A loop, made of straight edges, has six corners at $A(0,0,0)$, $B(L,0,0)$, $C(L,L,0)$, $D(0,L,0)$, $E(0,L,L)$ and $F(0,0,L)$. A magnetic field $\mathbf{B} = B_0(\hat{i} + \hat{k})\ \text{T}$ is present in the region. The flux passing through the loop $ABCDEFA$ (in that order) is (a) B₀L² Wb. (b) 2 B₀L² Wb. (c) √2 B₀L² Wb. (d) 4 B₀L² Wb.

A loop, made of straight edges, has six corners at A(0,0,0), B(L,0,0), C(L,L,0), D(0,L,0), E(0,L,L) and F(0,0,

NCERT Exemplar · Q2

A(0,0,0)

(a) B₀L² Wb.

(b) 2 B₀L² Wb.

(d) 4 B₀L² Wb.

Gujarat GsebMCQ· 1mImportance★★★★★

40% · 20/50 Questions

✓ Free question

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 $\Phi_B$ through a surface of area $A$ in a field $\vec{B}$ :

\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta

where $\theta$ is the angle between $\vec{B}$ and the area's normal. Its SI unit is the weber (Wb), where $1\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2$ .

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 $\theta$ 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:

\mathcal{E} = -N\,\frac{d\Phi_B}{dt}

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

\mathcal{E} = Blv

Here the emf arises because the free charges in the rod experience a magnetic force $q\vec{v}\times\vec{B}$ , which drives them along the rod.

Note

Induction does not require physical contact or a battery. It is the change of flux that matters, not its value. A loop sitting in a huge but constant field has zero induced emf.

Where It Leads

Once a coil's own changing current induces an emf in itself, we call it self-inductance ( $L$ ); when one coil's changing current induces emf in a neighbour, that is mutual inductance ( $M$ ). Both are direct consequences of Faraday's law. Rotate a coil steadily in a magnetic field and the sinusoidal emf it produces is exactly the alternating voltage that runs the AC circuits studied in this chapter.

Faraday's and Lenz's laws of electromagnetic induction form one of the highest-weightage chapters in NCERT Class 12 Physics, tested extensively in CBSE boards, JEE Main and NEET. Anyone searching "Faraday's law of electromagnetic induction formula and examples class 12 physics" will find this changing-flux explanation, including the motional emf case, is exactly how NCERT presents the chapter.

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:

\Phi_B = \int \vec{B} \cdot d\vec{A} = BA\cos\theta

It can change three ways: by changing $B$ , by changing the area $A$ , or by rotating the loop (changing $\theta$ ).

Faraday's law

The induced EMF equals the rate of change of flux:

\mathcal{E} = -\frac{d\Phi_B}{dt}

For a coil of $N$ turns, $\mathcal{E} = -N\,\dfrac{d\Phi_B}{dt}$ . 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

The negative sign is 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 opposition is required by energy conservation — you must do work against the induced current, and that work is what becomes electrical energy. If the current instead aided the change, energy would be created from nothing.

A worked idea

A rod of length $l$ slides at speed $v$ along rails in a field $B$ . In time $dt$ it sweeps area $l\,v\,dt$ , so the flux changes by $d\Phi_B = B\,l\,v\,dt$ , giving a motional EMF:

\mathcal{E} = \frac{d\Phi_B}{dt} = Blv

The same result follows from the magnetic force $q(\vec{v}\times\vec{B})$ pushing free electrons to one end of the rod — a direct check that Faraday's law and the Lorentz force tell one consistent story.

The loop bounds two square faces — $ABCD$ in the plane $z=0$ and $ADEF$ in the plane $x=0$ — with total area vector $L^2(\hat i+\hat k)$ . With $\vec B=B_0(\hat i+\hat k)$ the flux is $\Phi=2B_0L^2$ .

The path $A\to B\to C\to D\to E\to F\to A$ is non-planar, so choose a convenient open surface bounded by it: two adjoining faces of the cube of side $L$ .

Face 1 — $ABCD$ in the plane $z=0$ . Traversed $A\to B\to C\to D$ (counterclockwise seen from $+z$ ), its area vector is

\vec A_1=L^2\,\hat k.

Face 2 — $ADEF$ in the plane $x=0$ . Along the loop this face is traversed $D\to E\to F\to A$ . Using two consecutive edges $\overrightarrow{DE}=L\hat k$ and $\overrightarrow{EF}=-L\hat j$ :

\overrightarrow{DE}\times\overrightarrow{EF}=(L\hat k)\times(-L\hat j)=L^2\,\hat i\Rightarrow \vec A_2=L^2\,\hat i.

The shared edge $AD$ is interior to this surface, so it is not part of the boundary.

Total area vector and flux. For a uniform field the flux is $\vec B\cdot\vec A$ summed over the planar pieces:

\vec A=\vec A_1+\vec A_2=L^2(\hat i+\hat k),

\Phi=\vec B\cdot\vec A=B_0(\hat i+\hat k)\cdot L^2(\hat i+\hat k)=B_0L^2(1+1)=2B_0L^2.

