Q.A long solenoid $S$ has $n$ turns per metre, with diameter $a$. At the centre of this coil, we place a smaller coil of $N$ turns and diameter $b$ (where $b < a$). If the current in the solenoid increases linearly with time, what is the induced emf appearing in the smaller coil? Plot a graph showing the nature of variation in emf, if the current varies as a function of $mt^2 + C$.

Showing the 12 most recent of 17 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.When a current of 2.5 A passes through the primary coil of a transformer of 200 number turns, the magnetic flux linked with the secondary coil having 400 turns is $600 \times 10^{-6} \mathrm{~T} \mathrm{~m}^2$. Find the induced emf in the secondary coil, when the current in the primary coil increases at a rate of $0.2 \mathrm{As}^{-1}$ (A) $1.92 \times 10^{-2} V$ (B) $1.92 \times 10^{-4} V$ (C) $0.92 \times 10^{-4} V$ (D) $0.92 \times 10^{-2} V$

   ›Reveal solutionSolution

   The induced emf in the secondary coil is found using mutual inductance, which relates the flux in the secondary to the current in the primary. The result is $1.92 \times 10^{-2} \, \text{V}$, corresponding to option (A).

   The key concept here is mutual inductance. When the current in the primary coil changes, it changes the magnetic flux through the secondary coil. By Faraday’s law, this changing flux induces an emf in the secondary. The mutual inductance $M$ links the two coils: it tells us how much flux in the secondary is produced per unit current in the primary. Once we know $M$, the induced emf is simply $M$ times the rate of change of primary current.

   Let’s work through it step by step.

   1. Understand the given data

      • Primary current: $I_p = 2.5 \, \text{A}$ (steady value, used to find flux linkage)
      • Primary turns: $N_p = 200$
      • Secondary turns: $N_s = 400$
      • Flux linked with secondary at that current: $\Phi_s = 600 \times 10^{-6} \, \text{T m}^2$
      • Rate of change of primary current: $\frac{dI_p}{dt} = 0.2 \, \text{A/s}$ We need the induced emf in the secondary.

   2. Find the mutual inductance $M$

      Mutual inductance is defined by:

   $N_s \Phi_s = M I_p$

   Here, $N_s \Phi_s$ is the total flux linkage in the secondary due to the primary current $I_p$. So:

   $M = \frac{N_s \Phi_s}{I_p} = \frac{400 \times 600 \times 10^{-6}}{2.5}$

   Calculate:

   $400 \times 600 = 240000, \quad \text{so } 240000 \times 10^{-6} = 0.24$

   Then:

   $M = \frac{0.24}{2.5} = 0.096 \, \text{H}$

   3. Apply Faraday’s law for the secondary emf The induced emf in the secondary is: $\mathcal{E}_s = -M \frac{dI_p}{dt}$ …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.An AC generator having 400 turns and an area of cross section of $2 \times 10^{-3} \mathrm{~m}^2$ rotates with an angular speed of $200 \pi \mathrm{rad} \mathrm{s}^{-1}$ in a uniform magnetic field of strength 0.4 T . The generator is connected to the primary of an ideal transformer having 500 turns in the primary and 2000 turns in the secondary. The secondary is connected to a $400 \Omega$ resistive load. What is the rms current in the secondary of the transformer? Assume, the transformer is ideal and the resistance of the coil is negligible (A) 2.84 A (B) 14.2 A (C) 1.41 A (D) 28.4 A

   ›Reveal solutionSolution

   Working from the generator's peak emf through the transformer's turns ratio to the secondary current gives about 1.42 A, matching option (C) (1.41 A).

