Q.A square loop of side $10\ \text{cm}$ and resistance $0.5\ \Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\ \text{T}$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $0.70\ \text{s}$ at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

Showing the 12 most recent of 45 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The current I in an induction coil is varying with time t as shown in the figure: [FIGURE: I–t graph — a symmetric triangular pulse; the current I rises linearly from 0 to a peak value and then decreases linearly back to 0]. Which one of the following graphs shows the variation of voltage in the coil with time t? (A) [FIGURE: V–t graph — a single symmetric triangular pulse lying entirely above the axis: V rises linearly from 0 to a peak value, then falls linearly back to 0 (the same triangular shape as the given I–t graph)] (B) [FIGURE: V–t graph — V rises linearly from 0 to a positive peak, falls linearly through zero to a negative peak of equal magnitude, then rises back to 0 (a positive triangular lobe immediately followed by a negative triangular lobe, i.e. a zig-zag waveform)] (C) [FIGURE: V–t graph — a rectangular (square) wave: V is constant and negative for the first half of the time interval, then jumps abruptly to a constant positive value of equal magnitude for the second half] (D) [FIGURE: V–t graph — one smooth rounded lobe: V rises from 0, curves up to a rounded positive peak, curves back down through zero, dips to a rounded negative trough of equal magnitude, then returns to zero (resembling one full cycle of a sine wave)]

   ›Reveal solutionSolution

   This tests the relation $V=-L\,dI/dt$ for a piecewise-linear current; a triangular I-t graph gives a rectangular (square-wave) V-t graph.

   Concept and Intuition

   Induced EMF depends only on the rate of change of current, not on the current's value itself. A triangular current pulse has a constant (but different-signed) slope during its rising and falling halves — the slope itself doesn't vary smoothly, it switches abruptly from one constant value to another (with a sign flip) right at the peak. Since voltage tracks the slope, not the current, the voltage must also be piecewise-constant: it jumps sharply at the current's peak, producing a rectangular wave rather than mirroring the current's triangular shape.

   Step-by-Step Solution

   1. Write the induced voltage as $V = -L\dfrac{dI}{dt}$.
   2. During the rising half of the triangular pulse, $I$ increases linearly, so $\dfrac{dI}{dt}$ is a positive constant; hence $V$ is a constant negative value.
   3. During the falling half, $I$ decreases linearly, so $\dfrac{dI}{dt}$ is a negative constant of the same magnitude (symmetric triangle); hence $V$ is a constant positive value of equal magnitude.
   4. At the exact peak, the slope switches abruptly, so $V$ jumps discontinuously from the negative constant to the positive constant. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In a coil the current varies from $-3A$ to $+3A$ in 4s, and induces an emf 0.2V. The self-inductance of the coil is (A) 0.133 H (B) 0.266 H (C) 0.65 H (D) 0.532 H

   ›Reveal solutionSolution

   This tests the self-inductance relation $\varepsilon = L\,dI/dt$ with a signed current change. Answer: 0.133 H.

   Concept and Intuition

   Self-induced emf is proportional to the rate of change of current through the coil, not the current's absolute value. Going from $-3$ A to $+3$ A is a total swing of 6 A (not 0 A — the current doesn't stay the same, it reverses and changes by the full 6 A), and dividing by the time taken gives the average $dI/dt$, which plugs directly into $\varepsilon = L\,dI/dt$.

   Step-by-Step Solution

   1. Change in current: $\Delta I = I_f - I_i = 3 - (-3) = 6$ A.
   2. Time taken: $\Delta t = 4$ s.
   3. Rate of change: $\dfrac{dI}{dt} = \dfrac{6}{4} = 1.5\ A/s$. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.In a coil of resistance $10\Omega$, the induced current developed by changing magnetic flux through it is shown in figure as a function of time. The magnitude of charge in flux through the coil in weber is (the graph shows current i, in amp, decreasing linearly with time t, in s, from $i=4$ A at $t=0$ to $i=0$ at $t=0.1$ s) (A) 8 (B) 6 (C) 4 (D) 2

   ›Reveal solutionSolution

   The charge that flows through a coil during a flux change is $q=\Delta\phi/R$, so $\Delta\phi = qR$. The charge is the area under the current–time graph. Answer: 2 Wb.

   Concept and Intuition

   Faraday's law gives the induced EMF as $\varepsilon = -d\phi/dt$, and with $\varepsilon = iR$, we get $i = -\dfrac{1}{R}\dfrac{d\phi}{dt}$. Integrating over time, the total charge that flows is $q = \int i\,dt = \dfrac{\Delta\phi}{R}$ (in magnitude) — so the charge equals the area under the $i$–$t$ graph, independent of exactly how $i$ varies with time.

