Q.Consider Experiment 6.2.

Showing the 12 most recent of 46 on this concept.

1. CBSE 2026Set 55/2/11 markMCQ
   Q.A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top, (A) the induced current will flow in the anticlockwise direction. (B) the induced current will flow in the clockwise direction. (C) no induced current will flow in the solenoid. (D) the magnet will fall with a constant velocity.

   ›Reveal solutionSolution

   As the magnet falls with its north pole down, the downward magnetic flux through the solenoid increases. By Lenz's law the induced current opposes this change — it must produce an upward field inside the solenoid, making the top face a north pole that repels the approaching magnet. That requires an anticlockwise current as seen from above. The correct option is (A).

   Why this approach works

   Electromagnetic induction is about change: a current is induced in the solenoid only because the flux through it is changing as the magnet falls. The direction of that current is fixed by Lenz's law — the induced current always flows so that its own magnetic field opposes the change in flux that produced it. This is not an arbitrary rule; it is energy conservation. If the induced current aided the magnet's fall, the magnet would speed up and generate ever more electrical energy from nothing.

   So the plan is: track what the flux is doing, decide what field the solenoid must create to oppose it, then convert that field direction into a current sense using the right-hand rule.

   Step-by-step reasoning

   1. Set up the situation.

      The solenoid stands vertically on the table. The magnet is dropped along its axis from above, north pole downward. The observer looks down from the top.

   2. What is the flux doing?

      Field lines emerge from the magnet's north pole — here, pointing downward toward the solenoid. As the magnet approaches, the downward flux through the solenoid's turns increases.

   3. What must the induced current do?

      By Lenz's law it must oppose the increase of downward flux — so it must produce an upward magnetic field inside the solenoid. Equivalently: the top face of the solenoid must behave as a north pole, repelling the incoming north pole of the magnet.

   4. Convert the field direction into a current sense.

      Use the right-hand rule for a coil: curl the fingers of the right hand along the current, and the thumb gives the field inside. For the thumb to point up (toward the observer looking down), the fingers must curl anticlockwise as seen from above. So the induced current is anticlockwise for the top observer.

   Tip

   Quick pole check: the solenoid must repel the approaching north pole, so its top face is a north pole. Looking at a face that is a north pole, the current always appears anticlockwise (a south-pole face appears clockwise — remember by writing N and S with arrowheads on the letter ends). Same conclusion.

   5. Eliminate the other options.
      • (B) Clockwise (from above) would make the top face a south pole, producing a downward field that aids the growing downward flux and attracts the magnet — the opposite of Lenz's law, and a violation of energy conservation.
      • (C) No induced current is impossible: the flux through the solenoid changes continuously while the magnet moves, so an emf — and, in a closed solenoid, a current — must exist.
      • (D) Constant velocity is wrong: the induced current exerts a retarding (upward) force on the magnet, so the magnet falls with a reduced acceleration, not at constant velocity.

   Watch out

   A common mistake is to mix up the viewing direction. "Clockwise seen from above" corresponds to a downward field inside the coil; "anticlockwise seen from above" corresponds to an upward field. Fix the observer first, then apply the right-hand rule — many wrong answers come from silently switching viewpoints midway.

   ✓Final answer

   The induced current flows in the anticlockwise direction when viewed from above, so the correct option is (A).

2. CBSE 2026Set 55/3/11 markMCQ
   Q.A square loop of side $50$ cm is placed in a uniform magnetic field of $3.0$ T acting perpendicular to the plane of the loop. If the loop is rotated through an angle of $90^\circ$ in $0.3$ s, the value of emf induced in the loop would be : (A) $0.25$ V (B) $0.50$ V (C) $0.75$ V (D) $1.0$ V

   ›Reveal solutionSolution

   The induced emf is found from Faraday’s law: the change in magnetic flux divided by the time taken. The flux changes from maximum to zero as the loop rotates by $90^\circ$, giving an average emf of $2.5$ V — but the options are in the range $0.25$–$1.0$ V, so we must check the calculation carefully. The correct value is $2.5$ V, which does not match any given option; however, if the side length is $50$ cm = $0.5$ m, area $= 0.25$ m², flux change $= 3.0 \times 0.25 = 0.75$ Wb, time $= 0.3$ s, emf $= 0.75/0.3 = 2.5$ V. None of the options are correct as stated.

   The core idea here is Faraday’s law of electromagnetic induction: whenever the magnetic flux through a loop changes, an emf is induced. The flux depends on three things — the field strength $B$, the area $A$ of the loop, and the angle $\theta$ between the field and the normal to the loop. When you rotate the loop, you change $\theta$, and that changes the flux. The induced emf is the rate of change of flux.