   Step-by-step solution

   1. Peak emf of the generator.

   $\mathcal{E}_0 = NBA\omega = 400 \times 0.4 \times (2\times10^{-3}) \times 200\pi = 64\pi\ \text{V} \approx 201.06\ \text{V}$

   2. RMS voltage at the primary.

   $V_{\text{rms,primary}} = \frac{\mathcal{E}_0}{\sqrt2} = \frac{64\pi}{\sqrt2} \approx 142.17\ \text{V}$

   3. RMS voltage at the secondary. Turns ratio $N_s/N_p = 2000/500 = 4$:

   $V_{\text{rms,secondary}} = 142.17 \times 4 \approx 568.69\ \text{V}$

   4. RMS current through the 400 Ω load. $I_{\text{rms}} = \frac{568.69}{400} \approx 1.42\ \text{A}$ …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.A current of 2 A is passed through the primary coil. The total flux linked with the secondary coil, which is closely wound over the primary is $2000 \times 10^{-6}$ weber. What is the induced emf in the secondary if the current through the primary increases at a rate of $0.2 \mathrm{As}^{-1}$ ? (A) $2 \times 10^{-4} V$ (B) $4 \times 10^{-4} V$ (C) $1 \times 10^{-4} V$ (D) $8 \times 10^{-4} V$

   ›Reveal solutionSolution

   The induced emf in the secondary is found using mutual inductance: $M = \frac{\Phi_s}{I_p}$ and then $\mathcal{E}_s = M \frac{dI_p}{dt}$. The result is $2 \times 10^{-4} \, \text{V}$, so the correct option is (A).

   The key idea here is mutual inductance. When current changes in the primary coil, the magnetic flux through the secondary changes, inducing an emf. The mutual inductance $M$ links the flux in the secondary to the current in the primary: $\Phi_s = M I_p$. Once we know $M$, the induced emf is simply $\mathcal{E}_s = M \frac{dI_p}{dt}$. No need for complicated integration — just direct proportionality.

   Let’s work through it step by step.

   1. Find the mutual inductance $M$. The total flux linked with the secondary coil when the primary carries a steady current $I_p = 2 \, \text{A}$ is given as $\Phi_s = 2000 \times 10^{-6} \, \text{Wb}$. By definition, $\Phi_s = M I_p$, so

   $M = \frac{\Phi_s}{I_p} = \frac{2000 \times 10^{-6}}{2} = 1000 \times 10^{-6} = 1 \times 10^{-3} \, \text{H}.$

   2. Use Faraday’s law for the secondary. The induced emf in the secondary is

   $\mathcal{E}_s = -M \frac{dI_p}{dt}.$

   (The sign indicates direction; we care about magnitude here.)

   The rate of change of current is $\frac{dI_p}{dt} = 0.2 \, \text{A/s}$.

   3. Calculate the magnitude of the induced emf. …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.The ratio of the number of turns of the primary coil to the secondary coil of an ideal transformer is $5: 1$. The primary is connected to 220 V supply. What is the output power of the transformer if the output resistance is $11 \Omega$ ? (A) 1760 W (B) 1100 W (C) 176 W (D) 100 W

   ›Reveal solutionSolution

   For an ideal transformer, the voltage ratio equals the turns ratio, so the secondary voltage is 44 V. With an 11 Ω load, the output power is $P = V^2/R = 44^2/11 = 176$ W. The correct option is (C).

   The key idea is that an ideal transformer conserves power and has a voltage ratio exactly equal to its turns ratio. Once we know the secondary voltage, the output power is simply the power dissipated in the load resistor.

   ---

   1. Understand the turns ratio and voltage relationship

   For an ideal transformer, the ratio of primary voltage $V_p$ to secondary voltage $V_s$ equals the ratio of primary turns $N_p$ to secondary turns $N_s$:

   $\frac{V_p}{V_s} = \frac{N_p}{N_s}$

   Here, $N_p : N_s = 5 : 1$, so $\frac{N_p}{N_s} = 5$. The primary is connected to 220 V, so:

   $\frac{220}{V_s} = 5 \quad \Rightarrow \quad V_s = \frac{220}{5} = 44\ \text{V}$

   2. Determine the output power

   The output power is the power dissipated in the load resistor $R = 11\ \Omega$ connected to the secondary. Using the formula $P = \frac{V^2}{R}$:

   $P = \frac{(44)^2}{11} = \frac{1936}{11} = 176\ \text{W}$

   3. Check consistency with ideal transformer properties …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 2000 turns. The output power is delivered at 220 V by the transformer. If the current in the primary of the transformer is 2 A and its efficiency is $80 \%$, the output current would be: (A) 32 A (B) 8 A (C) 16 A (D) 4 A

   ›Reveal solutionSolution

   The key is to use the transformer’s power relationship, accounting for efficiency: output power = efficiency × input power. Input power is $2200\,\text{V} \times 2\,\text{A} = 4400\,\text{W}$, so output power is $0.8 \times 4400 = 3520\,\text{W}$. At 220 V output, current is $3520 / 220 = 16\,\text{A}$. The correct option is (C).

   Concept & Intuition

   A transformer changes voltage and current, but the ideal transformer conserves power (ignoring losses). Real transformers have efficiency $\eta < 1$, meaning output power is a fraction of input power. Here, we are given primary voltage, current, turns, and efficiency — but the turns ratio is actually a red herring because we already have the primary current and voltage directly. The output voltage is given, so we can find output current from the power balance: $P_{\text{out}} = \eta \times P_{\text{in}}$, and $P_{\text{out}} = V_{\text{out}} \times I_{\text{out}}$.

   Step-by-step solution

   1. Find input power Input power to the transformer is simply the product of primary voltage and primary current:

   $P_{\text{in}} = V_p \times I_p = 2200\,\text{V} \times 2\,\text{A} = 4400\,\text{W}.$

   2. Account for efficiency Efficiency $\eta = 80\% = 0.8$ means only 80% of input power is delivered as output power:

   $P_{\text{out}} = \eta \times P_{\text{in}} = 0.8 \times 4400 = 3520\,\text{W}.$

   3. Relate output power to output current Output is at $V_s = 220\,\text{V}$, so:

   $P_{\text{out}} = V_s \times I_s \quad \Rightarrow \quad I_s = \frac{P_{\text{out}}}{V_s} = \frac{3520}{220} = 16\,\text{A}.$

   4. Check the turns ratio (optional sanity check) …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.A long solenoid has 400 turns. When a current of 100 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 mWb . The self-inductance of the solenoid is (A) 1.6 mH (B) 16 mH (C) 16 H (D) 0.16 mH

   ›Reveal solutionSolution

   Self-inductance is the ratio of total flux linkage to current. Here, total flux linkage = (400 turns) × (4 mWb) = 1.6 Wb, and current = 100 A, so $L = 1.6 / 100 = 0.016 \, \text{H} = 16 \, \text{mH}$. The correct option is (B).

   The key concept is self-inductance — it measures how much magnetic flux a coil “links” with itself per unit current. For a solenoid, the total flux linkage (flux through one turn times the number of turns) is directly proportional to the current. The constant of proportionality is $L$.

   Why this approach works:

   We are given the flux per turn and the number of turns, so we can find the total flux linkage. Then, using the definition $L = \frac{N \Phi}{I}$, we get the inductance directly — no need for geometry or permeability.

   1. Identify the given quantities

      • Number of turns: $N = 400$
      • Current: $I = 100 \, \text{A}$
      • Magnetic flux through each turn: $\Phi = 4 \, \text{mWb} = 4 \times 10^{-3} \, \text{Wb}$

   2. Compute the total flux linkage

      Flux linkage $\lambda = N \Phi = 400 \times (4 \times 10^{-3}) = 1.6 \, \text{Wb}$ (or weber-turns).

      This is the total magnetic flux “linking” the entire solenoid.

