Q.Figure 6.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz's law.

Showing the 12 most recent of 27 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A metallic circular loop is placed with its plane perpendicular to a uniform magnetic field of 0.3 T . If the radius of the loop decreases at a constant rate of $2 \mathrm{~mm} \mathrm{~s}^{-1}$, what will be the induced emf in the loop, when the radius of the loop becomes 5 cm . (A) $1.5 \times 10^{-6} V$ (B) $0.75 \times 10^{-6} V$ (C) $1.84 \times 10^{-6} V$ (D) $1.89 \times 10^{-4} V$

   ›Reveal solutionSolution

   The induced emf is found from Faraday’s law: the rate of change of magnetic flux through the loop equals $B \cdot dA/dt$. As the radius shrinks, the area decreases, giving an emf of $1.884 \times 10^{-5}\,\text{V}$, which matches option (D).

   The key concept is Faraday’s law of electromagnetic induction: the induced emf in a loop equals the negative rate of change of magnetic flux through it. Here, the magnetic field is uniform and perpendicular to the loop, so the flux is simply $\Phi = B \cdot A$, where $A = \pi r^2$ is the area. Since $B$ is constant, the only change comes from the shrinking radius. The induced emf is therefore $|\mathcal{E}| = B \cdot \frac{dA}{dt}$. We don’t need to worry about sign for magnitude.

   1. Identify the given data

      • Magnetic field: $B = 0.3\ \text{T}$ (constant, perpendicular to loop)
      • Rate of change of radius: $\frac{dr}{dt} = -2\ \text{mm/s} = -2 \times 10^{-3}\ \text{m/s}$ (negative because radius decreases)
      • Radius at the instant of interest: $r = 5\ \text{cm} = 0.05\ \text{m}$

   2. Relate area change to radius change

      Area of the loop: $A = \pi r^2$

      Differentiate with respect to time:

   $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$

   This is a straightforward chain rule — the area changes because the radius changes.

   3. Plug in the numbers At $r = 0.05\ \text{m}$ and $\frac{dr}{dt} = -2 \times 10^{-3}\ \text{m/s}$:

   $\frac{dA}{dt} = 2\pi (0.05)(-2 \times 10^{-3}) = -2\pi \times 10^{-4}\ \text{m}^2/\text{s}$

   The negative sign indicates area is decreasing.

   4. Apply Faraday’s law Induced emf magnitude:

   $|\mathcal{E}| = \left| -\frac{d\Phi}{dt} \right| = \left| -B \frac{dA}{dt} \right| = B \left| \frac{dA}{dt} \right|$

   So:

   $|\mathcal{E}| = 0.3 \times (2\pi \times 10^{-4}) = 0.6\pi \times 10^{-4}\ \text{V}$

   Using $\pi \approx 3.1416$:

   $|\mathcal{E}| = 0.6 \times 3.1416 \times 10^{-4} = 1.88496 \times 10^{-4}\ \text{V}$

   Rounding gives $1.88 \times 10^{-4}\ \text{V}$, which is $1.89 \times 10^{-4}\ \text{V}$ in the options.

   Watch out

   A common mistake is to forget that the area depends on $r^2$, so the rate of change of area is $2\pi r\, dr/dt$, not $\pi (dr/dt)^2$. Also, watch units: convert mm to m and cm to m consistently.

   Tip

   Notice that the induced emf is proportional to the instantaneous radius $r$, not the initial radius. So you must use $r = 0.05\ \text{m}$ at the moment you want the emf.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A square loop of wire 2.2 cm on each side contains 2 turns and has a total resistance of $0.0002 \Omega$. It is located 22 cm from a long straight current carrying wire. If the current in the straight wire is increased at a steady rate from 20 A to 50 A in 5 s , determine the magnitude of the current induced in the squared loop. (A) 0.26 mA (B) 2.64 mA (C) 0.026 mA (D) 0.0026 mA

   ›Reveal solutionSolution

   The changing current in the straight wire changes the flux through the loop, inducing an emf (Faraday's law); dividing by the resistance gives the current. The result is about $0.026\ \text{mA}$ — option (C).

   Concept & Intuition

   A long straight wire carrying a current creates a magnetic field that circles the wire, with strength $B = \frac{\mu_0 I}{2\pi r}$ at a perpendicular distance $r$. The square loop near the wire has magnetic flux through it. When the wire's current changes, the flux changes, inducing an emf (Faraday's law); the emf drives a current found from Ohm's law. Because the field varies across the loop, the flux is obtained by integration (a uniform-field estimate at the mean distance gives essentially the same answer here).