   In this problem, the field is uniform and perpendicular to the loop initially. That means the initial angle between the field and the normal is $0^\circ$, so the flux is maximum. After a $90^\circ$ rotation, the plane of the loop is parallel to the field, so the flux becomes zero. The change in flux is simply the initial flux minus zero.

   Let’s work it out step by step.

   1. Find the area of the loop.

      Side length $= 50$ cm $= 0.5$ m.

      Area $A = (0.5)^2 = 0.25$ m².

   2. Initial magnetic flux.

      Flux $\Phi = B A \cos\theta$.

      Initially $\theta = 0^\circ$, so $\cos 0 = 1$.

      $\Phi_i = 3.0 \times 0.25 \times 1 = 0.75$ Wb.

   3. Final magnetic flux.

      After $90^\circ$ rotation, $\theta = 90^\circ$, $\cos 90 = 0$.

      $\Phi_f = 3.0 \times 0.25 \times 0 = 0$ Wb.

   4. Change in flux.

      $\Delta\Phi = \Phi_f - \Phi_i = 0 - 0.75 = -0.75$ Wb.

      The magnitude of the change is $0.75$ Wb.

   5. Average induced emf.

      By Faraday’s law, $|\mathcal{E}| = \left|\frac{\Delta\Phi}{\Delta t}\right|$.

      $\Delta t = 0.3$ s.

      $|\mathcal{E}| = \frac{0.75}{0.3} = 2.5$ V.

   Watch out

   A common mistake is to forget that the side is given in cm and not convert to metres. If you use $50$ cm as $50$ m, you get an absurdly large area and emf. Another pitfall: using the angle between the field and the plane of the loop instead of the normal. Here, the field is perpendicular to the plane initially, so the normal is parallel to the field — that’s $\theta = 0^\circ$, not $90^\circ$.

   Tip

   For a $90^\circ$ rotation from alignment to perpendicular, the flux goes from $BA$ to $0$, so the change is always $BA$ regardless of the shape of the loop. The induced emf depends only on $B$, $A$, and the time taken.

   Now, the options given are $0.25$ V, $0.50$ V, $0.75$ V, and $1.0$ V. Our calculated value is $2.5$ V, which is not among them. Let’s double-check: if the side were $50$ cm = $0.5$ m, area $= 0.25$ m², $B=3$ T, flux change $= 0.75$ Wb, time $= 0.3$ s, emf $= 2.5$ V. That is correct.

   If the side were $25$ cm, area would be $0.0625$ m², flux change $= 0.1875$ Wb, emf $= 0.625$ V — still not matching. If the time were $1$ s, emf $= 0.75$ V, which matches option (C), but the problem clearly states $0.3$ s.

   Important

   The numbers in the problem lead to $2.5$ V, which is not among the choices. This suggests either a misprint in the options or an intended different interpretation (e.g., instantaneous emf at some angle). But for the average emf over $90^\circ$ rotation, the answer is $2.5$ V.

   ✓Final answer

   The induced emf is $\boxed{2.5\ \text{V}}$, which does not match any of the given options (A)–(D).

3. CBSE 2026Set V11 markMCQ
   Q.The working principle of an A.C. generator is :

   (a) mutual induction

   (b) eddy currents

   (c) self induction

   (d) electromagnetic induction

   ›Reveal solutionSolution

   (d) electromagnetic induction

   ✓Final answer

   (d) electromagnetic induction

   An A.C. generator works on the principle of electromagnetic induction: a coil rotating in a magnetic field experiences a continuously changing magnetic flux, which induces an alternating e.m.f. $\varepsilon = -N\dfrac{d\Phi}{dt} = NBA\omega\sin\omega t$ (Faraday's law).

4. CBSE 2026Set A1 markMCQ
   Q.An example of natural electromagnetic induction is (A) radio (B) television (C) battery charging (D) lightning strike

   ›Reveal solutionSolution

   Lightning involves huge, rapidly changing currents/fields that induce emf in nearby conductors — natural electromagnetic induction.

   Electromagnetic induction is the production of emf by a changing magnetic flux (Faraday's law). A lightning strike carries an enormous, rapidly varying current, producing a fast-changing magnetic field that induces emf/current in nearby loops and conductors — a naturally occurring example of electromagnetic induction.

   Radio, television and battery charging are man-made devices, not natural phenomena.

   ✓Final answer

   (D) lightning strike.

5. CBSE 2026Set A1 markMCQ
   Q.If magnetic field is same but the area of the loop is increased, then the flux (A) increases (B) decreases (C) becomes zero (D) remains unchanged

   ›Reveal solutionSolution

   Magnetic flux Φ = BA cosθ; with B constant, a larger area gives greater flux.

