Q.A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Electromagnetic Induction
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 …
Concept: Electromagnetic Induction — the induced emf is given by Faraday’s law: E=−dtdΦ, where Φ=B⋅A.
Step 1: Find the initial magnetic flux.
Area of the loop: A=(0.10 m)2=0.01 m2.
The field is in the north-east direction, and the loop is in the east-west plane. The normal to the loop (east-west plane) is along the north-south direction. The angle between the field (NE) and the normal (N–S) is 45∘.
So, Φi=BAcos45∘=(0.10)(0.01)(21)=20.001 Wb.
Step 2: Final flux and change in flux.
Final field is zero, so Φf=0.
Change in flux: ∣ΔΦ∣=20.001 Wb.
Step 3: Induced emf and current.
Time interval: Δt=0.70 s. …
The induced emf is found from Faraday’s law using the change in magnetic flux through the loop. The flux changes because the field strength decreases to zero, while the area and orientation stay fixed. The magnitude of induced emf is 1.0×10−3 V and the induced current is 2.0×10−3 A.
The core idea here is electromagnetic induction: a changing magnetic flux through a loop induces an emf. The flux depends on three things — the field strength B, the area A of the loop, and the angle between the field and the normal to the loop. In this problem, only B changes, and it does so uniformly.
Let’s unpack the geometry first. The loop is vertical and lies in the east-west plane. That means its plane contains the east-west direction and the vertical direction. The normal to the loop (the direction perpendicular to its plane) therefore points north-south. The magnetic field is given as 0.10 T in the north-east direction. So the field is at an angle to the normal.
We need the component of the field that actually passes through the loop — that is, the component along the normal. That’s Bcosθ, where θ is the angle between the field direction and the normal.
The normal to a vertical east-west plane points either north or south. Since the field is north-east, the angle between north and north-east is 45∘. So θ=45∘ and cos45∘=21.
Now let’s go step by step.
-
Find the area of the loop.
Side length =10 cm=0.10 m.
Area A=(0.10)2=1.0×10−2 m2.
-
Find the initial magnetic flux through the loop.
Flux Φ=BAcosθ.
Here B=0.10 T, A=1.0×10−2 m2, cos45∘=1/2.
So
Φi=(0.10)(1.0×10−2)⋅21=21.0×10−3 Wb.
-
Find the final flux.
The field is decreased to zero, so Bf=0 and therefore Φf=0.
-
Calculate the change in flux.
ΔΦ=Φf−Φi=0−21.0×10−3=−21.0×10−3 Wb.
The magnitude is ∣ΔΦ∣=21.0×10−3 Wb.
- Apply Faraday’s law to find induced emf. Faraday’s law:
∣E∣=ΔtΔΦ.
The time interval is Δt=0.70 s.
So
∣E∣=0.701.0×10−3/2=0.7021.0×10−3.
Compute: 0.70×2≈0.70×1.414=0.9898≈0.99.
So
Method: Faraday’s Law of Electromagnetic Induction
This problem is solved using Faraday’s Law, which states that the induced emf in a loop equals the negative rate of change of magnetic flux through the loop.
Step 1: Identify the given data
- Side of square loop, a=10 cm=0.10 m
- Area of loop, A=a2=(0.10)2=0.01 m2
- Resistance, R=0.5 Ω
- Initial magnetic field, Bi=0.10 T
- Final magnetic field, Bf=0 T
- Time interval, Δt=0.70 s
- Field direction: north-east (at 45∘ to the east-west plane)
Step 2: Find the angle between field and area vector
The loop is in the east-west vertical plane.
The area vector is perpendicular to the loop — pointing north (or south).
The magnetic field is north-east — at 45∘ to north.
So, the angle between B and area vector A is:
θ=45∘
Step 3: Calculate initial magnetic flux
Magnetic flux:
Φi=BiAcosθ
Φi=(0.10)(0.01)cos45∘
Φi=0.001×21=20.001 Wb
Step 4: Calculate final flux
Since Bf=0:
Φf=0
Step 5: Apply Faraday’s Law for induced emf
Magnitude of induced emf: …
Here are the common mistakes students make on this Electromagnetic Induction problem, along with how to avoid each.
Mistake 1: Getting the area vector direction wrong
The error:
Students often take the area vector as simply "up" or "perpendicular to the loop" without checking the orientation relative to the magnetic field. Here, the loop is in the east-west vertical plane, so its area vector is perpendicular to that plane — pointing either north or south.
How to avoid:
- Draw a clear diagram.
- For a loop in the east-west vertical plane, the normal is horizontal and points north (or south).
- The magnetic field is given as north-east, so the angle between the area vector (north) and the field (north-east) is 45∘.
Key: Always find the angle θ between B and the area vector (not the plane of the loop).
Mistake 2: Using the wrong formula for flux change
The error:
Some students directly use emf=Blv (motional emf) instead of Faraday’s law for a changing magnetic field.