   Step-by-Step Solution

   1. The $i$–$t$ graph is a straight line from $i=4$A at $t=0$ to $i=0$ at $t=0.1$s — a right triangle. …

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4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A current carrying loop is placed perpendicular to the direction of a uniform magnetic field of 40 mT. If the radius of the loop decreases at constant rate of $1.4\ mm\,s^{-1}$, the induced emf when the radius of the loop becomes 2.5 cm is (A) $6.6\ \mu V$ (B) $2.2\ \mu V$ (C) $4.4\ \mu V$ (D) $8.8\ \mu V$

   ›Reveal solutionSolution

   A shrinking loop in a uniform field induces an emf from the changing enclosed area; the answer is $8.8\ \mu V$.

   Concept and Intuition

   The flux through the loop is $\Phi=B\pi r^2$ (loop perpendicular to $B$). As the radius shrinks, the enclosed area — and hence the flux — changes with time, inducing an emf $\varepsilon=-\dfrac{d\Phi}{dt}=-B\cdot 2\pi r\dfrac{dr}{dt}$ (using the chain rule on $r^2$).

   Step-by-Step Solution

   1. $|\varepsilon| = B\cdot 2\pi r\cdot\left|\dfrac{dr}{dt}\right|$.
   2. Substitute $B=40\times10^{-3}$ T, $r=2.5\times10^{-2}$ m, $\left|\dfrac{dr}{dt}\right|=1.4\times10^{-3}$ m/s.
   3. $2\pi r = 2\pi(0.025)=0.15708$ m. …

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5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A metal sheet is placed in a magnetic field whose magnitude changes from zero to maximum. The direction of eddy currents produced in the plate is shown in the figure. Then the direction of magnetic field is [FIGURE: a square metal plate with N marked at top, S at bottom, W at left, E at right; concentric circular eddy-current loops are shown inside the plate with arrows indicating a counter-clockwise sense] (A) Normally inwards (B) Normally outwards (C) From West to East (D) From North to South

   ›Reveal solutionSolution

   Reading the eddy-current sense with the right-hand rule and applying Lenz's law (opposing the increasing flux) shows the external field points into the page. Answer: (A).

   Concept and Intuition

   Eddy currents are induced only by a flux change normal to the sheet — a field lying in the plane of the sheet (W-to-E or N-to-S) cannot drive circular loops confined to that plane. So the answer must be one of the two 'normal' options. Lenz's law says the induced current always opposes the change in flux that caused it; since the field is increasing from zero to a maximum, the induced current's own field must point opposite to the external field, partially cancelling the increase.

   Step-by-Step Solution

   1. The loops are confined to the plane of the sheet, so the driving flux must be along the sheet's normal (into or out of the page) — this rules out the W→E and N→S options.
   2. Trace the eddy-current sense as described: current rises along the E (right) side toward N, crosses the top moving toward W, and falls along the W (left) side toward S — tracing this out is a counter-clockwise sense as seen by someone looking at the page.
   3. Apply the right-hand rule to this counter-clockwise loop: curl the right-hand fingers along the current's direction (counter-clockwise); the thumb points out of the page, toward the viewer. So the eddy current's own magnetic field points out of the page at the centre. …

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6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A conducting circular loop with area $3.5 \times 10^{-3}\ \mathrm{m^2}$ and resistance $10\Omega$ is placed normally in a magnetic field $B(t) = 0.4\sin(50\pi t)$ tesla. The net charge flowing through the loop during $t = 0$ to $t = 10\ \mathrm{ms}$ is (Assume the field is uniform over the loop) (A) 0.14 $mC$ (B) 21 $mC$ (C) 6 $mC$ (D) 7 $mC$

   ›Reveal solutionSolution

   Charge through a loop from changing flux is $q=\Delta\Phi/R$, independent of how $B$ varied in between — only the endpoint values of $B$ matter.

   Concept and Intuition

   The induced charge $q=\int I\,dt = \int \dfrac{1}{R}\dfrac{d\Phi}{dt}dt = \dfrac{\Delta\Phi}{R}$ depends only on the net change in flux between the initial and final instants, not on the details of the time-variation in between (as long as we integrate over the correct interval). This makes the calculation depend only on $B(0)$ and $B(10\,ms)$.

