Q.The mutual inductance $M_{12}$ of coil 1 with respect to coil 2

Showing the 12 most recent of 24 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A 2 m long solenoid with diameter 2 cm and 2000 turns has a secondary coil of 1000 turns wound closely near its midpoint. The mutual inductance between the two coils is (A) $2.4 \times 10^{-4}\ H$ (B) $3.9 \times 10^{-4}\ H$ (C) $1.28 \times 10^{-3}\ H$ (D) $3.14 \times 10^{-3}\ H$

   ›Reveal solutionSolution

   A tightly-wound secondary near the midpoint of a long solenoid links essentially the full flux of the primary; using $M=\mu_0(N_1/l)N_2A$ gives $M\approx 3.9\times10^{-4}$ H.

   Concept and Intuition

   A long solenoid carrying current $I_1$ produces a nearly uniform field $B = \mu_0 n_1 I_1$ inside it, where $n_1=N_1/l$ is turns per unit length. If a secondary coil of $N_2$ turns is wound tightly around the solenoid near its middle (away from the ends, where the field is most uniform), essentially all of that field threads through each of its turns. The mutual inductance is then just "flux linkage per unit primary current":

   $M = \frac{N_2\Phi_1}{I_1} = N_2\cdot\frac{\mu_0 N_1 I_1}{l}\cdot\frac{A}{I_1} = \mu_0\frac{N_1 N_2}{l}A.$

   Step-by-Step Solution

   1. Solenoid length $l = 2$ m, diameter $2$ cm so radius $r = 0.01$ m, cross-sectional area $A = \pi r^2 = \pi\times10^{-4} \approx 3.1416\times10^{-4}\ m^2$.
   2. $N_1 = 2000$ turns (primary), $N_2 = 1000$ turns (secondary).
   3. $M = \mu_0\dfrac{N_1 N_2}{l}A = (4\pi\times10^{-7})\times\dfrac{2000\times1000}{2}\times3.1416\times10^{-4}$. …

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2. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.The mutual inductance of a pair of adjacent coils is 5 H. If the current in one coil changes from 0 to 12A in a time of 0.75s, then the change of flux linkage with the other coil is (A) 45 Wb (B) 80 Wb (C) 60 Wb (D) 30 Wb

   ›Reveal solutionSolution

   Mutual inductance directly relates the change in flux linkage in one coil to the change in current in the other — no need for the time interval unless EMF itself is asked.

   Concept and Intuition

   By definition, mutual inductance $M$ is given by $\Delta\Phi_2 = M\,\Delta I_1$, where $\Delta\Phi_2$ is the change in flux linkage in the second coil due to a change $\Delta I_1$ in the current of the first coil. The time over which this change happens only matters if we want the induced EMF ($\varepsilon=M\,dI/dt$); the flux linkage change itself depends only on $\Delta I$.

   Step-by-Step Solution

   1. Given $M=5\,H$, and current changes from 0 to 12 A, so $\Delta I=12\,A$.
   2. Change in flux linkage: $\Delta\Phi = M\Delta I = 5\times12=60\,Wb$. …

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3. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.The coefficient of mutual induction between the primary and secondary coil of a transformer is 0.4 H. When the current in the primary coil changes at the rate of 10 As$^{-1}$, then the induced emf in the secondary will be (A) 1.0 V (B) 4.0 V (C) 2.2 V (D) 3.1 V

   ›Reveal solutionSolution

   Direct use of $\varepsilon=M\,dI/dt$ gives an induced emf of 4.0 V in the secondary.

   Concept and Intuition

   Mutual inductance $M$ relates the rate of change of current in one coil (primary) to the emf induced in a nearby coil (secondary), via $\varepsilon_2=M\dfrac{dI_1}{dt}$ — a direct statement of Faraday's law applied to two magnetically-coupled coils.

