Q.Consider Experiment 6.2.

Showing the 12 most recent of 19 on this concept.

1. 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)

2. 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.

3. GUJCET 2025Set 031 markMCQ
   Q.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).

   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)

   

4. GUJCET 2025Set 031 markMCQ
   Q.In an AC generator, induced emf $\varepsilon = 0$ at $t = 0$, then its value ______. (A) minimum at time $\frac{2\pi}{3\omega}$ (B) minimum at time $\frac{\pi}{2\omega}$ (C) maximum at time $\frac{2\pi}{\omega}$ (D) maximum at time $\frac{\pi}{2\omega}$

   ›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)

5. GUJCET 2024Set 131 markMCQ
   Q.A square loop of side 10 cm and resistance $0.5\,\Omega$ 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) $8.0 \times 10^{-3}$ A (B) $4.0 \times 10^{-3}$ A (C) $6.0 \times 10^{-3}$ A (D) $2.0 \times 10^{-3}$ A

   ›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

   Option (D) $2.0\times10^{-3}\,\text{A}$

   ANSWER: (D)

6. 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

   Option (A) $5.0\,\text{A}$

   ANSWER: (A)

7. GUJCET 2023Set 091 markMCQ
   Q.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 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)

   

8. GUJCET 2023Set 091 markMCQ
   Q.A circular coil of area 2 $cm^2$ is placed in a magnetic field of 3T perpendicularly. The coil has 10 turns and 5 $\Omega$ 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]

   $q=N\,BA/R = 1.2$ 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

9. 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.

10. 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.

11. 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)

12. 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.