The $\hat k$ -component of $\vec B$ threads face 1 and the $\hat i$ -component threads face 2, each contributing $B_0L^2$ .

✓Final answer

ABCDEFA

Method: Flux Through a Non-Planar (3D) Loop — Split Into Flat Faces

When a closed loop's corners don't all lie in one plane, you can't write down a single area vector for it directly. This method shows how to still compute the flux linked by such a loop, using the fact that flux depends only on the loop's boundary, not on which surface you choose to span it with.

Steps

Step 1: Recognise that any surface bounded by the loop gives the same flux

For a uniform field, the flux linked by a closed loop equals $\vec B \cdot \vec A_{\text{total}}$ for any surface bounded by that loop — so choose the most convenient one. The easiest choice is almost always the set of flat, coordinate-plane faces that the loop's straight edges naturally trace out (here, two adjoining faces of a cube).

Step 2: Break the loop into its flat, planar segments

Trace the loop corner by corner and group consecutive edges into flat faces — e.g. four corners that all share the same coordinate value (all $z=0$ , or all $x=0$ ) form one flat rectangular/square face each.

Step 3: Find each face's area vector, respecting the loop's own traversal direction

For each flat face, use the right-hand rule on the loop's stated direction of travel around that face (e.g. $A\to B\to C\to D$ ) to fix which way its area vector points — this consistency matters because a face traversed the "wrong way" flips the sign of its contribution.

\vec A_{\text{face}} = (\text{edge}_1) \times (\text{edge}_2)

using two consecutive edge vectors of that face (or simply "area $\times$ outward normal" if the orientation is obvious by inspection).

Step 4: Add the face area vectors, then dot with $\vec B$

\vec A_{\text{total}} = \sum \vec A_{\text{face}}, \qquad \Phi_B = \vec B \cdot \vec A_{\text{total}}

Any edge shared between two chosen faces (interior to your surface, not part of the outer loop boundary) doesn't need separate treatment — it's already accounted for once each face is added.

This "span the loop with convenient flat pieces, then add area vectors" trick works for any polygonal 3D loop, not just cube edges — the loop's own corner coordinates always tell you how to choose the flat faces.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

GUJCET 2026Set x1 markMCQ
$Q. In an ideal step up transformer, the number of turns in primary coil and secondary coil are 100 and 200 respectively. If output current is found to be 5A, then input current will be ______. (A) 2.5 A (B) 100 A (C) 5.0 A (D) 10 A$
›Reveal solutionSolution

$I_p=I_s\,N_s/N_p = 5\times2 = 10$ A.

For an ideal transformer $\dfrac{I_p}{I_s}=\dfrac{N_s}{N_p}$ . With $N_p=100$ , $N_s=200$ , output (secondary) current $I_s=5$ A:
$I_p = I_s\frac{N_s}{N_p}=5\times\frac{200}{100}=10\ \text{A}.$
(Step-up transformer → larger primary current.)

✓Final answer
$Option (D) 10 A$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. A.C. generator converts ___.$
(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 magnetic braking torque is converted into the electrical energy delivered - it is not created from heat, light, or electrical energy.

✓Final answer
$(a) mechanical energy into electrical energy.$
GUJCET 2025Set 031 markMCQ
$r$
(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).

Ring (b): The field is out of the page (.) and decreasing, so out-of-page flux is shrinking. The induced current opposes this by trying to maintain the out-of-page field, i.e. it also creates an out-of-page field inside the ring -> anticlockwise.

Therefore ring (a) is anticlockwise and ring (b) is anticlockwise.

[!ANSWER] Anticlockwise, Anticlockwise.

ANSWER: (B)

Two identical conducting rings of radius r shown in a magnetic field, viewed from the top
GUJCET 2025Set 031 markMCQ
$\varepsilon = 0$
›Reveal solutionSolution
[!TLDR]

With $\varepsilon=\varepsilon_0\sin\omega t$ (zero at $t=0$ ), the emf is maximum at $t=\dfrac{\pi}{2\omega}$ — option (D).

Concept

In an AC generator the induced emf is sinusoidal. The condition $\varepsilon=0$ at $t=0$ selects the sine form:
$\varepsilon=\varepsilon_0\sin(\omega t).$
A sine function reaches its maximum value when its argument equals $\frac{\pi}{2}$ .

Solution

Set $\omega t=\dfrac{\pi}{2}$ :
$t=\frac{\pi}{2\omega}.$
So the emf first becomes maximum at $t=\dfrac{\pi}{2\omega}$ .

[!ANSWER]

(D)
GUJCET 2024Set 131 markMCQ
$0.5\,\Omega$
›Reveal solutionSolution

Induced $I=\dfrac{\Delta\Phi}{R\,\Delta t}$ , with the NE field making $45^\circ$ to the loop's normal.