   The magnetic flux through a loop is:

   $\Phi=BA\cos\theta$

   Here the magnetic field $B$ (and orientation) is unchanged while the area $A$ of the loop is increased. Since $\Phi$ is directly proportional to $A$, the flux increases.

   ✓Final answer

   (A) increases.

6. CBSE 2026Set ANNUAL1 mark
   Q.What is electromagnetic induction?

   ›Reveal solutionSolution

   Any change of magnetic flux through a circuit produces an EMF in that circuit - this is electromagnetic induction.

   Electromagnetic induction is the phenomenon in which an electromotive force (emf) is induced in a coil or conductor whenever the magnetic flux linked with it changes with time - whether the change is caused by a changing magnetic field, relative motion between the conductor and the field source, or a changing orientation/area of the loop. If the circuit is closed, this induced emf drives an induced current. It is quantitatively described by Faraday's law, EMF = -d(phi)/dt, with the negative sign (Lenz's law) showing that the induced effects oppose the change producing them.

   ✓Final answer

   The generation of an induced EMF/current in a conductor due to a change of magnetic flux linked with it (Faraday's law, EMF = -d(phi)/dt).

7. CBSE 2026Set ANNUAL1 mark
   Q.When will the magnetic flux linked with a coil held in the magnetic field be zero?

   ›Reveal solutionSolution

   Flux is zero whenever the field lines lie entirely in the plane of the coil.

   Magnetic flux linked with a coil is $\Phi = BA\cos\theta$, where $\theta$ is the angle between the coil's area vector (normal) and the magnetic field $\vec{B}$. This is zero when $\cos\theta=0$, i.e. $\theta=90°$ — meaning the normal to the coil is perpendicular to $\vec{B}$, which is the same as saying the field lines lie entirely within (parallel to) the plane of the coil, passing along it rather than through it.

   ✓Final answer

   The flux is zero when the plane of the coil is oriented parallel to $\vec{B}$ (i.e. the coil's normal is perpendicular to the field, $\theta=90°$).

8. CBSE 2026Set ANNUAL1 mark
   Q.Match the Column A with Column B and write the correct pair. Column A: Magnetic flux. Column B:

   (i) μ₀nI,

   (ii) Volt × second,

   (iii) μ₀nI/2,

   (iv) Volt × meter,

   (v) Volt × meter⁻¹,

   (vi) Volt.

   ›Reveal solutionSolution

   Magnetic flux unit = weber = Volt × second (from Faraday's law emf = dΦ/dt), option (ii).

   Faraday's law states that the induced emf equals the rate of change of magnetic flux: emf = −dΦ/dt. Rearranging, Φ = emf × time (dimensionally). Since emf is in volts and time in seconds, magnetic flux has the unit volt × second, which is the weber (Wb). This matches Column B entry (ii).

   ✓Final answer

   (ii) Volt × second.

9. CBSE 2025Set X11 markMCQ
   Q.Consider the following statements : Statement – 1: A.C. Generator works on the principle of electromagnetic induction Statement – 2: In an A.C. Generator, as the armature is rotated in a uniform magnetic field, the magnetic flux linked with the coil changes which induces an emf in the coil. Among the above two statements :

   (a) Both Statements are true

   (b) Both Statements are false

   (c) Statement-1 is true and Statement-2 is false

   (d) Statement-1 is false and Statement-2 is true

   ›Reveal solutionSolution

   (a) Both Statements are true. An A.C. generator works on electromagnetic induction (Statement 1). As the armature coil rotates in a uniform magnetic field, the flux $\phi = NBA\cos\omega t$ linked

   ✓Final answer

   (a) Both Statements are true.

   An A.C. generator works on electromagnetic induction (Statement 1). As the armature coil rotates in a uniform magnetic field, the flux $\phi = NBA\cos\omega t$ linked with it changes continuously, inducing an emf $\varepsilon = NBA\omega\sin\omega t$ (Statement 2). Statement 2 correctly explains Statement 1.

10. CBSE 2025Set D1 markMCQ
    Q.Which of the following devices is based on the principle of electromagnetic induction? (A) Voltmeter (B) Electric motor (C) Electric generator (D) Ammeter

    ›Reveal solutionSolution

    The electric generator is based on electromagnetic induction.

    An electric generator rotates a coil in a magnetic field. The continuous change of magnetic flux linked with the coil induces an emf by Faraday's law of electromagnetic induction, converting mechanical energy into electrical energy.

    • An electric motor is the reverse device (electrical → mechanical, using the force on a current in a field).

    • Voltmeters and ammeters (galvanometer-based) work on the torque on a current loop, not on induction.

    ✓Final answer

    (C) Electric generator.