   Step-by-Step Solution

   1. $B(t) = 0.4\sin(50\pi t)$.
   2. At $t=0$: $B(0) = 0.4\sin(0) = 0$.
   3. At $t=10\,ms=0.01\,s$: argument $= 50\pi\times0.01 = 0.5\pi = \pi/2$, so $B(0.01) = 0.4\sin(\pi/2) = 0.4\,T$.
   4. $\Delta B = 0.4 - 0 = 0.4\,T$; $\Delta\Phi = A\Delta B = 3.5\times10^{-3}\times0.4 = 1.4\times10^{-3}\,Wb$. …

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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A circular coil of radius 8 cm, 400 turns and resistance $2\Omega$ is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180°$ in 0.30 sec. Horizontal component of the earth's magnetic field at the place is $3\times10^{-5}$ T. The magnitude of current induced in the coil is approximately (A) $4\times10^{-2}\ A$ (B) $8\times10^{-4}\ A$ (C) $8\times10^{-2}\ A$ (D) $1.92\times10^{-3}\ A$

   ›Reveal solutionSolution

   This tests average EMF/current from a flux reversal (coil flipped $180°$ in a uniform field) — answer is $8\times10^{-4}\ A$.

   Concept and Intuition

   When a coil's plane is perpendicular to a magnetic field, the field lines pass straight through it, so the flux linkage is maximum: $\Phi = NBA$. Flipping the coil by $180°$ about a diameter doesn't change the magnitude of flux through it, but the normal now points the opposite way, so the flux becomes $-NBA$. The coil has therefore swept through a flux change of $2NBA$, not zero — this is the classic trick in this problem type. Since we're only given a total rotation time (not the instantaneous $\omega t$ dependence), we use the average-EMF form of Faraday's law over that interval.

   Step-by-Step Solution

   1. Area of coil: $A = \pi r^2 = \pi (0.08)^2 = 3.14159\times 0.0064 = 0.02011\ m^2$.
   2. Initial flux (normal $\parallel B_H$): $\Phi_i = N B_H A$.
   3. After $180°$ rotation, normal reverses: $\Phi_f = -N B_H A$.
   4. Magnitude of flux change: $|\Delta\Phi| = |\Phi_f-\Phi_i| = 2NB_HA = 2\times400\times(3\times10^{-5})\times0.02011$. $= 2\times400\times3\times10^{-5}\times0.02011 = 4.826\times10^{-4}\ Wb$.
   5. Average induced EMF: $\varepsilon = \dfrac{|\Delta\Phi|}{\Delta t} = \dfrac{4.826\times10^{-4}}{0.30} = 1.609\times10^{-3}\ V$. …

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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A coil of 45 turns and radius 4 cm is placed in a uniform magnetic field such that its plane is perpendicular to the direction of the field. If the magnetic field increases from 0 to 0.70 T at a constant rate in a time interval of 220 s, then the induced emf in the coil is (A) 0.32 mV (B) 0.50 mV (C) 0.72 mV (D) 0.96 mV

   ›Reveal solutionSolution

   Faraday's law for a multi-turn coil in a linearly changing field gives $\varepsilon = NA\,dB/dt$; substituting the given numbers yields 0.72 mV.

   Concept and Intuition

   Faraday's law states that an EMF is induced in a coil whenever the magnetic flux linked with it changes: $\varepsilon = -N\dfrac{d\Phi}{dt}$. Since the coil's plane is perpendicular to $B$, the flux through one turn is simply $\Phi = BA$ (no angle factor needed), and because the field changes at a constant rate, $\dfrac{dB}{dt}$ is just $\Delta B/\Delta t$. Multiplying by the number of turns $N$ accounts for the fact that each turn contributes its own EMF, and these add in series.

   Step-by-Step Solution

   1. Area of the coil: $A = \pi r^2 = \pi (0.04\ \mathrm{m})^2 = \pi\times1.6\times10^{-3} \approx 5.027\times10^{-3}\ \mathrm{m^2}$.
   2. Rate of change of field: $\dfrac{dB}{dt} = \dfrac{0.70\ \mathrm{T} - 0}{220\ \mathrm{s}} = 3.1818\times10^{-3}\ \mathrm{T/s}$. …

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9. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.When the current in a coil decreases from 10 A to I in a time of 2 seconds, the induced emf in the coil is $e_1$. When the current in the coil decreases from I to zero in 4 seconds, the induced emf in the coil is $e_2$. If the ratio of the induced emfs $e_1 : e_2$ is 2 : 3, then the value of I is (A) 3.75 A (B) 7.5 A (C) 5 A (D) 2.5 A

   ›Reveal solutionSolution

   Using $e=L\,\Delta I/\Delta t$ for both intervals and the given ratio $e_1:e_2=2:3$ gives $I=7.5\ \text{A}$.

