl⃗=−dϕmdt⇒E=−s2dBdt\oint \vec{E}\cdot\vec{dl} = -\dfrac{d\phi_m}{dt}\qquad\Rightarrow\qquad E = -\dfrac{s}{2}\dfrac{dB}{dt}
  • Induced electric field at the rim of a circular region of radius ss filled by a uniform field B(t)B(t). A changing magnetic field generates an electric field even where there is no conductor.

Motional electromotive force

FormulaSymbols, units and conditions
∣e∣=dϕdt=ddt(Blx)=Bldxdt=Blv\lvert e\rvert = \dfrac{d\phi}{dt} = \dfrac{d}{dt}(Blx) = Bl\dfrac{dx}{dt} = Blv
  • Straight conductor of length ll sliding with velocity vv on a C-shaped frame, with B⃗\vec{B}, l⃗\vec{l} and v⃗\vec{v} mutually perpendicular.
e=vBlsin⁡θ,emax⁡=Blv (sin⁡θ=1)e = vBl\sin\theta,\qquad e_{\max} = Blv\ (\sin\theta = 1)
  • Same rod when v⃗\vec{v} makes an angle θ\theta with B⃗\vec{B}. If v⃗\vec{v} is parallel to B⃗\vec{B}, then sin⁡θ=0\sin\theta = 0 and the emf is zero.
de=Bv dr=Bωr dr⇒e=∫0lBωr dr=12Bωl2=Bπfl2de = Bv\,dr = B\omega r\,dr\qquad\Rightarrow\qquad e = \int_0^l B\omega r\,dr = \dfrac{1}{2}B\omega l^2 = B\pi f l^2
  • Rod of length ll rotating in a plane perpendicular to B⃗\vec{B} about a pivot at one end, at angular velocity ω=2πf\omega = 2\pi f. Emf appears between the pivot and the free end.
e=12BωR2e = \dfrac{1}{2}B\omega R^2
  • Faraday disc dynamo: metal disc of radius RR spinning at ω\omega about its axis in a field BB parallel to the axis; emf between the rim and the axis.
Area swept per second=πR2f,flux cut per second=BπR2f=e\text{Area swept per second} = \pi R^2 f,\qquad \text{flux cut per second} = B\pi R^2 f = e
  • The same disc result read as "flux cut per second". Numerically the flux cut per second in Wb and the emf in volt are equal.

Induced emf in a stationary coil in a changing magnetic field

FormulaSymbols, units and conditions
∣e∣=Nd(BA)dt=NAdBdt\lvert e\rvert = N\dfrac{d(BA)}{dt} = NA\dfrac{dB}{dt}
  • Stationary coil of NN turns and fixed area AA in a field changing with time.
e=−dϕdt=−dϕdθ⋅dθdte = -\dfrac{d\phi}{dt} = -\dfrac{d\phi}{d\theta}\cdot\dfrac{d\theta}{dt}
  • Magnet-coil (oscillating magnet) system. The first factor is fixed by the geometry of magnet and coil; the second comes from the oscillation.
θ=θ0sin⁡2πtT⇒dθdt=2πθ0Tcos⁡2πtT\theta = \theta_0\sin\dfrac{2\pi t}{T}\qquad\Rightarrow\qquad \dfrac{d\theta}{dt} = \dfrac{2\pi\theta_0}{T}\cos\dfrac{2\pi t}{T}
  • Angular displacement of the oscillating magnet, amplitude θ0\theta_0, period TT.
e0=(dϕdθ)max⁡(dθdt)max⁡=(dϕdθ)max⁡2πθ0Te_0 = \left(\dfrac{d\phi}{d\theta}\right)_{\max}\left(\dfrac{d\theta}{dt}\right)_{\max} = \left(\dfrac{d\phi}{d\theta}\right)_{\max}\dfrac{2\pi\theta_0}{T}
  • Peak induced emf of the magnet-coil system: e0∝θ0e_0 \propto \theta_0 and e0∝1Te_0 \propto \dfrac{1}{T}.

Generators

FormulaSymbols, units and conditions
ϕB=B⃗⋅A⃗=BAcos⁡ωt\phi_B = \vec{B}\cdot\vec{A} = BA\cos\omega t
  • Flux through a coil of area AA rotating with constant angular velocity ω\omega in a uniform field, with θ=ωt\theta = \omega t measured from the position where the area vector is along B⃗\vec{B}.
e=−NdϕBdt=NBAωsin⁡ωt=e0sin⁡ωte = -N\dfrac{d\phi_B}{dt} = NBA\omega\sin\omega t = e_0\sin\omega t
  • Sinusoidal (alternating) emf of an AC generator with NN turns.
e0=NBAω=2πfNBAe_0 = NBA\omega = 2\pi f NBA
  • Peak value of the generated emf. ff is the frequency of revolution of the coil in Hz. Domestic AC in India has f=50 Hzf = 50\ \text{Hz}.
e0=π2BR2fe_0 = \pi^2 BR^2 f
  • Amplitude for a stiff semicircular wire of radius RR rotated at frequency ff about the axis through its two ends: put A=πR22A = \dfrac{\pi R^2}{2} and ω=2πf\omega = 2\pi f in e0=BAωe_0 = BA\omega.

Induction and energy transfer

FormulaSymbols, units and conditions
ϕB=BLx,∣e∣=BLv,i=∣e∣R=BLvR\phi_B = BLx,\qquad \lvert e\rvert = BLv,\qquad i = \dfrac{\lvert e\rvert}{R} = \dfrac{BLv}{R}
  • Rectangular loop of width LL and total resistance RR being pulled out of a field region at constant speed vv; xx is the length still inside the field.
∣F∣=iLB=B2L2vR\lvert F\rvert = iLB = \dfrac{B^2L^2v}{R}
  • External force needed to keep the speed constant; it balances the retarding magnetic force on the segment still in the field.
P=F⃗⋅v⃗=B2L2v2R=i2RP = \vec{F}\cdot\vec{v} = \dfrac{B^2L^2v^2}{R} = i^2R
  • Mechanical power supplied equals the rate of Joule heating in the loop: all the work done in pulling the loop appears as heat.