   3. Apply the definition of self-inductance

   $L = \frac{\lambda}{I} = \frac{1.6 \, \text{Wb}}{100 \, \text{A}} = 0.016 \, \text{H}$

   4. Convert to millihenries …

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7. KCET 2024Set D-21 markMCQ
   Q.A coil having 9 turns carrying a current produces magnetic field $B_1$ at the centre. Now the coil is rewound into 3 turns carrying same current. Then the magnetic field at the centre $B_2 =$ ——— (A) $\dfrac{B_1}{9}$ (B) $9B_1$ (C) $3B_1$ (D) $\dfrac{B_1}{3}$

   ›Reveal solutionSolution

   The wire length is fixed (it's REWOUND, not re-supplied), so fewer turns means a proportionally bigger radius per turn — and field at the centre depends on both N and r.

   Step 1 — Field at the centre of a coil.

   $B = \dfrac{\mu_0 N I}{2r}$

   Step 2 — Conserve the wire length.

   Total wire length $\ell = N \cdot 2\pi r$ stays constant when rewinding, so $r \propto \dfrac{1}{N}$.

   Step 3 — Substitute.

   $B = \dfrac{\mu_0 N I}{2r} \propto \dfrac{N}{1/N} = N^2$

   Step 4 — Apply the ratio. …

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8. KCET 2024Set D-21 markMCQ
   Q.In the figure, a conducting ring of certain resistance is falling towards a current carrying straight long conductor. The ring and conductor are in the same plane. Then the (A) Induced electric current is zero (B) Induced electric current is anticlockwise (C) Induced electric current is clockwise (D) Ring will come to rest

   ›Reveal solutionSolution

   As the ring falls, the magnetic flux through it changes, inducing a current. The induced current is clockwise, and the ring does not come to rest — it continues to fall. The correct option is (C).

   The key here is Lenz's law and the geometry of the magnetic field around a long straight current-carrying wire. The wire carries a steady current (say upward, by convention), so its magnetic field circles around it. In the plane of the ring and wire, the field lines are perpendicular to the radial direction — they go into or out of the page depending on which side of the wire you're on.

   Let’s set the scene: the wire is vertical, carrying current upward. To the right of the wire (where the ring is), the magnetic field points into the page. As the ring falls downward, it moves through a region where the field strength changes — it gets stronger as it gets closer to the wire.

   1. Magnetic flux through the ring

      The field is into the page, so the flux through the ring is negative (or positive, depending on sign convention — what matters is the change). As the ring falls, it moves closer to the wire, so the field strength at every point of the ring increases. Hence the magnitude of the flux (into the page) increases.

   2. Lenz’s law — opposing the change

      Lenz’s law says the induced current will create its own magnetic field that opposes the change in flux. Since the flux into the page is increasing, the induced field must point out of the page to oppose that increase.

   3. Direction of induced current

      To produce a field out of the page inside the ring, the induced current must be clockwise (right-hand rule: curl your fingers in the direction of current, thumb points in the direction of the field inside the loop). Clockwise current gives an outward field.

   4. Is the induced current zero? …

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9. COMEDK 2024Set 2024-A1 markMCQ
   Q.A transformer has 400 turns in its primary winding and 800 turns in its secondary winding. The primary voltage is $20 \mathrm{~V}$ and the load in the secondary is 4 ohm. The current in the primary, assuming it to be an ideal transformer, is (A) 40 A (B) 20 A (C) 4 A (D) 2 A

   ›Reveal solutionSolution

   Ideal transformer conserves power. The 1:2 turns ratio steps voltage up to $V_s = 40$ V, so $I_s = 10$ A, and the primary current steps up inversely to $20$ A — option (B).

   For an ideal transformer the turns ratio fixes the voltage ratio and, by power conservation, the inverse current ratio.

   1. Secondary voltage from the turns ratio ($N_p=400$, $N_s=800$, $V_p=20$ V):

   $V_s = V_p\,\frac{N_s}{N_p} = 20\times\frac{800}{400} = 40\text{ V}.$

   2. Secondary current through the $4\ \Omega$ load: $I_s = \frac{V_s}{R} = \frac{40}{4} = 10\text{ A}.$ …

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10. COMEDK 2024Set 2024-E1 markMCQ
    Q.A transformer of $100 \%$ efficiency has 200 turns in the primary and 40000 turns in the secondary. It is connected to a $220 \mathrm{~V}$ main supply and secondary feeds to a $100 \mathrm{~K} \Omega$ resistance. The potential difference per turn is (A) 11 V (B) 18 V (C) 25 V (D) 1.1 V

    ›Reveal solutionSolution

    Potential difference per turn is the same for both windings of an ideal transformer, $V/N = 220/200 = 1.1$ V/turn.