   Step-by-Step Solution

   1. Given $M=0.4$ H, $\dfrac{dI}{dt}=10\ \text{A/s}$. …

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4. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.Two concentric loops of radii, 'r' and 'R' are in same plane such that $R \gg r$. If a current I is maintained in the loop of radius 'r', then the magnetic flux associated with the loop of radius 'R' due to this current is (A) $\dfrac{\pi\mu_0 I r^2}{R}$ (B) $\dfrac{\pi\mu_0 I r}{2}$ (C) $\dfrac{\pi\mu_0 I R^2}{2r}$ (D) $\dfrac{\pi\mu_0 I R}{2}$

   ›Reveal solutionSolution

   The flux through the large loop varies as $\mu_0 I r^2 / R$, which corresponds to option (A).

   Concept and Intuition

   Directly integrating the small loop's non-uniform field over the large loop is hard, so use mutual inductance and its reciprocity: the mutual inductance $M$ between the two loops is the same whichever loop carries the current. It is easiest to imagine the current in the LARGE loop, whose field near the centre is uniform, $B=\dfrac{\mu_0 I}{2R}$, and link it through the tiny inner loop.

   Step-by-Step Solution

   1. Field at the common centre due to the large loop: $B=\dfrac{\mu_0 I}{2R}$ (roughly uniform over the small loop since $R\gg r$).
   2. Flux through the small loop: $\Phi_{\text{small}}=B\cdot \pi r^2=\dfrac{\mu_0 I \pi r^2}{2R}$, giving $M=\dfrac{\mu_0 \pi r^2}{2R}$.
   3. By reciprocity, the flux through the LARGE loop when the same current $I$ flows in the small loop is $\Phi=MI=\dfrac{\mu_0 \pi I r^2}{2R}$.
   4. So the flux scales as $\dfrac{\mu_0 \pi I r^2}{R}$ in form — the only option with this $r^2/R$ dependence is (A).

   Common Mistakes …

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5. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.A pair of adjacent coils has a mutual inductance of 80 mH. If the current in one coil changes from 10 A to 25 A in a time of 1.5 s, the change of flux linked with the other coil (A) 0.6 Wb (B) 1.8 Wb (C) 1.2 Wb (D) 0.8 Wb

   ›Reveal solutionSolution

   Mutual inductance directly relates flux linkage to current: $\Delta\Phi = M\Delta I = 1.2$ Wb, independent of how long the change took.

   Concept and Intuition

   Mutual inductance is defined through $\Phi_2 = M I_1$ — the flux linked with coil 2 due to current in coil 1 is directly proportional to that current, with M as the constant of proportionality. So a change in current directly gives a change in linked flux: $\Delta\Phi_2=M\Delta I_1$. Time only enters if we want the induced EMF ($\varepsilon=-\dfrac{d\Phi}{dt}=-M\dfrac{dI}{dt}$), not the flux change itself.

   Step-by-Step Solution

   1. $M=80\ \text{mH}=0.08$ H.
   2. $\Delta I = 25 - 10 = 15$ A.
   3. $\Delta\Phi = M\Delta I = 0.08\times15 = 1.2$ Wb. …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.When two coaxial coils having same current in same direction are brought to each other, then the value of current in both the coils (A) Increases (B) Decreases (C) Remains same (D) Increases in one coil and decreases in other coil

   ›Reveal solutionSolution

   Bringing two same-direction-current coils closer increases the mutual flux linking each coil in

   the same sense as its own flux; by Lenz's law the induced effect opposes this increase, so the

   current in each coil decreases.

   Concept and Intuition

   Two coaxial coils carrying current in the same direction reinforce each other's magnetic flux.

   Moving them closer increases the mutual inductance M, and hence increases the flux each coil

   receives from the other — in addition to its own self-flux, since the currents (and hence the

   fields) point the same way. By Faraday's/Lenz's law, any increase in linked flux induces an EMF

   that opposes the increase, meaning it acts to reduce the current that is producing that flux.