Self inductance

FormulaSymbols, units and conditions
ϕ∝i,ϕ=Li,L=ϕi,L=NϕBi\phi \propto i,\qquad \phi = Li,\qquad L = \dfrac{\phi}{i},\qquad L = \dfrac{N\phi_B}{i}
  • Self-inductance as flux (or flux linkage, for NN turns) per unit current. LL depends only on the geometry of the coil and the material of its core, never on ii or on ee.
e=−ddt(Li)=−Ldidt,L=∣e∣∣di/dt∣e = -\dfrac{d}{dt}(Li) = -L\dfrac{di}{dt},\qquad L = \dfrac{\lvert e\rvert}{\lvert di/dt\rvert}
  • Self-induced (back) emf, and self-inductance as emf per unit rate of change of current. Unit: voltA/s=henry\dfrac{\text{volt}}{\text{A/s}} = \text{henry}.
B=μ0niB = \mu_0 n i
  • Field inside a long solenoid with nn turns per unit length, needed for every inductance calculation below.
L=μ0n2Al=μ0N2Al,Ll=μ0n2A=μ0n2πd24L = \mu_0 n^2 A l = \dfrac{\mu_0 N^2 A}{l},\qquad \dfrac{L}{l} = \mu_0 n^2 A = \dfrac{\mu_0 n^2 \pi d^2}{4}
  • Long solenoid of length ll, cross-sectional area AA, diameter dd, N=nlN = nl turns. So L∝n2L \propto n^2 and L∝d2L \propto d^2.
L=μ0N2r22R(r≪R)L = \dfrac{\mu_0 N^2 r^2}{2R}\qquad (r \ll R)
  • Toroid of NN turns, circular cross-section of radius rr, major radius RR.
L=μ0l2πln⁡(ba)L = \dfrac{\mu_0 l}{2\pi}\ln\left(\dfrac{b}{a}\right)
  • Coaxial cable of length ll; current goes down the inner cylinder of radius aa and returns along the outer cylinder of radius bb.
L=μ0lπln⁡(da)L = \dfrac{\mu_0 l}{\pi}\ln\left(\dfrac{d}{a}\right)
  • Pair of parallel wires of length ll and radius aa in the same circuit, axes a distance dd apart.
Ltotal=L1+L2+L3+⋯(series)L_{\text{total}} = L_1 + L_2 + L_3 + \cdots \qquad\text{(series)}—
1Ltotal=1L1+1L2+1L3+⋯(parallel)\dfrac{1}{L_{\text{total}}} = \dfrac{1}{L_1}+\dfrac{1}{L_2}+\dfrac{1}{L_3}+\cdots \qquad\text{(parallel)}
  • Combination of inductances, exactly like resistances. The parallel value is always smaller than the smallest coil.

Energy stored in a magnetic field

FormulaSymbols, units and conditions
dW=−e dq=Ldidt dq=L i didW = -e\,dq = L\dfrac{di}{dt}\,dq = L\,i\,di—
W=∫0iL i di=12Li2=UBW = \int_0^{i} L\,i\,di = \dfrac{1}{2}Li^2 = U_B
  • Work done against the back emf in building the current up to ii; it is stored as magnetic energy UBU_B in joule. Analogous to UE=12CV2U_E = \dfrac{1}{2}CV^2 for a capacitor and to KE=12mv2KE = \dfrac{1}{2}mv^2.
L=2W(numerically, when i0=1 A)L = 2W \quad\text{(numerically, when } i_0 = 1\ \text{A)}
  • Self-inductance is numerically equal to twice the work done in establishing the flux for unit current.

Energy density of a magnetic field

FormulaSymbols, units and conditions
uB=UBAl=12(Ll)i2A=12μ0n2i2u_B = \dfrac{U_B}{Al} = \dfrac{1}{2}\left(\dfrac{L}{l}\right)\dfrac{i^2}{A} = \dfrac{1}{2}\mu_0 n^2 i^2
  • Energy per unit volume inside a long solenoid of interior volume AlAl.
uB=B22μ0u_B = \dfrac{B^2}{2\mu_0}
  • Energy density at any point where the field is BB, in J m−3\text{J m}^{-3}. It holds for every magnetic field, however produced. Compare the electric case uE=12ε0E2u_E = \dfrac{1}{2}\varepsilon_0 E^2.

Mutual inductance

FormulaSymbols, units and conditions
ϕ21=∫S2B1⃗⋅da⃗,ϕ21=M21I1,ϕ12=M12I2,M12=M21=M\phi_{21} = \int_{S_2}\vec{B_1}\cdot\vec{da},\qquad \phi_{21} = M_{21}I_1,\qquad \phi_{12} = M_{12}I_2,\qquad M_{12} = M_{21} = M
  • Flux linked with one coil per unit current in the other. The two coefficients are always equal, by symmetry.
e21=−M21dI1dt,e12=−M12dI2dte_{21} = -M_{21}\dfrac{dI_1}{dt},\qquad e_{12} = -M_{12}\dfrac{dI_2}{dt}—
M=ϕsIp=∣es∣∣dIp/dt∣M = \dfrac{\phi_s}{I_p} = \dfrac{\lvert e_s\rvert}{\lvert dI_p/dt\rvert}
  • Mutual inductance as flux in the secondary per unit primary current, or as secondary emf per unit rate of change of primary current. M=1 HM = 1\ \text{H} when 1 A s−11\ \text{A s}^{-1} in the primary induces 1 V1\ \text{V} in the secondary.
M=KL1L2,K=ML1L2M = K\sqrt{L_1L_2},\qquad K = \dfrac{M}{\sqrt{L_1L_2}}
  • KK is the coefficient of coupling, K≤1K \le 1. K=1K = 1 is perfect coupling (M=L1L2M = \sqrt{L_1L_2}); K>0.5K > 0.5 tight, K<0.5K < 0.5 loose. Expressed as a percentage it is the percentage of flux from one coil that reaches the other.
M=μ0N1N2AlM = \dfrac{\mu_0 N_1 N_2 A}{l}
  • Long solenoid of length ll, area AA, N1N_1 turns, with a small coil of N2N_2 turns wound around its centre. The same form, M=μ0NBNHAlM = \dfrac{\mu_0 N_B N_H A}{l}, gives the mutual inductance of a wireless charger's base and handle coils.
W=MI1I2,M=W when I1=I2=1 AW = MI_1I_2,\qquad M = W \text{ when } I_1 = I_2 = 1\ \text{A}
  • Mutual potential energy of two circuits.