How to avoid:
- Here, the field is decreasing uniformly — no motion, no velocity.
- Use Faraday’s law:
E=−dtdΦ
- For a uniform field and steady rate of change:
E=ΔtΔΦ=ΔtAΔBcosθ
Mistake 3: Forgetting the cosθ factor in flux
The error:
Students compute flux as BA directly, ignoring the angle between B and the area vector.
How to avoid:
- Always write:
Φ=BAcosθ
- Here, θ=45∘, so cos45∘=21.
- The flux is not BA — it’s BA/2.
Mistake 4: Using the wrong area or units
The error:
Using side length in cm without converting to metres, or using perimeter instead of area.
How to avoid:
- Side =10 cm=0.1 m
- Area A=(0.1)2=0.01 m2
- Always convert to SI units before plugging into formulas.
Mistake 5: Sign errors or ignoring magnitude
The error:
Students carry the negative sign from Faraday’s law into the final answer, or get confused about direction when only magnitude is asked.
How to avoid:
- The question asks for magnitudes of emf and current.
- Use:
∣E∣=ΔtA∣ΔB∣cosθ
- Ignore the negative sign — it only indicates direction (Lenz’s law).
Mistake 6: Using ΔB=Bf−Bi incorrectly
The error:
Some write ΔB=0−0.10=−0.10 T and then get confused about sign.
How to avoid: …
Showing the 12 most recent of 19 on this concept.
- GUJCET 2026Set x1 markMCQQ.In an ideal step up transformer, the number of turns in primary coil and secondary coil are 100 and 200 respectively. If output current is found to be 5A, then input current will be ______. (A) 2.5 A (B) 100 A (C) 5.0 A (D) 10 A
›Reveal solutionSolution
Ip=IsNs/Np=5×2=10 A.
For an ideal transformer IsIp=NpNs. With Np=100, Ns=200, output (secondary) current Is=5 A: …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.A.C. generator converts ___.(a) mechanical energy into electrical energy(b) mechanical energy into heat energy(c) mechanical energy into light energy(d) electrical energy into mechanical energy
›Reveal solutionSolution
An AC generator (alternator) converts the mechanical energy used to rotate its coil/armature into electrical energy, via electromagnetic induction.
A coil is mechanically rotated in a magnetic field (or a field is rotated past a stationary coil). The changing flux linkage induces an alternating emf (Faraday's law), which drives current in an external circuit. The mechanical work done to keep the coil turning against the magne …
- GUJCET 2025Set 031 markMCQQ.As shown in figure two identical conducting rings of radius r are placed in magnetic field. In figure(a) magnetic field increasing at the rate of 0.3 T/s and in figure(b) magnetic field decreasing at the rate of 0.2 T/s. The direction of current in ring(a) and ring(b), when observe from top are ______. [FIGURE:(a) a ring in a magnetic field directed into the page (crosses);(b) a ring in a magnetic field directed out of the page (dots).] (A) Clockwise, Anticlockwise (B) Anticlockwise, Anticlockwise (C) Clockwise, Clockwise (D) Anticlockwise, Clockwise
›Reveal solutionSolution
[!TLDR] Ring (a): into-page flux increasing -> induced current opposes it by making an out-of-page field -> anticlockwise. Ring (b): out-of-page flux decreasing -> induced current opposes by maintaining the out-of-page field -> anticlockwise. Both anticlockwise.
Apply Lenz's law, viewing each ring from the top (the reader's side), where x = into page (away from viewer) and . = out of page (toward viewer).
Ring (a): The field is into the page (x) and increasing, so into-page flux is growing. The induced current opposes this change and must create a magnetic field OUT of the page inside the ring. By the right-hand rule, a current producing an out-of-page field flows anticlockwise (as seen from the top). …
- GUJCET 2025Set 031 markMCQQ.In an AC generator, induced emf ε=0 at t=0, then its value ______. (A) minimum at time 3ω2π (B) minimum at time 2ωπ (C) maximum at time ω2π (D) maximum at time 2ωπ
›Reveal solutionSolution
[!TLDR]
With ε=ε0sinωt (zero at t=0), the emf is maximum at t=2ωπ — option (D).
Concept
In an AC generator the induced emf is sinusoidal. The condition ε=0 at t=0 selects the sine form:
ε=ε0sin(ωt). …
- GUJCET 2024Set 131 markMCQQ.A square loop of side 10 cm and resistance 0.5Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is setup across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 S at a steady rate. Then the magnitude of induced current during this time interval will be ________. (A) 8.0×10−3 A (B) 4.0×10−3 A (C) 6.0×10−3 A (D) 2.0×10−3 A
›Reveal solutionSolution
Induced I=RΔtΔΦ, with the NE field making 45∘ to the loop's normal.
Concept. EMF =ΔtΔΦ; the normal to the east–west vertical plane points N–S, so a NE field makes 45∘ with it.