   Step-by-Step Solution

   1. Flux linked with coil 1: $\Phi_1 = L_1 I_1 + M I_2$. As the coils approach, $M$ increases while $I_2$ (same-direction current) stays positive, so the mutual term $MI_2$ rises.
   2. This rising flux induces an EMF $-\dfrac{d\Phi_1}{dt}$ in coil 1 that, by Lenz's law, opposes …

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7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.X and Y are two circuits having coefficient of mutual inductance 3 mH and resistances 10 Ω and 4 Ω respectively. To have induced current $60\times10^{-4}$ A in circuit Y, the amount of current to be changed in circuit X in 0.02 sec is (A) 1.6 A (B) 0.16 A (C) 0.32 A (D) 3.2 A

   ›Reveal solutionSolution

   The induced EMF in Y is found from its current and resistance, then mutual inductance converts that EMF into the rate of current change in X, which is scaled by the given time to get the actual current change.

   Concept and Intuition

   Mutual inductance links a changing current in one circuit to an induced EMF in a nearby circuit: $\varepsilon_Y = M\,dI_X/dt$. We're told the resulting current in Y (and its resistance), which via Ohm's law gives us the EMF in Y. From there, mutual inductance directly gives the rate of change of current in X, and multiplying by the time interval gives the total change in $I_X$.

   Step-by-Step Solution

   1. EMF in Y: $\varepsilon_Y = I_Y R_Y = (60\times10^{-4})(4) = 0.024$ V.
   2. Mutual inductance relation: $\varepsilon_Y = M\dfrac{dI_X}{dt} \Rightarrow \dfrac{dI_X}{dt} = \dfrac{0.024}{3\times10^{-3}} = 8$ A/s. …

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8. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.A changing current in a coil can induce an emf in a nearby coil. This process is (A) self-induction (B) eddy currents (C) mutual induction (D) emf

   ›Reveal solutionSolution

   Induction between two different coils (one coil's changing current inducing emf in another) is mutual induction, as distinct from self-induction (a coil inducing emf in itself).

   Concept and Intuition

   Self-induction refers to a coil opposing changes in its own current by inducing an emf in itself. When the changing current is in one coil and the induced emf appears in a different, nearby coil (via their shared changing magnetic flux), that is mutual induction — the principle behind transformers.

   Step-by-Step Solution

   1. Identify: current changes in coil 1.
   2. Emf is induced in a separate, nearby coil (coil 2), due to the changing mutual flux linkage.
   3. This process, by definition, is mutual induction. …

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9. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
   Q.The correct statement among the following is (A) Inductance is independent of the geometry of the coil and depends on the intrinsic material properties. (B) Inductance depends only on the geometry of the coil and independent of intrinsic material properties. (C) Inductance depends on both the geometry of the coil and intrinsic material properties. (D) Inductance does not depend on the geometry of the coil and intrinsic material properties.

   ›Reveal solutionSolution

   Self-inductance depends on both the coil's shape/turns (geometry) and the permeability of its core material (an intrinsic property) — neither factor alone fully determines it.

   Concept and Intuition

   For a long solenoid, $L=\mu_0\mu_r n^2 A l$ (with $n$ turns per unit length, area $A$, length $l$). This formula makes both dependences explicit: the geometric factors ($n^2$, $A$, $l$) fix how much flux links the coil for a given current, while the material factor ($\mu_r$, the relative permeability of whatever core the coil is wound on — air, iron, ferrite, etc.) scales how much magnetic flux is actually produced for that current. Winding the identical geometry on an iron core instead of an air core dramatically increases $L$, showing the material dependence is real and significant — not just the geometry.

   Step-by-Step Solution

   1. Recall the solenoid inductance formula: $L=\mu_0\mu_r n^2 A l$.
   2. The geometric part ($n^2 A l$) depends purely on the coil's construction — number of turns, cross-sectional area, length.
   3. The material part ($\mu_r$) depends on what the core is made of (air, soft iron, ferrite, etc.) — this is an intrinsic material property.
   4. Since $L$ is a product of both factors, changing either one changes $L$ — so inductance depends on both. …

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10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The self-inductance of a long solenoid of cross-sectional area A, length l and n turns per unit length is given by (A) $\mu_0 nAl$ (B) $\mu_0 n^2Al$ (C) $\mu_0 n^2A^2l$ (D) $\mu_0 n^2\pi A^2l$

    ›Reveal solutionSolution

    This tests deriving the standard formula for the self-inductance of a long solenoid from first principles. The result is $L=\mu_0n^2Al$.

