Q.An infinitely long straight wire lies in the plane of the paper and carries a current I(t) whose rate of change dI/dt = λ is constant. A rectangular loop ABCD of wire, of resistance R, lies in the same plane with its two sides AB and DC (each of length l) parallel to the long wire. The nearer side DC is at a perpendicular distance x₀ from the wire and the farther side AB at a perpendicular distance x₀ + r (so the loop's breadth measured away from the wire is r). Find the magnitude of the current induced in the loop.
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Electromagnetic Induction
Electromagnetic induction is the phenomenon in which a changing magnetic flux through a circuit produces an electromotive force (emf) — and hence a current, if the circuit is closed. It is the single idea behind generators, transformers, inductors, and the entire AC power grid.
The Central Discovery
Michael Faraday found (1831) that a current is induced in a coil not when a magnet sits still near it, but only while the magnet moves — that is, only while the magnetic flux linked with the coil is changing. A steady magnet, however strong, induces nothing.
Magnetic Flux
The key quantity is magnetic flux ΦB through a surface of area A in a field B:
ΦB=B⋅A=BAcosθ
where θ is the angle between B and the area's normal. Its SI unit is the weber (Wb), where 1 Wb=1 T⋅m2.
Flux can change in three distinct ways, and any of them induces an emf:
- the field strength B changes,
- the area A of the loop changes,
- the orientation θ changes (a coil rotating in a field — the basis of the generator).
Faraday's Law
The induced emf equals the negative rate of change of flux. For a coil of N turns:
E=−NdtdΦB
The faster the flux changes, the larger the emf. This is why a magnet dropped quickly through a coil gives a bigger deflection than one moved slowly.
Lenz's Law — the Minus Sign
The negative sign expresses Lenz's law: the induced current flows in the direction that opposes the change producing it. Push a magnet's north pole toward a coil, and the coil's near face becomes a north pole to repel it; pull it away, and the face becomes a south pole to attract it. This is simply energy conservation — you must do work against this opposition, and that work becomes the electrical energy of the induced current.
Motional emf
A special, very useful case: a conducting rod of length l moving with speed v perpendicular to a field B sweeps out area and develops an emf
E=Blv
Here the emf arises because the free charges in the rod experience a magnetic force qv×B, which drives them along the rod. …
Why this formula?
Electromagnetic Induction
Electromagnetic induction is the effect discovered by Faraday: a changing magnetic flux through a circuit drives an induced EMF (and hence a current). The key word is changing — a steady field, however strong, induces nothing.
Magnetic flux
Flux measures how many field lines thread a surface bounded by the loop:
ΦB=∫B⋅dA=BAcosθ
It can change three ways: by changing B, by changing the area A, or by rotating the loop (changing θ).
Faraday's law
The induced EMF equals the rate of change of flux:
E=−dtdΦB
For a coil of N turns, E=−NdtdΦB. The EMF depends on how fast the flux changes, not on the flux itself — a slow change gives a small EMF, a rapid change a large one.
Lenz's law — the minus sign …
Integrate the 1/x field of the wire across the loop: ϕ=2πμ0Illnx0x0+r; with dI/dt=λ, the induced current is 2πRμ0lλlnx0x0+r. …
The long wire's field falls off as 1/x, so the flux through the loop is found by integrating across its breadth: ϕ=2πμ0Illnx0x0+r. Since dI/dt=λ is constant, the induced emf is constant and Iloop=2πRμ0lλlnx0x0+r.
Concept
The magnetic field of an infinitely long straight wire at perpendicular distance x is
B(x)=2πxμ0I,
directed perpendicular to the loop's plane. Because B varies across the loop, the flux must be obtained by integration.
Flux through the loop
Take a strip of width dx at distance x, parallel to the wire and of length l; its area is ldx. Integrating from the near side x0 to the far side x0+r:
ϕ=∫x0x0+r2πxμ0Ildx=2πμ0Illn(x0x0+r).
Induced emf and current …
Method: Flux by Integration When the Field Is Non-Uniform Across the Loop
Use this whenever a loop sits near a source (typically a long straight current-carrying wire) whose field strength changes across the loop's own area — Φ=BA no longer works because B is not one single value over the loop.
Steps
Step 1: Write the field as a function of position
For an infinite straight wire, the field at perpendicular distance x is
B(x)=2πxμ0I,
directed perpendicular to the loop's plane. Since x varies across the loop's breadth, B is not constant there.
Step 2: Set up a differential strip
Slice the loop into thin strips parallel to the wire, each at distance x, of width dx and length equal to the loop's side parallel to the wire (l here). Each strip has area dA=ldx and (because it's thin) a nearly-uniform field B(x) on it.
Step 3: Integrate to get the total flux
Sum (integrate) B(x)dA across the loop's full breadth, from the near edge to the far edge:
Φ=∫nearfarB(x)ldx=2πμ0Illn(near distancefar distance).
The logarithm is the signature of integrating a 1/x field — expect it whenever this method applies. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.A circular coil of radius 10 cm with 200 turns is placed perpendicular to a uniform magnetic field of 0.4 T. If the magnetic field is reduced to zero uniformly in 0.2 s, then the average induced emf (in volt) is (A) 4π (B) 120 (C) 20 (D) 15 (E) 40π
›Reveal solutionSolution
Average induced emf =ΔtNΔΦ=ΔtNAB=4π V.