    For a transformer $\dfrac{V_p}{N_p} = \dfrac{V_s}{N_s}$, so the emf induced per turn is identical in the primary and secondary.

    Primary: $\dfrac{V_p}{N_p} = \dfrac{220}{200} = 1.1$ V per turn. …

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11. COMEDK 2024Set 2024-M1 markMCQ
    Q.A transformer which steps down $330 \mathrm{~V}$ to $33 \mathrm{~V}$ is to operate a device having impedance $110 \Omega$. The current drawn by the primary coil of the transformer is : (A) 0.3 A (B) 0.03 A (C) 3 A (D) 1.5 A

    ›Reveal solutionSolution

    The key idea is that for an ideal transformer, power in the primary equals power in the secondary. Using the turns ratio from the voltage step-down, we find the secondary current, then the primary current. The result is 0.03 A.

    Concept & Intuition

    A transformer doesn’t create power — it transfers it. If it steps voltage down, it must step current up (for an ideal, lossless transformer). The ratio of voltages equals the ratio of turns, and the ratio of currents is the inverse. Here, we know the secondary voltage and the load impedance, so we can find the secondary current. Then, using the voltage ratio, we find the primary current.

    Step-by-step solution

    1. Find the turns ratio from the voltage step-down. The primary voltage is $V_p = 330\ \mathrm{V}$, the secondary voltage is $V_s = 33\ \mathrm{V}$. For an ideal transformer:

    $\frac{V_p}{V_s} = \frac{N_p}{N_s}$

    So the turns ratio is

    $\frac{N_p}{N_s} = \frac{330}{33} = 10$

    This means the primary has 10 times the turns of the secondary.

    2. Find the secondary current using Ohm’s law. The load impedance is $Z = 110\ \Omega$ connected across the secondary.

    $I_s = \frac{V_s}{Z} = \frac{33}{110} = 0.3\ \mathrm{A}$

    3. Relate primary and secondary currents. For an ideal transformer, the power in equals power out:

    $V_p I_p = V_s I_s$

    Therefore,

    $$$$

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12. COMEDK 2024Set 2024-M1 markMCQ
    Q.Select the unit of the coefficient of mutual induction from the following. (A) volt. second / ampere (B) weber. ampere (C) ampere / weber (D) volt. ampere / second

    ›Reveal solutionSolution

    The coefficient of mutual induction (mutual inductance) has units of henry, which is equivalent to volt·second per ampere; thus the correct choice is (A).

    The concept here is mutual inductance — a measure of how effectively a change in current in one coil induces an electromotive force (emf) in a nearby coil. The defining equation is:

    $\mathcal{E}_2 = -M \frac{dI_1}{dt}$

    where $\mathcal{E}_2$ is the induced emf (in volts), $I_1$ is the current in the first coil (in amperes), and $t$ is time (in seconds). Rearranging gives:

    $M = -\frac{\mathcal{E}_2}{dI_1/dt}$

    So the units of $M$ are volts divided by (amperes per second), i.e., volt·second per ampere. This combination is called the henry (H). Now let’s check each option.

    1. Option (A): volt·second / ampere

       This matches exactly the derived unit. So (A) is correct.

    2. Option (B): weber·ampere

       A weber is the unit of magnetic flux (volt·second). Multiplying by ampere gives volt·second·ampere, which is not the same as volt·second/ampere. So (B) is wrong.

    3. Option (C): ampere / weber

       This is the reciprocal of the correct unit. It would correspond to $1/M$, not $M$. So (C) is wrong.

    4. Option (D): volt·ampere / second …

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