    Concept and Intuition

    Self-inductance is defined via total flux linkage per unit current: $L=\dfrac{N\Phi}{I}$. For a long solenoid, the magnetic field inside is uniform, $B=\mu_0 nI$ (where $n$ is turns per unit length), so the flux through one turn of cross-sectional area $A$ is $\Phi=BA=\mu_0nIA$. Multiplying by the total number of turns $N=nl$ (turns density times length) accounts for the fact that the changing current links every turn, giving an extra factor of $n$.

    Step-by-Step Solution

    1. Magnetic field inside a long solenoid: $B=\mu_0nI$.
    2. Flux through one turn: $\Phi = BA = \mu_0nIA$.
    3. Total number of turns: $N = nl$.
    4. Total flux linkage: $N\Phi = nl\times\mu_0nIA = \mu_0n^2AlI$. …

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11. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
    Q.A circular loop of area $5\ \mathrm{mm^2}$ is placed coaxially inside a long solenoid that has 1000 turns/cm and carries a sinusoidally varying current of amplitude 1 A and angular frequency $\omega$. If the amplitude of the emf induced in the loop is $4.4\times10^{-4}$ V, then the value of '$\omega$' is (A) $350\ \mathrm{rad\,s^{-1}}$ (B) $700\ \mathrm{rad\,s^{-1}}$ (C) $300\pi\ \mathrm{rad\,s^{-1}}$ (D) $200\pi\ \mathrm{rad\,s^{-1}}$

    ›Reveal solutionSolution

    Uses Faraday's law inside a long solenoid to solve for the driving angular frequency from the induced-emf amplitude.

    Concept and Intuition

    Deep inside a long solenoid, the magnetic field is uniform over the loop's small cross-section, $B(t)=\mu_0 n I(t)$. The flux linking the coaxial loop is simply $\Phi=BA$, and Faraday's law gives the induced emf as $-d\Phi/dt$. Differentiating a sinusoidal current brings down a factor of $\omega$, which is what lets us solve for it.

    Step-by-Step Solution

    1. $I(t)=I_0\sin\omega t \Rightarrow B(t)=\mu_0 n I_0\sin\omega t$.
    2. $\Phi(t)=A\mu_0 n I_0\sin\omega t$.
    3. $\varepsilon(t)=-\dfrac{d\Phi}{dt}=-A\mu_0 n I_0\omega\cos\omega t$; amplitude $\varepsilon_0=A\mu_0 n I_0\omega$.
    4. Convert $n=1000\,\mathrm{turns/cm}=10^5\,\mathrm{turns/m}$, $A=5\,\mathrm{mm^2}=5\times10^{-6}\,\mathrm{m^2}$. …

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12. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.A power transmission line feeds input voltage 2500 V to a step down transformer with its primary winding having 3000 turns. The number of turns in the secondary in order to get output voltage of 220 V is (A) 264 (B) 288 (C) 312 (D) 244

    ›Reveal solutionSolution

    This tests the turns-ratio relation for an ideal transformer. $N_s/N_p = V_s/V_p$, giving $N_s = 264$.

    Concept and Intuition

    An ideal transformer conserves power and the ratio of secondary to primary turns equals the ratio of secondary to primary voltages: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$. A step-down transformer has fewer turns on the secondary than the primary, exactly matching the fact that $V_s < V_p$ here (220 V < 2500 V).

    Step-by-Step Solution

    1. Given: $V_p = 2500$ V, $N_p = 3000$, $V_s = 220$ V.
    2. Use $\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}$.
    3. $N_s = N_p \times \dfrac{V_s}{V_p} = 3000 \times \dfrac{220}{2500}$. …

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