Area A=(0.10)2=0.01m2.
Φi=BAcos45∘=0.10×0.01×0.707=7.07×10−4Wb, Φf=0. …
- GUJCET 2024Set 131 markMCQQ.If the primary coil of a transformer has 100 turns and the secondary has 200 turns. Then for a input of 220 V at 10 A find output current, in step up transformer. (A) 5.0 A (B) 50.0 A (C) 0.5 A (D) 0.05 A
›Reveal solutionSolution
Ideal transformer: Is/Ip=Np/Ns (current steps down when voltage steps up).
Concept. Power conserved ⇒VpIp=VsIs and NpNs=VpVs=IsIp. …
- GUJCET 2023Set 091 markMCQQ.As shown in the figure a bar magnet is moving towards a stationary coil with constant speed v. The direction of induced current in the coil observed by the observer on R.H.S. is ______. [FIGURE: bar magnet with S on left and N on right moving right (velocity v) towards a coil; a galvanometer G is connected; observer stands to the right of the coil] (A) Anticlockwise (B) Clockwise (C) Current changes its direction randomly (D) Induced current will not be produced
›Reveal solutionSolution
[!TLDR] The approaching N pole makes the coil's near face a North pole (Lenz's law); its far face - the one the R.H.S. observer sees - is therefore a South pole, which appears as a clockwise current.
The magnet's N pole moves toward the coil, so the coil flux directed away from the magnet increases. By Lenz's law the induced current opposes this change, so the coil's face nearest the magnet (its left face) becomes a North pole to repel the incoming N pole. Consequently the coil's opposite face - the right-hand face, which the observer on the R.H.S. is looking at - behaves as a South pole. For an observer facing a magnetic South pole of a current loop, the induced current circulates clockwise (the loop's magnetic moment points toward the m …
- GUJCET 2023Set 091 markMCQQ.A circular coil of area 2 cm2 is placed in a magnetic field of 3T perpendicularly. The coil has 10 turns and 5 Ω resistance. Now the coil is removed from magnetic field in 0.2 s. The value of induced charge flowing through the coil is ______. (A) 1.1 mC (B) 1.9 mC (C) 1.2 mC (D) Zero
›Reveal solutionSolution
[!TLDR]
q=NBA/R=1.2 mC (time is irrelevant); option (C).
Concept
When flux through a coil changes, the charge that flows is
q=∫idt=RNΔΦ,
which depends only on the total flux change and resistance, not on how fast it happens. Here ΔΦ=BA (flux goes from BA to 0).
Solution
Data: N=10, B=3 T, A=2 cm2=2×10−4 m2, R=5Ω.
q=RNBA=510×3×2×10−4=56×10−3×10= ... …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.A square of side L meter lies in the x-y plane in a region where the magnetic field is given by B (vector) = B_0 (2 i + 4 j + 3 k) T, where B_0 is constant. The magnitude of flux passing through the square is ___.(a) 4 B_0 L^2 Wb(b) 3 B_0 L^2 Wb(c) 2 B_0 L^2 Wb(d) sqrt(29) B_0 L^2 Wb
›Reveal solutionSolution
The square lies in the x-y plane, so its area vector is along z; only the k-component (3 B_0) of B contributes: flux = 3 B_0 L^2.
The square (side L) lies in the x-y plane, so its area vector is A = L^2 k-hat (along z).
…
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.Magnetic flux linked with the coil is given by phi(t) = (2t^2 + 2t + 1) Wb and its resistance is 10 ohm. The current passing through the coil at t = 2 s is ___ A.(a) 1.5(b) 1(c) 0.5(d) 2
›Reveal solutionSolution
Induced emf = -dphi/dt = -(4t+2); at t = 2, magnitude = 10 V, giving I = 10/10 = 1 A.
Flux phi(t) = 2t^2 + 2t + 1. Induced emf:
e = -dphi/dt = -(4t + 2).
…
- GUJCET 2022Set 171 markMCQQ."The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it." This statement is known as ________. (A) Faraday (B) Maxwell (C) Kirchhoff (D) Lenz
›Reveal solutionSolution
The induced current opposing the change in flux that produced it is Lenz's law.
Concept: Lenz's law states that the polarity of the induced emf is such that the induced current opposes the change in magnetic …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.A current-carrying coil with N turns and cross-sectional area A is placed in a uniform magnetic field of magnitude B such that its plane remains perpendicular to the magnetic field. If the number of turns of the coil is now doubled, the magnetic flux linked with it = ______.(a) N^2 B A(b) N B A / 2(c) N B A(d) 2 N B A
›Reveal solutionSolution
Flux linkage scales directly with the number of turns; doubling N doubles the flux linked with the coil.
Magnetic flux through a single loop with plane perpendicular to B: Φ=BA (since normal is parallel to B, cosθ=1).
…
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