Concept. EMF $=\dfrac{\Delta\Phi}{\Delta t}$ ; the normal to the east–west vertical plane points N–S, so a NE field makes $45^\circ$ with it.

Area $A=(0.10)^2=0.01\,\text{m}^2$ .

$\Phi_i=BA\cos45^\circ=0.10\times0.01\times0.707=7.07\times10^{-4}\,\text{Wb}$ , $\Phi_f=0$ .

$\varepsilon=\dfrac{\Delta\Phi}{\Delta t}=\dfrac{7.07\times10^{-4}}{0.70}=1.01\times10^{-3}\,\text{V}.$

$I=\dfrac{\varepsilon}{R}=\dfrac{1.01\times10^{-3}}{0.5}\approx2.0\times10^{-3}\,\text{A}.$

✓Final answer
$2.0\times10^{-3}\,\text{A}$

ANSWER: (D)
GUJCET 2024Set 131 markMCQ
$Q. If the primary coil of a transformer has 100 turns and the secondary has 200 turns. Then for a input of 220 V at 10 A find output current, in step up transformer. (A) 5.0 A (B) 50.0 A (C) 0.5 A (D) 0.05 A$
›Reveal solutionSolution

Ideal transformer: $I_s/I_p=N_p/N_s$ (current steps down when voltage steps up).

Concept. Power conserved $\Rightarrow V_pI_p=V_sI_s$ and $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}=\dfrac{I_p}{I_s}$ .

$I_s=I_p\dfrac{N_p}{N_s}=10\times\dfrac{100}{200}=5.0\,\text{A}.$

✓Final answer
$5.0\,\text{A}$

ANSWER: (A)
GUJCET 2023Set 091 markMCQ
$v$
›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 magnet, i.e. away from the observer, so the current appears clockwise from the observer's side).

[!ANSWER] The observer on the R.H.S. sees the induced current flowing clockwise.

ANSWER: (B)

A bar magnet lies horizontally on the left with its S pole on the left end and N pole on t
GUJCET 2023Set 091 markMCQ
$cm^2$
›Reveal solutionSolution
[!TLDR]

$q=N\,BA/R = 1.2$ mC (time is irrelevant); option (C).

Concept

When flux through a coil changes, the charge that flows is
$q=\int i\,dt=\frac{N\,\Delta\Phi}{R},$
which depends only on the total flux change and resistance, not on how fast it happens. Here $\Delta\Phi = B A$ (flux goes from $BA$ to $0$ ).

Solution

Data: $N=10$ , $B=3$ T, $A=2\ \text{cm}^2=2\times10^{-4}\ \text{m}^2$ , $R=5\,\Omega$ .
$q=\frac{N\,B\,A}{R}=\frac{10\times3\times2\times10^{-4}}{5}=\frac{6\times10^{-3}}{5}\times10=\ ...$ $=\frac{10\times3\times2\times10^{-4}}{5}=\frac{60\times10^{-4}}{5}=12\times10^{-4}=1.2\times10^{-3}\ \text{C}=1.2\ \text{mC}.$
The 0.2 s is a distractor since $q$ is time-independent. Hence option (C).

[!ANSWER]

(C) 1.2 mC
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. A square of side L meter lies in the x-y plane in a region where the magnetic field is given by B (vector) = B_0 (2 i + 4 j + 3 k) T, where B_0 is constant. The magnitude of flux passing through the square is ___.$
(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).

Flux Phi = B . A = [B_0(2 i + 4 j + 3 k)] . (L^2 k) = 3 B_0 L^2 (only the k-component survives the dot product).

✓Final answer
$(b) 3 B_0 L^2 Wb.$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Magnetic flux linked with the coil is given by phi(t) = (2t^2 + 2t + 1) Wb and its resistance is 10 ohm. The current passing through the coil at t = 2 s is ___ A.$
(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).

At t = 2 s: |e| = 4(2) + 2 = 10 V.

Current: I = |e|/R = 10/10 = 1 A.

✓Final answer
$(b) 1 A.$
GUJCET 2022Set 171 markMCQ
$Q. "The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it." This statement is known as ________. (A) Faraday (B) Maxwell (C) Kirchhoff (D) Lenz$
›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 flux causing it — a consequence of conservation of energy.

✓Final answer
$(D) Lenz$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. A current-carrying coil with N turns and cross-sectional area A is placed in a uniform magnetic field of magnitude B such that its plane remains perpendicular to the magnetic field. If the number of turns of the coil is now doubled, the magnetic flux linked with it = ______.$
(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: $\Phi = BA$ (since normal is parallel to B, $\cos\theta=1$ ).

For N turns, total flux linkage is $N\Phi = NBA$ .

If turns are doubled to $2N$ , the new flux linkage is $2N \cdot BA = 2NBA$ .

✓Final answer
$(d) 2NBA.$

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