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AITS 03Physics · Chapter 4 of 6

Physics · last-minute revision

Electromagnetic Induction

Flux, Faraday and Lenz, motional emf from sliding and rotating conductors, eddy currents, self and mutual inductance, and the AC generator — numericals verified.

ε = −N dΦ/dta changing flux, always opposed

Magnetic flux and Faraday's lawMust

Magnetic flux
Φ = B · A = BA cosθ (θ between B and normal); scalar; SI unit weber, 1 Wb = 1 T m2
Faraday's law
ε = −N dΦ/dt
Induced current
I = ε/R
Induced charge
q = NΔΦ/R — independent of how fast the flux changes
Flux can change by
changing B, changing area A, or changing angle θ (rotation)
Bar magnet moving into and out of a coil with a galvanometer showing induced currentSNGNSnear faceLenz's lawN approaches:near face → Ncoil repels magnetN recedes:near face → Scoil attracts itAt rest: no emf
Needle deflects only while the magnet moves; direction reverses on withdrawal. The coil face letter shows Lenz's opposition.
Faraday's and Henry's observations (NCERT)

Relative motion between magnet and coil induces current; a changing current in a nearby coil induces current; pressing or releasing a key in a primary circuit gives a momentary deflection in the secondary; inserting an iron rod increases the effect.

Lenz's law and energy conservationMust

The induced emf drives a current whose magnetic effect opposes the change in flux that produces it — the minus sign in Faraday's law. It follows from conservation of energy: if the induced current aided the change, energy would appear from nowhere.

N pole of magnet approaches a coil
Near face becomes N (anticlockwise as seen from the magnet). Coil repels the magnet.
N pole moves away
Near face becomes S (clockwise as seen from the magnet). Coil attracts the magnet.
Magnet dropped through a ring or copper tube
Acceleration less than g, both while entering and while leaving.
Current in one loop increasing
Induced current in nearby coaxial loop is opposite; loops repel. Decreasing current → same direction, attract.
Loop being pulled out of a field
Induced current tries to keep the flux inside: force on loop opposes the pull.

Motional emfMust

Conducting rod sliding on rails in a magnetic field with induced current and opposing force××××××××××××××××××××××××××××××××××××××××××××××××××××××××RvFB into page · rod length l · current anticlockwise (flux into page increasing)
Sliding rod: area and flux increase, induced current opposes it, and the force BIl on the rod points against v. Pushing at constant speed means supplying B²l²v²/R as heat in R.
Straight rod, v ⊥ B ⊥ l
ε = Blv
Current and force
I = Blv/R; F = BIl = B2l2v/R (opposes motion)
Power
P = Fv = B2l2v2/R = ε2/R
Rod falling on vertical rails (terminal speed)
mg = B2l2vt/R → vt = mgR/B2l2
Rod of length L rotating about one end
ε = ½ BωL2
Metal disc of radius R rotating (Faraday disc)
ε = ½ BωR2 between centre and rim
Rod rotating about its centre
ε = 0 between the two ends (each half gives ½Bω(L/2)2 in opposite sense)
Rectangular loop entering field
emf = Blv while partly inside; zero when fully inside; reverses while leaving

Eddy currentsGap content

Gap content

Removed from the rationalised NCERT text but named in the NTA syllabus for this chapter.

A bulk conductor in changing flux carries circulating eddy currents. They oppose the change (Lenz) and dissipate energy as heat.

Solid and slotted metal plates swinging through a magnet showing eddy current dampingNSsolid plate: large eddy currents, stops fastNSslotted plate: eddy paths cut, swings longer
The solid plate loses energy quickly to eddy currents. Slots break the current loops, so damping is much smaller.
  • Useful: magnetic braking in electric trains; electromagnetic damping in galvanometers (coil on a metallic frame); induction furnace; electric power meters (analogue).
  • Harmful: heating and energy loss in transformer and motor cores → reduced by using laminated cores (thin insulated sheets) or slots.