Transformer

FormulaSymbols, units and conditions
ep=−Npdϕdt,es=−Nsdϕdte_p = -N_p\dfrac{d\phi}{dt},\qquad e_s = -N_s\dfrac{d\phi}{dt}
  • ϕ\phi is the flux linked per turn, the same for both coils in an ideal transformer with a common soft-iron core.
esep=NsNp=K\dfrac{e_s}{e_p} = \dfrac{N_s}{N_p} = K
  • Equation of the transformer. K=NsNpK = \dfrac{N_s}{N_p} is the turns ratio (transformer ratio): K>1K>1 step-up, K<1K<1 step-down.
epip=esis⇒esep=ipise_p i_p = e_s i_s \qquad\Rightarrow\qquad \dfrac{e_s}{e_p} = \dfrac{i_p}{i_s}
  • Ideal (100% efficient) transformer: output power equals input power.
esep=NsNp=ipis\dfrac{e_s}{e_p} = \dfrac{N_s}{N_p} = \dfrac{i_p}{i_s}
  • The complete transformer relation. Note that the current ratio is inverted: stepping the voltage up steps the current down.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.13 subtopics
  1. 1Introduction
  2. 2Faraday's Laws of Electromagnetic Induction2 board questions
  3. 3Lenz's Law2 board questions
  4. 4Flux of the Field
  5. 5Motional Electromotive Force5 board questions
  6. 6Induced emf in a Stationary Coil in a Changing Magnetic Field3 board questions
  7. 7Generators1 board question
  8. 8Back emf and back torque
  9. 9Eddy Currents1 board question
  10. 10Self Inductance1 board question
  11. 11Energy Stored in a Magnetic Field1 board question
  12. 12Mutual Inductance (M)5 board questions
  13. 13Transformer6 board questions
The whole chapter in brief

12.1 Introduction

  • Electromagnetic induction is the phenomenon of producing an induced emf in a conductor or conducting coil because of a changing magnetic flux. Discovered by Faraday (1831); Joseph Henry demonstrated the same effect independently in the same year.
  • Faraday's experiments with a bar magnet and a coil connected to a galvanometer (G):
  • A magnet approaching the closed coil produces a current in it (the induced current); taking the magnet away produces a current in the opposite direction.
  • Moving the coil instead of the magnet gives the same effect — only relative motion matters.
  • Reversing the polarity of the moving magnet reverses the induced current.
  • The magnitude of induced current depends on the relative speed of coil and magnet and on the number of turns of the coil.
  • The induced current exists only so long as there is relative motion.
  • Two-coil experiment: a primary coil with battery, rheostat and key faces a secondary coil connected to a galvanometer.
  • A momentary deflection appears at make and at break of the primary circuit, in opposite directions.
  • Relative motion between the two closed coils induces a current in the secondary as long as the motion lasts.
  • Changing the primary current with the rheostat also deflects the galvanometer, for as long as the current is changing.
  • Conclusion: whenever the flux through a coil changes — by a magnet or by a current-carrying coil — an emf is induced, and its direction reverses when the flux decreases instead of increasing.

12.2 Faraday's laws of electromagnetic induction

  • First law (qualitative): whenever there is a change of magnetic flux in a closed circuit, an induced emf is produced in the circuit. Also, if a conductor cuts the lines of magnetic field, an emf is induced between its ends.
  • Second law (quantitative): the magnitude of the induced emf produced in the circuit is directly proportional to the rate of change of magnetic flux linked with the circuit.
  • Mathematically e∝dϕdte \propto \dfrac{d\phi}{dt}, so e=Kdϕdte = K\dfrac{d\phi}{dt}; with all quantities in the same system of units K=1K=1, giving e=dϕdte = \dfrac{d\phi}{dt} and, with Lenz's law, e=−dϕdte = -\dfrac{d\phi}{dt}.
  • For nn turns each linking flux ϕ′\phi', the total flux is ϕ=nϕ′\phi = n\phi' and e=−ndϕ′dte = -n\dfrac{d\phi'}{dt}. This is the flux rule.
  • SI: ee in volt, dϕdt\dfrac{d\phi}{dt} in weber per second.
  • A moving magnet (a time-varying magnetic field) exerts a force on the stationary charges inside a conductor — this is the converse of a current exerting a force on a magnet.

12.3 Lenz's law

  • Statement: the direction of the induced current in a circuit is such that the magnetic field produced by the induced current opposes the change in the magnetic flux that induces the current. The direction of the induced emf is the same as that of the induced current.
  • Short form: the induced emf tends to set up a current whose action opposes the change that causes it. "Every effect of induction acts in opposition to the cause that produces it."
  • Magnet approaching a loop: the loop develops a north pole facing the approaching north pole so as to repel it; the induced current is anticlockwise as seen from the magnet side. When the magnet is pulled away, the loop develops a south pole to attract the retreating north pole and the current reverses to clockwise.
  • Jumping-ring experiment: a metal ring sits on the iron core of a vertical coil; when the current is switched on, the sudden upward flux induces a current in the ring opposite in direction to the coil current, opposite currents repel, and the ring flies into the air.
  • Energy conservation: if the induced current helped the change that produced it, the current and the flux would grow without limit, giving Joule heating for free — a perpetual motion machine. Lenz's law is therefore a necessary consequence of the law of conservation of energy.
  • Two cases of opposition — (i) a conductor moving in a field: the thrust on it from the field is opposite to its motion; (ii) a stationary circuit in a changing field: the induced current sets up a field opposite to the original field if that field is increasing, and in the same direction as it if it is decreasing.
  • It is the change in flux, not the flux itself, that is opposed.
  • For an open conductor, mentally complete the circuit, apply Lenz's law, and the polarity of the free ends follows.
  • The negative sign in e=−dϕdte = -\dfrac{d\phi}{dt} is exactly Lenz's law written into Faraday's law: it says the emf opposes the change of flux linked with the coil.

12.4 Flux of the field

  • Give a small element of area a direction: if the bounding curve is traversed in the sense of the arrow, the normal comes out towards the reader — the right-handed screw rule.
  • dϕ=B⃗⋅da⃗=B dacos⁡θd\phi = \vec{B}\cdot\vec{da} = B\,da\cos\theta, where θ\theta is the angle between B⃗\vec{B} and the assigned direction of da⃗\vec{da}.
  • Over a finite surface SS, ϕ=∫SB⃗⋅da⃗\phi = \int_S \vec{B}\cdot\vec{da}. B⃗\vec{B} can be pulled out of the integral only when it is the same at every point of SS.
  • If the field varies with time, ϕ(t)=∫SB⃗(t)⋅da⃗\phi(t) = \int_S \vec{B}(t)\cdot\vec{da}, and Faraday's law reads e=−dϕdt=−ddt∫SB⃗⋅da⃗e = -\dfrac{d\phi}{dt} = -\dfrac{d}{dt}\int_S \vec{B}\cdot\vec{da}.
  • The rate of change of flux equals the work done in taking unit positive charge once round the bounding contour in the reverse sense — that work is the emf.
  • Even if B⃗\vec{B} is constant in time, the flux still changes if the area SS changes — this is the source of motional emf.
  • SI: ee in volt, ϕ\phi in weber, tt in second.