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.The laws of electromagnetic induction have been used in the construction of :

    (a) voltmeter

    (b) ammeter

    (c) electric motor

    (d) generator

    ›Reveal solutionSolution

    An electric generator converts mechanical energy into electrical energy by rotating a coil in a magnetic field, directly applying Faraday's law of electromagnetic induction.

    Faraday's and Lenz's laws of electromagnetic induction state that a changing magnetic flux through a coil induces an emf in it. A generator (dynamo) is built on exactly this principle: mechanical energy rotates a coil within a magnetic field (or a magnet within a coil), continuously changing the flux linked with the coil and inducing an alternating emf. A voltmeter and ammeter are measuring instruments based on the magnetic effect of current, and an electric motor works on the reverse effect (force on a current-carrying conductor in a field) — only the generator is built directly on electromagnetic induction.

    ✓Final answer

    The laws of electromagnetic induction are used in the construction of a generator — option (d).

12. CBSE 2024Set 55/2/11 markMCQ
    Q.A circular coil of radius 10 cm is placed in a magnetic field $\vec{B} = (1{\cdot}0\,\hat{i} + 0{\cdot}5\,\hat{j})$ mT such that the outward unit vector normal to the surface of the coil is $(0{\cdot}6\,\hat{i} + 0{\cdot}8\,\hat{j})$. The magnetic flux linked with the coil is : (A) $0{\cdot}314\ \mu\text{Wb}$ (B) $3{\cdot}14\ \mu\text{Wb}$ (C) $31{\cdot}4\ \mu\text{Wb}$ (D) $1{\cdot}256\ \mu\text{Wb}$

    ›Reveal solutionSolution

    Magnetic flux is the dot product of the field and the area vector; here $\Phi = \vec{B} \cdot \vec{A} = \vec{B} \cdot (A\,\hat{n})$, which gives $31.4\,\mu\text{Wb}$.

    Why the dot product?

    Magnetic flux measures how much of the magnetic field "threads through" a surface. Not all field lines contribute equally: only the component of $\vec{B}$ perpendicular to the surface matters. When the field is at an angle, we project it onto the surface normal using the dot product.

    The flux through a flat surface is

    $\Phi = \vec{B} \cdot \vec{A}$

    where $\vec{A} = A\,\hat{n}$ is the area vector—magnitude $A$ (the area) pointing along the outward normal $\hat{n}$.

    Step-by-step calculation

    1. Find the area of the coil. The coil is circular with radius $r = 10\,\text{cm} = 0.1\,\text{m}$.

    $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi\,\text{m}^2$

    2. Write the area vector. The outward normal is $\hat{n} = 0.6\,\hat{i} + 0.8\,\hat{j}$ (already a unit vector since $0.6^2 + 0.8^2 = 1$), so

    $\vec{A} = A\,\hat{n} = 0.01\pi\,(0.6\,\hat{i} + 0.8\,\hat{j})\,\text{m}^2$

    3. Express the magnetic field in SI units.

       Given $\vec{B} = (1.0\,\hat{i} + 0.5\,\hat{j})\,\text{mT} = (1.0\,\hat{i} + 0.5\,\hat{j}) \times 10^{-3}\,\text{T}$.

    4. Compute the dot product $\vec{B} \cdot \vec{A}$.

    $\Phi = \vec{B} \cdot \vec{A} = (1.0\,\hat{i} + 0.5\,\hat{j}) \times 10^{-3} \cdot 0.01\pi\,(0.6\,\hat{i} + 0.8\,\hat{j})$

    The dot product of the unit vectors:

    $(1.0\,\hat{i} + 0.5\,\hat{j}) \cdot (0.6\,\hat{i} + 0.8\,\hat{j}) = 1.0 \times 0.6 + 0.5 \times 0.8 = 0.6 + 0.4 = 1.0$

    So

    $\Phi = 10^{-3} \times 0.01\pi \times 1.0 = 10^{-5}\pi\,\text{Wb}$

    5. Evaluate numerically.

    $\Phi = \pi \times 10^{-5}\,\text{Wb} \approx 3.14159 \times 10^{-5}\,\text{Wb} = 31.4 \times 10^{-6}\,\text{Wb} = 31.4\,\mu\text{Wb}$

    Tip

    Always check that the normal vector is a unit vector (magnitude 1) before using it. Here $\sqrt{0.6^2 + 0.8^2} = 1$, so no normalization is needed.

    Watch out

    A common mistake is forgetting to convert mT to T, or cm to m. Dimensional consistency is essential: area in $\text{m}^2$, field in T, flux in Wb.

    ✓Final answer

    The magnetic flux linked with the coil is $\boxed{31.4\,\mu\text{Wb}}$, so the correct option is (C).