   Concept and Intuition

   The emf induced in a coil due to a changing current is $e = -L\dfrac{dI}{dt}$, and for a uniform rate of change over an interval $\Delta t$, the magnitude is $e = L\dfrac{|\Delta I|}{\Delta t}$. Since the same coil (same self-inductance $L$) is used in both cases, $L$ cancels out when we take the ratio of the two emfs, leaving an equation purely in terms of $I$.

   Step-by-Step Solution

   1. First interval: current falls from $10\,\text{A}$ to $I$ in $2\,\text{s}$:

   $e_1 = L\frac{10-I}{2}$

   2. Second interval: current falls from $I$ to $0$ in $4\,\text{s}$:

   $e_2 = L\frac{I}{4}$

   3. Given $\dfrac{e_1}{e_2} = \dfrac{2}{3}$:

   $\frac{L(10-I)/2}{L\,I/4} = \frac{2}{3}$

   4. Simplify the left side:

   $\frac{(10-I)/2}{I/4} = \frac{4(10-I)}{2I} = \frac{2(10-I)}{I}$

   5. So: …

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10. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
    Q.The device constructed based on the laws of electromagnetic induction is (A) galvanometer (B) electric motor (C) ohm meter (D) electric generator

    ›Reveal solutionSolution

    Among the listed devices, the electric generator is the one whose operating principle is Faraday's law of electromagnetic induction — a changing magnetic flux through a rotating coil induces an emf.

    Concept and Intuition

    Electromagnetic induction states that a changing magnetic flux through a circuit induces an emf in it. An electric generator exploits this directly: a coil is mechanically rotated within a magnetic field, continuously changing the flux linked with it, which induces an alternating emf — converting mechanical energy into electrical energy.

    Step-by-Step Solution

    1. Galvanometer: works on the torque on a current-carrying coil in a magnetic field (motor effect), not induction.
    2. Electric motor: converts electrical energy to mechanical energy via the force on a current-carrying conductor in a field — also the motor effect, not induction. …

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11. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.The current in a coil decreases from 5 A to zero in a time of 0.1 s. If an average emf of 200 V is induced, then the self inductance of the coil is (A) 20 H (B) 4 H (C) 2 H (D) 40 H

    ›Reveal solutionSolution

    Self-inductance is found from $\varepsilon=L\,dI/dt$; with the given rate of current change and induced emf, $L=4$ H.

    Concept and Intuition

    Self-induction opposes any change in current through a coil, producing an induced emf proportional to the rate of change of current: $\varepsilon=-L\dfrac{dI}{dt}$. The magnitude of this relationship lets us solve for $L$ given the emf and the current's rate of change.

    Step-by-Step Solution

    1. Rate of change of current: $\dfrac{dI}{dt}=\dfrac{|0-5|}{0.1}=\dfrac{5}{0.1}=50\ \mathrm{A/s}$.
    2. Magnitude of induced emf: $\varepsilon=L\dfrac{dI}{dt}$, so $200=L(50)$. …

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12. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.A coil of resistance 200 Ω is placed in a magnetic field. If the magnetic flux $\phi$ (in weber) linked with the coil varies with time 't' (in second) as per the equation $\phi = 50t^2 + 4$, then the current induced in the coil at a time $t = 2$ s is (A) 2 A (B) 1 A (C) 0.5 A (D) 0.1 A

    ›Reveal solutionSolution

    Faraday's law gives the induced EMF as the time-derivative of flux; dividing by the coil resistance gives an induced current of 1 A at $t=2$ s.

    Concept and Intuition

    Faraday's law of electromagnetic induction states that the EMF induced in a coil equals the (negative of the) rate of change of magnetic flux linked with it: $\varepsilon = -\dfrac{d\phi}{dt}$. Once you know the EMF, Ohm's law across the coil's own resistance gives the induced current directly (there's no other source in this circuit).

    Step-by-Step Solution

    1. Given $\phi(t) = 50t^2+4$ (in Wb).
    2. Differentiate: $\dfrac{d\phi}{dt} = 100t$.
    3. At $t=2$ s: $\left|\dfrac{d\phi}{dt}\right| = 100\times2 = 200$ V (this is the induced EMF). …

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