12.5 Motional electromotive force

  • Translation: a rod of length ll slides on a C-shaped frame with velocity vv, increasing the enclosed area. ϕ=Blx\phi = Blx, so ∣e∣=Bldxdt=Blv\lvert e\rvert = Bl\dfrac{dx}{dt} = Blv.
  • The same result from the Lorentz force: a charge qq in the moving rod feels F⃗=q(v⃗×B⃗)\vec{F} = q(\vec{v}\times\vec{B}), which is parallel to the rod; the work done over the length ll is W=qvBsin⁡θ⋅lW = qvB\sin\theta\cdot l, so e=Wq=vBlsin⁡θe = \dfrac{W}{q} = vBl\sin\theta, and emax⁡=Blve_{\max} = Blv when sin⁡θ=1\sin\theta = 1.
  • The two routes — flux rule and Lorentz force — agree. In general, for any circuit whose parts move in a fixed field, the induced emf is the time derivative of the flux, whatever the shape of the circuit.
  • The flux rule also holds when the loop is stationary and the field changes. Then the force on the charges comes from the E⃗\vec{E} term of F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E}+\vec{v}\times\vec{B}): in a region where the magnetic field is changing with time, electric fields are generated, and it is this electric field that drives the current.
  • Rotating bar: an element drdr at distance rr from the pivot moves with v=ωrv = \omega r and contributes de=Bωr drde = B\omega r\,dr; the elements are in series, so e=∫0lBωr dr=12Bωl2e = \int_0^l B\omega r\,dr = \dfrac{1}{2}B\omega l^2. Compare with e=Blve = Blv for a sliding bar.
  • Faraday disc dynamo: a disc of radius RR spinning at ω\omega in an axial field BB gives e=12BωR2e = \dfrac{1}{2}B\omega R^2 between rim and axis.

12.6 Induced emf in a stationary coil in a changing magnetic field

  • Apparatus: a bar magnet mounted on an arc of a semicircle of radius 50 cm, part of a rigid aluminium frame suspended at the centre of the arc so the whole system swings freely in its plane. A coil of about 10000 turns of copper wire loops the arc so the magnet passes through it freely.
  • Measurement: a capacitor CC and a diode DD are connected across the coil; the induced emf charges the capacitor through the diode and the voltage across CC is measured. Because the time constant RCRC can exceed the duration of the emf pulse, several oscillations may be needed; a milliammeter shows when the charging current stops, i.e. when the peak has been reached.
  • The field at the coil is appreciable only when the magnet is near it, and the magnet is fastest at the mean position, so BB against tt is a hump 'bcde' whose flat top 'cd' corresponds to the finite length of the magnet.
  • The induced emf is proportional to the slope of that curve. The slope is largest at t1t_1 (flux increasing) and t2t_2 (flux decreasing), so ee is largest there, and the minus sign makes ee negative at t1t_1 and positive at t2t_2: one negative and one positive pulse in half a cycle, repeated on the return swing.
  • Only the positive pulse is passed by the diode, so the capacitor charges in steps e1,e2,…e_1, e_2, \ldots up to the peak e0e_0.
  • Writing e=−dϕdθdθdte = -\dfrac{d\phi}{d\theta}\dfrac{d\theta}{dt} with θ=θ0sin⁡2πtT\theta = \theta_0\sin\dfrac{2\pi t}{T} gives dθdt=2πθ0Tcos⁡2πtT\dfrac{d\theta}{dt} = \dfrac{2\pi\theta_0}{T}\cos\dfrac{2\pi t}{T}, and near the mean position the cosine is nearly 1, so e0=(dϕdθ)max⁡2πθ0Te_0 = \left(\dfrac{d\phi}{d\theta}\right)_{\max}\dfrac{2\pi\theta_0}{T}.
  • Result: for a given magnet-coil system, the peak emf is directly proportional to the angular amplitude θ0\theta_0 and inversely proportional to the period TT.

12.7 Generators

  • A generator has the same construction as a motor, but the armature is turned by an external torque instead of by a current. As the armature wires cut the lines of force, an emf e=Blve = Blv appears across the terminals; it is proportional to the speed of rotation ω\omega.
  • Following one conductor of the armature: (i) moving up across the lines of force — emf maximum; (ii) moving parallel to the field — e=0e=0; (iii) moving down across the lines — emf reversed; (iv) the graph of current against time is a sinusoid.
  • For a coil rotating at constant ω\omega, θ=ωt\theta = \omega t, ϕB=BAcos⁡ωt\phi_B = BA\cos\omega t, so e=−NdϕBdt=NBAωsin⁡ωt=e0sin⁡ωte = -N\dfrac{d\phi_B}{dt} = NBA\omega\sin\omega t = e_0\sin\omega t with e0=NBAω=2πfNBAe_0 = NBA\omega = 2\pi f NBA.
  • sin⁡ωt\sin\omega t swings between +1+1 and −1-1, so the polarity reverses periodically: this is alternating current. The emf is extremum at θ=90∘\theta = 90^\circ and 270∘270^\circ, where the flux changes fastest. Domestic supply is 50 cycles per second.
  • A split-ring commutator reverses the connections to the armature in step with the reversals of the current, converting the AC waveform into pulsating DC — used for charging a storage battery.
  • Modern AC motors are more compact and rugged than DC motors.

12.8 Back emf and back torque

  • Emf can be produced in many ways: chemical force in a battery, mechanical pressure in a piezoelectric crystal, a temperature gradient in a thermocouple, light in a photoelectric cell, mechanical transport of charge in a Van de Graaff generator, and motion of a wire in a magnetic field in a generator.
  • A DC generator and a DC motor are physically the same machine: connect a generator to a battery and it runs as a motor; turn a motor by an external agency and it acts as a generator. So a running motor cannot switch its generator action off.
  • By Lenz's law the emf so induced opposes its cause — here the armature current — and therefore reduces it. This unavoidable emf is called the back emf.
  • At start-up the armature is not turning, there is no back emf, and the current is large. As the motor speeds up, the back emf grows and the armature current falls. That is why the starting current of a motor is much larger than its running current.
  • Application: in hybrid and electric vehicles the car acts as a generator while braking, and back emf is used to charge the battery for a longer drive.