Area. A=πr2=π(0.1)2=0.01π m2.
emf. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.The plane of a circular loop of area 150cm2 is perpendicular to a uniform magnetic field of 0.5T. If the loop is turned such that its plane is in the direction of the field in 0.5s, then the induced emf produced is (A) 25mV (B) 10mV (C) 2.5mV (D) 15mV (E) 7.5mV
›Reveal solutionSolution
Flux changes from BA to 0; ε=15mV.
Initially the plane is perpendicular to B, so the area vector is along B and flux is maximum: Φi=BA. Finally the plane is along the field, so Φf=0. …
- KEAM 2026Set pha-2026-0419F4 marksMCQQ.If an emf of 2 V is induced in a coil in 6 s, then the change of flux (in Wb) in the coil during the time is (A) 3 (B) 24 (C) 6 (D) 18 (E) 12
›Reveal solutionSolution
dΦ=εdt=2 V×6 s=12 Wb.
Faraday's law relates the induced emf to the rate of change of flux (for a single turn):
ε=dtdΦ.
Hence the change of flux over the interval is …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The mismatched pair regarding the induced emf is (A) Eddy current : Induction furnace (B) Transformer : Laminated core (C) Induced emf : Biot – Savart law (D) Ac generator : Electromagnetic induction (E) Coaxial coils : Mutual inductance
›Reveal solutionSolution
Induced emf comes from Faraday's law (changing flux); the Biot–Savart law gives the magnetic field of a current element. Pairing induced emf with Biot–Savart law is incorrect.
Checking the associations:
- (A) Eddy currents power an induction furnace — correct.
- (B) Transformers use a laminated core to cut eddy losses — correct. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.A coil having 100 turns and an area of 0.02 m2 is placed with its plane perpendicular to the magnetic field of 1 Wbm−2. The magnetic flux linked with the coil is (A) zero (B) 1 Wb (C) 2 Wb (D) 3 Wb (E) 5 Wb
›Reveal solutionSolution
With the coil plane perpendicular to the field, the field is fully along the area-normal, so flux is maximum: ϕ=NBA=2 Wb.
Magnetic flux linked with an N-turn coil is ϕ=NB⋅A=NBAcosθ, where θ is the angle between B and the coil's normal. …
- KEAM 2025Set eng-2025-04294 marksMCQQ.If the flux linked with the coil of area of cross-section 0.5 m2 placed in a magnetic field of 16 T is 4 Wb, then the angle between the magnetic field and the area vector of the coil is (A) 0° (B) 30° (C) 45° (D) 60° (E) 90°
›Reveal solutionSolution
Flux Φ=BAcosθ. With Φ=4 Wb, B=16 T, A=0.5 m2: cosθ=84=0.5, so θ=60∘. …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.An iron ring is held horizontally and a bar magnet is dropped gently through the ring with its length coinciding with the axis of the ring. The acceleration of the freely falling magnet through the ring (g = acceleration due to gravity) (A) is less than g (B) is equal to g (C) is greater than g (D) depends on the radius of the ring (E) depends on the length of the magnet
›Reveal solutionSolution
By Lenz's law the ring's induced current opposes the falling magnet's changing flux, producing a retarding force; the net acceleration is therefore less than g.
As the magnet falls through the ring, the magnetic flux through the ring changes, inducing an EMF and current in it. By Lenz's law this induced current opposes the change — it repels the approaching magnet and attracts the receding one, always retarding t …
- KEAM 2025Set pha-2025-0429F4 marksMCQQ.The flux linked with a coil at any instant is given by ϕ=5t2−25t−150 (in SI unit). The emf induced in the coil at t=2s is (A) +5 V (B) +3 V (C) -1 V (D) -5 V (E) -3 V
›Reveal solutionSolution
Faraday's law: induced emf =−dϕ/dt. Differentiating ϕ=5t2−25t−150 and evaluating at t=2s gives +5V.
By Faraday's law,
ε=−dtdϕ.
With ϕ=5t2−25t−150,
dtdϕ=10t−25. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.When a current passing through a coil changes at a rate of 30 As−1 the emf induced in the coil is 12 V. If the current passing through this coil changes at a rate of 20 As−1 the emf induced in this coil is (A) 8 V (B) 10 V (C) 2.5 V (D) 3 V (E) 5 V
›Reveal solutionSolution
Induced emf is proportional to the rate of change of current: ε=LdI/dt.
From the first case the self-inductance is
L=dI/dtε=3012=0.4 H. …
- KEAM 2023Set eng-2023-P1-A14 marksMCQQ.In a current carrying coil of inductance 60 mH, the current is changed from 2.5 A in one direction to 2.5 A in the opposite direction in 0.10 sec. The average induced EMF in the coil will be: (A) 1.2 V (B) 2.4 V (C) 3.0 V (D) 1.8 V (E) 0.6 V
›Reveal solutionSolution
emf = L(Delta_I/Delta_t) with Delta_I = 5 A gives 0.06 x 5/0.10 = 3.0 V.
Concept and Intuition
The self-induced EMF is L times the rate of change of current. Reversing the current from +2.5 A to -2.5 A is a total change of 5 A, not 2.5 A.
Step-by-Step Solution
- Change in current: Delta_I = 2.5 - (-2.5) = 5 A.
- emf = L Delta_I/Delta_t = 0.060 x 5 / 0.10. …
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