Self and mutual inductanceMust

Self inductance
NΦ = LI; ε = −L dI/dt; unit henry (H)
Long solenoid
L = μ0 n2 A l = μ0 N2 A / l; with core: μ0 μr N2 A / l
Energy stored
U = ½ L I2; energy density u = B2/2μ0
Mutual inductance
N2Φ2 = M I1; ε2 = −M dI1/dt; M12 = M21
Two coaxial long solenoids
M = μ0 n1 n2 π r12 l (r1 = inner radius)
Small loop (r) at centre of large loop (R), coplanar
M ≈ μ0 π r2 / 2R
Coupling
M = k√(L1L2), 0 ≤ k ≤ 1
Series, no coupling
L = L1 + L2; with coupling L1 + L2 ± 2M
Parallel, no coupling
1/L = 1/L1 + 1/L2

Inductance depends only on geometry (turns, area, length) and the medium — not on the current. An inductor opposes change of current: it is the electrical analogue of mass (inertia).

Current growth and decay in an LR circuit with time constant markedτ63%I₀Growth: I = I₀(1 − e^(−t/τ))τ37%I₀Decay: I = I₀ e^(−t/τ)τ = L/R · after 5τ the change is practically complete
Coaching-level extension: in an LR circuit current reaches 63% of I₀ after one time constant L/R on switch-on and falls to 37% after one time constant on switch-off.

AC generatorHigh yield

Rotating coil generator with sinusoidal emf graphBε₀tε = NBAω sin ωtplane ⊥ B: Φ maximum, ε zeroplane ∥ B: Φ zero, ε maximum = NBAωbrass line = normal to coil
Coil turns at angular speed ω. emf is zero when the coil faces B (maximum flux) and maximum when its plane lies along B (flux changing fastest).
Flux
Φ = NBA cos ωt
Emf
ε = NBAω sin ωt, ε0 = NBAω, ω = 2πf
Doubling ω (rev/s)
doubles both ε0 and frequency

Standard question patternsMust

Φ given as function of t; find emf at instant
Differentiate, then substitute t. Do not substitute into Φ itself.
Charge through coil when flux changes
q = NΔΦ/R — ignore the time given.
Rod rotating about one end
ε = ½BωL2; if ω is in rev/s convert with ω = 2πf.
Terminal velocity on vertical rails
vt = mgR/B2l2.
Energy in inductor / emf across it
U = ½LI2; |ε| = L|ΔI/Δt|.
Mutual inductance from L1, L2 and k
M = k√(L1L2).
Direction of induced current
Decide: is flux increasing or decreasing and in which direction? Induced B opposes the change; then use right-hand rule.

Traps that cost marksMust

Trap · Value versus rate

For Φ = 5t2 + 3t + 2 at t = 2 s, emf = 10t + 3 = 23 (mV if Φ in mWb). Substituting into Φ gives 28 — a planted option.

Trap · ω versus f

A coil making 50 rev/s has ω = 314 rad s−1. Using 50 in NBAω gives an answer 2π times too small.

Trap · Squares in B²l²

Terminal speed vt = mgR/B2l2. Dropping one square changes 10 m s−1 into 5 m s−1 in the standard question.

Trap · Direction reversal

Approaching N → near face N (repulsion). Many students write S because 'opposite attracts'. Lenz always opposes the motion: approach is repelled, withdrawal is attracted.

Trap · Assertion–reason

'Magnet falls slower through a copper tube' — reason must be induced (eddy) currents opposing relative motion, not 'copper is magnetic' (copper is not ferromagnetic).

Ten-question checkMust

Numericals are Python-verified. Aim for 9/10 in under 12 minutes.

Q1.A metal rod of length 0.5 m rotates about one end with angular speed 20 rad s−1 in a uniform field of 0.4 T perpendicular to its plane of rotation. The emf between its ends is

  1. (A)2.0 V
  2. (B)4.0 V
  3. (C)0.5 V
  4. (D)1.0 V
Show answer

Answer (D). ε = ½BωL2 = ½ × 0.4 × 20 × 0.25 = 1.0 V. Forgetting the ½ gives 2.0 V.

Q2.Flux through a coil is Φ = (5t2 + 3t + 2) mWb. The magnitude of induced emf at t = 2 s is

  1. (A)28 mV
  2. (B)13 mV
  3. (C)23 mV
  4. (D)10 mV
Show answer

Answer (C). ε = dΦ/dt = 10t + 3 = 23 mV at t = 2 s. 28 mV is the flux value, not its rate of change.