12.9 Induction and energy transfer

  • A rectangular loop of width LL and total resistance RR is pulled out of a uniform field region with constant velocity vv. The area inside the field, and hence the flux ϕB=BLx\phi_B = BLx, decreases.
  • Induced emf ∣e∣=BLv\lvert e\rvert = BLv; induced current i=BLvRi = \dfrac{BLv}{R}, clockwise, with its direction given by the right-hand rule.
  • Three segments of the loop are still in the field. The forces F2F_2 and F3F_3 on the two side arms are equal and opposite and cancel; the force F1=iLBF_1 = iLB on the leading arm opposes the motion, so the applied force must be F=−F1F = -F_1 in direction and iLBiLB in magnitude.
  • ∣F∣=iLB=B2L2vR\lvert F\rvert = iLB = \dfrac{B^2L^2v}{R}, so the mechanical power is P=Fv=B2L2v2RP = Fv = \dfrac{B^2L^2v^2}{R}.
  • The rate of Joule heating is P=i2R=(BLvR)2R=B2L2v2RP = i^2R = \left(\dfrac{BLv}{R}\right)^2 R = \dfrac{B^2L^2v^2}{R} — exactly the same.
  • Conclusion: the work done in pulling the loop through the field appears entirely as heat energy in the loop.

12.10 Eddy currents

  • Definition: the circulating currents induced in a metal block (a bulk conductor) when it is placed or moved in a changing magnetic field are called eddy currents.
  • Replace the loop of 12.9 by a solid plate and the electrons no longer follow one path — they swirl within the plate like eddies in water. Work must still be done to move the plate, and that work becomes heat.
  • Pendulum demonstration: a conducting plate free to swing about a pivot loses a part of its mechanical energy to heat each time it enters and leaves the field; after several swings it stops and the plate is warm.
  • Reduction: cut slots in the plate (a discontinuous, laminated structure) so the eddy paths are broken.
  • Applications the board accepts (any two):
  • Dead beat galvanometer — the coil is wound on a light aluminium frame; eddy currents in the frame oppose its motion and bring the coil quickly to rest.
  • Induction motor / speedometer — a magnet rotating at the vehicle's speed inside a spring-pivoted aluminium drum sets up eddy currents; the drum turns with the magnet and a pointer reads the speed on a calibrated scale.
  • Electric (induction) brake — with the drive power off, a stationary magnetic field applied to the rotating drum sets up strong eddy currents whose torque opposes the rotation and stops the train quickly and smoothly.
  • Induction furnace — a high-frequency alternating current in a coil around a crucible produces a rapidly varying field and very strong eddy currents whose heat melts the metal; used for making alloys in vacuum.

12.11 Self inductance

  • The production of an induced emf in a circuit (coil) itself, because of a change of current in it, is called self-induction.
  • ϕ∝i\phi \propto i, so ϕ=Li\phi = Li and L=ϕiL = \dfrac{\phi}{i}; for a closely wound coil of NN turns the flux linkage is NϕB=LiN\phi_B = Li, so L=NϕBiL = \dfrac{N\phi_B}{i}.
  • LL depends only on the geometry of the coil and the material properties of its core.
  • Three definitions the board accepts:
  • The magnetic flux (linkage) produced per unit current in the circuit.
  • The induced emf produced per unit rate of change of current: L=∣e∣∣di/dt∣L = \dfrac{\lvert e\rvert}{\lvert di/dt\rvert}, since e=−Ldidte = -L\dfrac{di}{dt}.
  • Numerically twice the work done in establishing the flux associated with unit current: W=12Li02W = \dfrac{1}{2}Li_0^2, so L=2WL = 2W when i0=1 Ai_0 = 1\ \text{A}.
  • Unit: henry=voltampere/second\text{henry} = \dfrac{\text{volt}}{\text{ampere/second}}; also 1 H=1 ohm⋅second=1 T m2A−11\ \text{H} = 1\ \text{ohm}\cdot\text{second} = 1\ \text{T m}^2\text{A}^{-1}.
  • Solenoid: NϕB=(nl)(BA)=(nl)(μ0ni)AN\phi_B = (nl)(BA) = (nl)(\mu_0 n i)A, so L=μ0n2Al=μ0N2AlL = \mu_0 n^2 A l = \dfrac{\mu_0 N^2 A}{l}, and Ll=μ0n2A=μ0n2πd24\dfrac{L}{l} = \mu_0 n^2 A = \dfrac{\mu_0 n^2\pi d^2}{4}. Hence L∝n2L \propto n^2 and L∝d2L \propto d^2, and μ0\mu_0 can be written in H m−1\text{H m}^{-1}.
  • LL plays the same role in a circuit — electrical inertia — that mass plays in mechanics: compare 12Li2\dfrac{1}{2}Li^2 with 12mv2\dfrac{1}{2}mv^2.
  • Combinations: Ltotal=L1+L2+L3+⋯L_{\text{total}} = L_1+L_2+L_3+\cdots in series; 1Ltotal=1L1+1L2+1L3+⋯\dfrac{1}{L_{\text{total}}} = \dfrac{1}{L_1}+\dfrac{1}{L_2}+\dfrac{1}{L_3}+\cdots in parallel.

12.12 Energy stored in a magnetic field

  • The induced emf opposes the growth of current, so work must be done to build up the magnetic field; that work is stored in the field and can be recovered as heat in the resistance of the circuit.
  • e=−Ldidte = -L\dfrac{di}{dt}; work done in moving charge dqdq against this emf is dW=−e dq=Ldidtdq=L i didW = -e\,dq = L\dfrac{di}{dt}dq = L\,i\,di, using dqdt=i\dfrac{dq}{dt} = i.
  • Total work W=∫0iL i di=12Li2=UBW = \int_0^i L\,i\,di = \dfrac{1}{2}Li^2 = U_B.
  • Analogy with electrostatics: UE=q22C=12CV2U_E = \dfrac{q^2}{2C} = \dfrac{1}{2}CV^2 for a capacitor; UB=12Li2U_B = \dfrac{1}{2}Li^2 for an inductor.

12.13 Energy density of a magnetic field

  • For a solenoid of length ll and area AA, the energy is confined to the interior volume AlAl (the field outside is almost zero) and is uniformly distributed there.
  • uB=UBAl=12Li2Al=12(Ll)i2Au_B = \dfrac{U_B}{Al} = \dfrac{1}{2}\dfrac{Li^2}{Al} = \dfrac{1}{2}\left(\dfrac{L}{l}\right)\dfrac{i^2}{A}; putting Ll=μ0n2A\dfrac{L}{l} = \mu_0 n^2 A gives uB=12μ0n2i2u_B = \dfrac{1}{2}\mu_0 n^2 i^2.
  • With B=μ0niB = \mu_0 n i this becomes uB=B22μ0u_B = \dfrac{B^2}{2\mu_0}, in J m−3\text{J m}^{-3}.
  • This holds at any point where the field is BB, no matter how the field is produced. It parallels uE=12ε0E2u_E = \dfrac{1}{2}\varepsilon_0 E^2.

12.14 Mutual inductance

  • Mutual induction: a change of current in one coil induces an emf in a neighbouring coil, through the flux it sends across.
  • ϕ21=∫S2B1⃗⋅da⃗\phi_{21} = \int_{S_2}\vec{B_1}\cdot\vec{da}; with the coils fixed in space ϕ21∝I1\phi_{21}\propto I_1, so ϕ21=M21I1\phi_{21} = M_{21}I_1 and e21=−M21dI1dte_{21} = -M_{21}\dfrac{dI_1}{dt}. Similarly ϕ12=M12I2\phi_{12} = M_{12}I_2 and e12=−M12dI2dte_{12} = -M_{12}\dfrac{dI_2}{dt}. By symmetry M12=M21=MM_{12} = M_{21} = M.
  • Definitions: M=ϕsIpM = \dfrac{\phi_s}{I_p}, the flux linked with the secondary per unit current in the primary; and M=∣es∣∣dIp/dt∣M = \dfrac{\lvert e_s\rvert}{\lvert dI_p/dt\rvert}, the emf induced in the secondary per unit rate of change of primary current. Also M=WI1I2M = \dfrac{W}{I_1I_2}, the mutual potential energy for unit current in each.
  • Unit: henry. M=1 HM = 1\ \text{H} if 1 A s−11\ \text{A s}^{-1} in the primary induces 1 V1\ \text{V} in the secondary; 1 henry=1 ohm⋅second1\ \text{henry} = 1\ \text{ohm}\cdot\text{second}.
  • Coefficient of coupling KK measures the fraction of coil 1's flux that reaches coil 2. Since e21∝N2Kϕ1e_{21}\propto N_2 K\phi_1, ϕ1∝N1\phi_1\propto N_1 and L∝N2L\propto N^2, one gets M=KL1L2M = K\sqrt{L_1L_2}.
  • K≤1K \le 1; K=1K = 1 is perfect coupling and M=L1L2M = \sqrt{L_1L_2}. K>0.5K>0.5 tightly coupled, K<0.5K<0.5 loosely coupled. If L1=L2=LL_1 = L_2 = L then M=LM = L for perfect coupling.
  • KK is largest when the coils are coaxial (parallel and in line) and smallest when their axes are at right angles; two coils on a common iron core have K≈1K\approx1; radio-frequency coils have KK between 0.001 and 0.05.
  • A large MM is wanted in a transformer but is dangerous in appliances: a clothes dryer has counter-wound heating coils so their fields cancel and no dangerous emf appears on the metal case.
  • Why two coils in parallel have less inductance than either coil: in parallel the current through each inductor is only a fraction of the total, while the voltage across each is the same. A given rate of change of total current therefore develops less voltage across the array than across any single coil — and less voltage for the same didt\dfrac{di}{dt} means less inductance.

12.15 Transformer

  • A transformer is an electrical device which converts low alternating voltage at high current into high alternating voltage at low current, and vice versa.
  • Principle: mutual induction — whenever the magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.
  • Construction: two coils, primary P (input) and secondary S (output), insulated from each other and wound separately on a laminated soft-iron core. In a step-up transformer NS>NPN_S > N_P.
  • Working: an alternating voltage applied to P makes the primary current, and hence the flux in the core, change sinusoidally; this changing flux links both coils and induces an emf in each. The emf depends on the number of turns.
  • With ϕ\phi the flux per turn, eP=−NPdϕdte_P = -N_P\dfrac{d\phi}{dt} and eS=−NSdϕdte_S = -N_S\dfrac{d\phi}{dt}, so eSeP=NSNP\dfrac{e_S}{e_P} = \dfrac{N_S}{N_P} — the equation of the transformer; NSNP\dfrac{N_S}{N_P} is the turns (transformer) ratio.
  • For an ideal transformer input power equals output power: ePiP=eSiSe_Pi_P = e_Si_S, so eSeP=iPiS\dfrac{e_S}{e_P} = \dfrac{i_P}{i_S}, and therefore eSeP=NSNP=iPiS\dfrac{e_S}{e_P} = \dfrac{N_S}{N_P} = \dfrac{i_P}{i_S}.
No.Step-up transformerStep-down transformer
i.Turns in secondary more than in primary, NS>NPN_S>N_PTurns in primary more than in secondary, NP>NSN_P>N_S
ii.Alternating voltage across secondary more than across primary, eS>ePe_S>e_PAlternating voltage across primary more than across secondary, eP>eSe_P>e_S
iii.Transformer ratio K>1K>1Transformer ratio K<1K<1
iv.Primary coil made of thick wireSecondary coil made of thick wire
v.Secondary coil made of thin wirePrimary coil made of thin wire
vi.Current through secondary less than through primaryCurrent through primary less than through secondary

Self induction against mutual induction

PointSelf inductionMutual induction
Coils involvedOne coilTwo neighbouring coils
CauseChange of current in the same coilChange of current in the neighbouring (primary) coil
CoefficientL=NϕBiL = \dfrac{N\phi_B}{i}M=ϕsIpM = \dfrac{\phi_s}{I_p}
Induced emfe=−Ldidte = -L\dfrac{di}{dt}es=−MdIpdte_s = -M\dfrac{dI_p}{dt}
Unithenryhenry