Q3.A 100-turn coil of resistance 10 Ω has flux per turn changed from 5 mWb to 1 mWb. Charge that flows through the coil is

  1. (A)0.4 C
  2. (B)0.04 C
  3. (C)4 × 10−4 C
  4. (D)0.06 C
Show answer

Answer (B). q = NΔΦ/R = 100 × (4 × 10−3)/10 = 0.04 C. Time taken does not matter.

Q4.Energy stored in a 2 H inductor carrying 3 A is

  1. (A)18 J
  2. (B)9 J
  3. (C)6 J
  4. (D)4.5 J
Show answer

Answer (B). U = ½LI2 = ½ × 2 × 9 = 9 J. 18 J omits the ½; 6 J uses LI.

Q5.Current in a 0.5 H coil falls uniformly from 5 A to 1 A in 0.2 s. Magnitude of self-induced emf is

  1. (A)15 V
  2. (B)2.5 V
  3. (C)5 V
  4. (D)10 V
Show answer

Answer (D). |ε| = L ΔI/Δt = 0.5 × 4/0.2 = 10 V.

Q6.Two coils have L1 = 4 mH and L2 = 9 mH with coefficient of coupling 0.5. Their mutual inductance is

  1. (A)6 mH
  2. (B)6.5 mH
  3. (C)3 mH
  4. (D)1.5 mH
Show answer

Answer (C). M = k√(L1L2) = 0.5 × √36 = 3 mH. 6 mH assumes perfect coupling.

Q7.The north pole of a bar magnet is moved towards a coil. Seen from the magnet's side, the induced current in the near face of the coil is

  1. (A)clockwise
  2. (B)zero
  3. (C)anticlockwise
  4. (D)alternately clockwise and anticlockwise
Show answer

Answer (C). The face must become a north pole to repel the approaching N pole; a north face looks anticlockwise to the observer facing it.

Q8.Assertion (A): A strong magnet dropped through a long vertical copper tube falls with acceleration less than g.
Reason (R): Eddy currents induced in the tube oppose the relative motion of the magnet.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (A). Both true, and the opposing eddy currents are precisely why the fall is slowed.

Q9.A coil of 100 turns and area 0.1 m2 rotates at 50 rev s−1 in a uniform field of 0.1 T. The peak emf is about

  1. (A)50 V
  2. (B)157 V
  3. (C)628 V
  4. (D)314 V
Show answer

Answer (D). ω = 2π × 50 = 314 rad s−1; ε0 = NBAω = 100 × 0.1 × 0.1 × 314 = 314 V. Using f = 50 instead of ω gives 50 V.

Q10.A rod of mass 20 g and length 0.4 m slides down frictionless vertical rails connected by a 2 Ω resistor in a horizontal field of 0.5 T. Its terminal speed is (g = 10 m s−2)

  1. (A)10 m s−1
  2. (B)5 m s−1
  3. (C)2 m s−1
  4. (D)20 m s−1
Show answer

Answer (A). vt = mgR/B2l2 = (0.02 × 10 × 2)/(0.25 × 0.16) = 0.4/0.04 = 10 m s−1. Forgetting to square B gives 5 m s−1.

60-second recap before the paperMust

  1. Φ = BA cosθ (Wb). ε = −N dΦ/dt. q = NΔΦ/R (time-independent).
  2. Lenz = energy conservation: approach repelled, withdrawal attracted; falling magnet a < g.
  3. Rod: ε = Blv, F = B2l2v/R, P = B2l2v2/R, vt = mgR/B2l2.
  4. Rotating rod or disc: ε = ½BωL2.
  5. Eddy currents: brakes, damping, induction furnace, meters; reduce with laminations.
  6. L = μ0N2A/l, U = ½LI2, u = B2/2μ0; ε = −L dI/dt.
  7. M = μ0n1n2πr12l; M = k√(L1L2); M12 = M21.
  8. Generator: ε = NBAω sin ωt; ε0 = NBAω; ω = 2πf.
  9. LR: τ = L/R; 63% rise, 37% left after one τ.