Traps that cost marks

  • Induced emf depends on the rate of change of flux, not on the flux itself. A coil sitting in a huge steady field has zero induced emf.
  • Forgetting the factor NN: for a coil of NN turns it is e=−Ndϕdte = -N\dfrac{d\phi}{dt}, and the flux linkage asked for in MCQs is NϕN\phi, not ϕ\phi.
  • When a field is reversed (or a coil is flipped through 180∘180^\circ), the change of flux is 2NBA2NBA, not NBANBA. A search coil pulled fully out of the field, however, changes by NBANBA.
  • "Plane of the coil perpendicular to B⃗\vec{B}" means θ=0∘\theta = 0^\circ and ϕ=BA\phi = BA; "plane of the coil parallel to B⃗\vec{B}" means θ=90∘\theta = 90^\circ and ϕ=0\phi = 0. Read which is which before writing cos⁡θ\cos\theta.
  • e=Blve = Blv only when B⃗\vec{B}, l⃗\vec{l} and v⃗\vec{v} are mutually perpendicular. In general e=Blvsin⁡θe = Blv\sin\theta with θ\theta the angle between v⃗\vec{v} and B⃗\vec{B} — a wire lying east-west and falling vertically moves along B⃗\vec{B}, so sin⁡θ=0\sin\theta = 0 and the emf is zero.
  • Rotating rod: e=12Bωl2e = \dfrac{1}{2}B\omega l^2, with the factor 12\dfrac{1}{2}; the sliding rod has no such factor. In terms of revolutions per second, e=Bπfl2e = B\pi f l^2.
  • Convert r.p.m. to ff in Hz by dividing by 60, and always use ω=2πf\omega = 2\pi f in e0=NBAωe_0 = NBA\omega. Writing e0=NBAfe_0 = NBAf loses the 2π2\pi.
  • LL is fixed by geometry and core material. It does not change when the current or the emf changes — a question giving ee and didt\dfrac{di}{dt} is asking you to compute LL, not claiming LL depends on them.
  • Inductances combine like resistances, not like capacitances: series adds, parallel adds reciprocals.
  • Energy stored is 12Li2\dfrac{1}{2}Li^2, not Li2Li^2; energy density is B22μ0\dfrac{B^2}{2\mu_0} (per unit volume), not the total energy.
  • M=KL1L2M = K\sqrt{L_1L_2} with the square root; KK is a pure number never greater than 1, and "what percentage of the flux reaches the other coil" is asking for K×100K\times100.
  • The transformer equation has the current ratio inverted: eSeP=NSNP=iPiS\dfrac{e_S}{e_P} = \dfrac{N_S}{N_P} = \dfrac{i_P}{i_S}. A step-up transformer raises voltage and lowers current; it never raises power.
  • A transformer works only on AC. With DC the flux is steady, so no emf is induced in the secondary.
  • Lenz's law opposes the change in flux, not the flux; and it follows from conservation of energy, which is what a question asking to "explain Lenz's law in the light of conservation of energy" wants.
  • Eddy currents are currents in a bulk conductor, not in a wire loop; the standard answer names dead beat galvanometer, speedometer, electric brake and induction furnace.
  • In numericals, work with magnitudes: the minus sign of Faraday's law is for direction only, so do not carry it into the arithmetic and then report a negative emf.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 1 have come up in two or more papers.4
  1. 1
    Motional emf induced in a straight conductor moving in a uniform magnetic field
    Derive an expression for emf (e) generated in a conductor of length (l) moving in a uniform magnetic field (B) with uniform velocity (v) along the x-axis.

    Asked in: March 2022 (Q.20, 3 marks)

    Show the derivation
    Step 1
    Consider a C shaped frame of wires PQRS of area (lx) situated in a constant magnetic field (B⃗)\left(\vec{B}\right). Here, QR is a conducting wire that slides on the frame parallel to PS.
    Step 2
    As the wire QR of length l is moved out with velocity v⃗\vec{v} to increase x, the area of the loop PQRS increases. Thus the flux of B⃗\vec{B} through the loop increases with time.
    Step 3
    According to the flux rule, the induced emf will be equal to the rate at which the magnetic flux through a conducting circuit changes.
    Step 4
    The induced emf will cause a current in the loop. It is assumed that there is enough resistance in the wire so that the induced currents are very small producing negligible magnetic field.
    Step 5
    As the flux ϕ\phi through the frame PQRS is Blx, magnitude of the induced emf can be written as
    ∣e∣=dϕdt=ddt(Blx)=Bldxdt=Blv....(1)|e|=\dfrac{d\phi}{dt}=\dfrac{d}{dt}(Blx)=Bl\dfrac{dx}{dt}=Blv\qquad....(1)
    where v is the velocity of wire QR increasing the length x of wires PQ and SR.
    Step 6
    Now, a charge q which is carried along by the moving wire QR, experiences Lorentz force F⃗=q(v⃗×B⃗)\vec{F}=q(\vec{v}\times\vec{B}); which is perpendicular to both v⃗\vec{v} and B⃗\vec{B} and hence is parallel to wire QR.
    Step 7
    The force F⃗\vec{F} is constant along the length l of the wire QR (as v and B are constant) and zero elsewhere (∵\because v = 0 for stationary part RSPQ of wire frame).
    Step 8
    When the charge q moves a distance l along the wire, the work done by the Lorentz force is
    W=F.l=qvBsin⁡θ⋅lW=F.l=qvB\sin\theta\cdot l
    where θ\theta = angle between B⃗\vec{B} and v⃗\vec{v}.
    Step 9
    The emf generated is,
    e=Workcharge=Wq=vBsin⁡θ⋅le=\dfrac{\text{Work}}{\text{charge}}=\dfrac{W}{q}=vB\sin\theta\cdot l
    Step 10
    For maximum induced emf, sin⁡θ=1\sin\theta=1
    emax=Blv....(2)e_{max}=Blv\qquad....(2)
    Step 11
    Thus, from equation (1) and (2), for any circuit whose parts move in a fixed magnetic field, the induced emf is the time derivative of flux (ϕ\phi) regardless of the shape of the circuit.
  2. 2
    Energy stored in the magnetic field in terms of induced current
    Derive an expression for energy stored in the magnetic field in terms of induced current.

    Asked in: February 2023 (Q.28(i), 4 marks)

    Show the derivation
    Step 1
    Changing magnetic flux in a coil causes an induced emf.
    Step 2
    The induced emf so produced opposes the change and hence the energy has to be spent to overcome it to build up the magnetic field.
    Step 3
    This energy may be recovered as heat in a resistance of the circuit.
    Step 4
    The induced emf is given as, e=−Ldidte=-L\dfrac{di}{dt}
    Step 5
    The work done in moving a charge dq against this emf is,
    dW=−e⋅dq=Ldidt⋅dq=L⋅di⋅dqdtdW=-e\cdot dq=L\dfrac{di}{dt}\cdot dq=L\cdot\dfrac{di\cdot dq}{dt}
    ∴dW=L⋅i⋅di[∵dqdt=i]\therefore dW=L\cdot i\cdot di\qquad\left[\because\dfrac{dq}{dt}=i\right]
    Step 6
    Therefore total work,
    W=∫dW=∫0iLi di=12Li2=UB....(1)W=\int dW=\int_0^i Li\,di=\dfrac{1}{2}Li^2=U_B\qquad....(1)
    Step 7
    Equation (1) gives the energy stored (UBU_B) in magnetic field and is analogous to the energy stored (UEU_E) in the electric field in a capacitor.
  3. 3
    Transformer -- construction, working, and the transformer equation Repeats
    Explain the construction and working of a transformer. Derive the equation for a transformer.

    Asked in: March 2014; March 2016; July 2017; February 2024 (Q.27, 4 marks)

    Show the derivation
    Principle: Working of a transformer is based on the principle of mutual induction i.e., whenever the magnetic flux linked with a coil changes, an e.m.f is induced in the neighbouring coil.
    Construction:
    Step 1
    A step-up transformer consists of two sets of coils primary P and secondary S insulated from each other. The number of turns in secondary coil NSN_S is greater than the number of turns in primary coil NPN_P. The coil P is called the input coil and coil S is called the output coil.
    Step 2
    The two coils are wound separately on a laminated soft iron core.
    Step 1
    Construction:
    a. A transformer consists of two sets of coils primary P and secondary S insulated from each other. The coil P is called the input coil and coil S is called the output coil.
    b. The two coils are wound separately on a laminated soft iron core.
    Step 2
    Working:
    a. When an alternating voltage is applied to the primary coil the current through the coil goes on changing. Hence, the magnetic flux through the core also changes.
    b. As this changing magnetic flux is linked with both the coils, an e.m.f is induced in each coil.
    c. The amount of the magnetic flux linked with the coil depends upon the number of turns of the coil.
    d. Let, 'ϕ\phi' be the magnetic flux linked per turn with both the coils at certain instant 't'.
    e. Let 'NPN_P' and 'NSN_S' be the number of turns of primary and secondary coil,
    NPϕN_P\phi = magnetic flux linked with the primary coil at certain instant 't'
    NSϕN_S\phi = magnetic flux linked with the secondary coil at certain instant 't'
    f. Induced e.m.f produced in the primary and secondary coil is given by,
    eP=−dϕPdt=−NPdϕdt....(1)e_P=-\dfrac{d\phi_P}{dt}=-N_P\dfrac{d\phi}{dt}\qquad....(1)
    eS=−dϕSdt=−NSdϕdt....(2)e_S=-\dfrac{d\phi_S}{dt}=-N_S\dfrac{d\phi}{dt}\qquad....(2)
    g. Dividing equation (2) by (1),
    ∴eSeP=NSNP....(3)\therefore\dfrac{e_S}{e_P}=\dfrac{N_S}{N_P}\qquad....(3)
    Equation (3) represents equation of transformer.
    The ratio NSNP\dfrac{N_S}{N_P} is called turns ratio (transformer ratio) of the transformer.
    h. For an ideal transformer,
    Input power = Output power
    ∴ePIP=eSIS\therefore e_PI_P=e_SI_S
    ∴eSeP=IPIS....(4)\therefore\dfrac{e_S}{e_P}=\dfrac{I_P}{I_S}\qquad....(4)
    Step 1
    From equation (3) and (4),
    eSeP=NSNP=IPIS\dfrac{e_S}{e_P}=\dfrac{N_S}{N_P}=\dfrac{I_P}{I_S}
  4. 4
    Motional emf in a rotating bar (constant angular velocity, uniform field)
    Obtain an expression for motional e.m.f. in a rotating bar with constant angular velocity 'ω\omega' in a uniform magnetic field perpendicular to the plane of rotation.

    Asked in: June 2025 (Q.29, 4 marks)

    Show the derivation

    The question is a real board question. This working is ours — no source printed a derivation for it.

    Diagram

    Draw a straight conducting bar OPOP of length ll pivoted at OO, rotating anticlockwise in the plane of the paper with constant angular velocity ω\omega, sweeping out a circle of radius ll. Show the uniform magnetic field B⃗\vec{B} perpendicular to that plane by a grid of crosses (into the page). Mark a small element of the bar of length dxdx at a distance xx from the pivot OO, and the linear velocity v=ωxv = \omega x of that element, drawn tangentially.

    Method 1 — integrating the motional emf along the bar

    Consider the small element of the bar of length dxdx at a distance xx from the pivot. Being rigidly attached, it moves with linear speed

    v=ωxv = \omega x

    Its velocity is tangential, and the field is perpendicular to the plane of rotation, so v⃗\vec{v}, B⃗\vec{B} and the element are mutually perpendicular. The motional emf across this element is

    de=Bv dx=B(ωx) dxde = B v \, dx = B(\omega x)\,dx

    Different elements move at different speeds, so the emfs must be added by integrating from the pivot (x=0x = 0) to the free end (x=lx = l):

    e=∫0lBωx dx=Bω[x22]0le = \int_{0}^{l} B\omega x\,dx = B\omega \left[\dfrac{x^{2}}{2}\right]_{0}^{l}

     e=12Bωl2 \boxed{\,e = \dfrac{1}{2} B \omega l^{2}\,}

    Method 2 — from Faraday's law (a useful check)

    In time tt the bar sweeps out a sector of angle θ=ωt\theta = \omega t, of area

    A=12l2θ=12l2ωtA = \dfrac{1}{2} l^{2}\theta = \dfrac{1}{2} l^{2}\omega t

    The flux through the swept area is Φ=BA=12Bl2ωt\Phi = BA = \tfrac{1}{2}B l^{2}\omega t, so by Faraday's law the magnitude of the induced emf is

    e=∣dΦdt∣=12Bωl2e = \left|\dfrac{d\Phi}{dt}\right| = \dfrac{1}{2} B \omega l^{2}

    the same result.

    Notes

    • Writing ω=2πn\omega = 2\pi n for nn rotations per second gives

    e=πBnl2e = \pi B n l^{2}.

    • The emf is steady, not alternating: the bar sweeps area at a constant rate,

    so the flux grows linearly and its rate of change is constant. The free end PP is at one polarity and the pivot OO at the other, the sign following from F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}.

    • The emf is proportional to l2l^{2}, not to ll — doubling the length of the

    bar quadruples the emf, because the outer parts of a longer bar also move faster.

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