   Step-by-step solution

   1. Geometry and field. Side $a = 2.2\ \text{cm} = 0.022\ \text{m}$, nearest side $d = 22\ \text{cm} = 0.22\ \text{m}$, $N = 2$ turns, $R = 0.0002\ \Omega$.

   $B(r) = \frac{\mu_0 I}{2\pi r}, \qquad \mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}.$

   2. Flux through one turn (integrating a strip of height $a$, width $dr$):

   $\Phi_1 = \int_{d}^{d+a} \frac{\mu_0 I}{2\pi r}\, a\, dr = \frac{\mu_0 I a}{2\pi} \ln\!\left(\frac{d+a}{d}\right).$

   3. Total flux for $N$ turns:

   $\Phi_{\text{total}} = N \Phi_1 = \frac{\mu_0 I a}{\pi} \ln\!\left(\frac{d+a}{d}\right).$

   4. Induced emf. With $\dfrac{dI}{dt} = \dfrac{50-20}{5} = 6\ \text{A/s}$:

   $|\mathcal{E}| = \frac{\mu_0 a}{\pi} \ln\!\left(\frac{d+a}{d}\right)\frac{dI}{dt}.$

   5. Plug in numbers.

   $\frac{d+a}{d} = \frac{0.242}{0.22} = 1.1, \qquad \ln(1.1) \approx 0.09531,$

   $|\mathcal{E}| = 4\times10^{-7}\times 0.022 \times 0.09531 \times 6 \approx 5.03\times10^{-9}\ \text{V}.$

   6. Induced current.

   $I_{\text{induced}} = \frac{|\mathcal{E}|}{R} = \frac{5.03\times10^{-9}}{0.0002} \approx 2.5\times10^{-5}\ \text{A} = 0.025\ \text{mA} \approx 0.026\ \text{mA}.$

   Watch out

   A common error is off-by-a-power-of-ten arithmetic — the answer is $0.026\ \text{mA}$, not $0.26\ \text{mA}$.

   Tip

   Notice the $\pi$ cancels neatly — simplify symbolic expressions before plugging numbers.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.A circular coil of radius 7 cm and 40 turns is rotated about its vertical diameter with an angular speed of 40 radians per second in a uniform horizontal magnetic field of $4 \times 10^{-2} T$. What is the maximum current induced in the coil if the resistance of the coil is $11 \Omega$ ? (A) 90 mA (B) 0.9 mA (C) 9 mA (D) 0.09 mA

   ›Reveal solutionSolution

   The maximum induced current is found from the peak emf generated by a rotating coil in a magnetic field: $\mathcal{E}_{\text{max}} = N B A \omega$, then $I_{\text{max}} = \mathcal{E}_{\text{max}} / R$. The result is 0.09 A = 90 mA, so option (A) is correct.

   Concept & Intuition

   When a coil rotates in a uniform magnetic field, the magnetic flux through it changes sinusoidally. Faraday’s law tells us that a changing flux induces an emf. The maximum emf occurs when the plane of the coil is parallel to the field (flux changing fastest). For a coil of $N$ turns, area $A$, rotating at angular speed $\omega$ in a field $B$, the peak induced emf is $\mathcal{E}_{\text{max}} = N B A \omega$. Then Ohm’s law gives the peak current.

   Step-by-step solution

   1. Identify the given quantities

      • Radius $r = 7\ \text{cm} = 0.07\ \text{m}$
      • Number of turns $N = 40$
      • Angular speed $\omega = 40\ \text{rad/s}$
      • Magnetic field $B = 4 \times 10^{-2}\ \text{T}$
      • Resistance $R = 11\ \Omega$

   2. Compute the area of the coil

      The coil is circular, so

   $A = \pi r^2 = \pi (0.07)^2 = \pi \times 0.0049 = 0.0049\pi\ \text{m}^2$

   3. Find the maximum induced emf For a coil rotating in a uniform field, the flux is $\Phi = N B A \cos(\omega t)$. The induced emf is $\mathcal{E} = -d\Phi/dt = N B A \omega \sin(\omega t)$. The maximum value is

   $\mathcal{E}_{\text{max}} = N B A \omega$

   Substitute:

   $\mathcal{E}_{\text{max}} = 40 \times (4 \times 10^{-2}) \times (0.0049\pi) \times 40$

   First, $40 \times 40 = 1600$, so

   $\mathcal{E}_{\text{max}} = 1600 \times 4 \times 10^{-2} \times 0.0049\pi$

   $1600 \times 4 \times 10^{-2} = 64$, so

   $\mathcal{E}_{\text{max}} = 64 \times 0.0049\pi = 0.3136\pi$

   Using $\pi \approx 3.1416$,

   $\mathcal{E}_{\text{max}} \approx 0.3136 \times 3.1416 \approx 0.985\ \text{V}$

   4. Calculate the maximum induced current By Ohm’s law,

   $I_{\text{max}} = \frac{\mathcal{E}_{\text{max}}}{R} = \frac{0.985}{11} \approx 0.0895\ \text{A}$

   That is about $0.09\ \text{A} = 90\ \text{mA}$.

   Tip

   Notice that the radius was given in cm — a classic trap is to forget converting to metres. Always check units before plugging into formulas.

   Watch out

   Some might mistakenly use the diameter instead of radius, or forget to multiply by $N$ (number of turns). Both errors lead to answers that match the distractors, so careful step-by-step work is essential.

   5. Match with the options $90\ \text{mA}$ corresponds to option (A).

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. KCET 2026Set C21 markMCQ
   Q.In Faraday-Henry's experiment, a coil is connected to a galvanometer. For the deflection of pointer in the galvanometer, which of the following statement/s is/are WRONG? The pointer in the galvanometer deflects -

   (a) When the bar magnet is moved towards the stationary coil along its axis

   (b) When the bar magnet is moved away from the stationary coil along its axis

   (c) When the coil is moved towards the stationary bar magnet along its axis

   (d) When the coil and the magnet are moved without relative motion between them (A) a and b (B) b and c (C) a, b and c (D) Only d

   ›Reveal solutionSolution

   By Faraday's law, an EMF (and hence a galvanometer deflection) is induced only when the magnetic flux through the coil changes — which requires relative motion between the coil and the magnet.

   Step 1 — Check (a) and (b): magnet moved relative to stationary coil

   When the bar magnet is moved towards or away from the stationary coil along its axis, the flux linked with the coil changes with time, inducing an EMF and deflecting the galvanometer. Both (a) and (b) correctly describe deflection.

   Step 2 — Check (c): coil moved relative to stationary magnet

   Moving the coil towards the stationary magnet also changes the relative position (and hence the flux through the coil) — the same relative motion that matters, whichever object physically moves. This correctly causes deflection too.

   Step 3 — Check (d): both moved together, no relative motion

   If the coil and the magnet are moved together with NO relative motion between them, the flux linked with the coil never changes (the coil's position relative to the magnet's field stays fixed). No change in flux means no induced EMF, so there is NO deflection. The statement in (d) — implying deflection still occurs — is therefore the WRONG one.

   Conclusion

   • (a) — correct, deflection occurs.
   • (b) — correct, deflection occurs.
   • (c) — correct, deflection occurs.
   • (d) — the wrong statement; no deflection occurs since there is no relative motion.

   ✓Final answer

   The correct option is (D) — Only (d) is wrong.

5. KCET 2025Set D-41 markMCQ
   Q.In the following circuit, the terminal voltage across the cell is (A) $1.68~\text{V}$ (B) $1.95~\text{V}$ (C) $2.71~\text{V}$ (D) $0.52~\text{V}$

   ›Reveal solutionSolution

   Find the loop current from $I = \dfrac{\varepsilon}{R+r}$, then subtract the internal drop: $V = \varepsilon - Ir$.

   Step 1 — What "terminal voltage" means.

   A real cell is an ideal emf $\varepsilon$ in series with an internal resistance $r$. When current $I$ flows, part of the emf is lost pushing that current through $r$ itself. What is left across the cell's terminals — and what the external circuit actually receives — is the terminal voltage:

   $\boxed{V = \varepsilon - I r}$

   Equivalently, $V$ is just the potential difference across the external resistor, $V = IR$. (Both routes must give the same number — a useful self-check.)

   Step 2 — Read the circuit values.

   From the diagram:

   $\varepsilon = 2\ \text{V}, \qquad r = 0.1\ \Omega, \qquad R = 3.9\ \Omega$

   The cell and the resistor form one closed series loop, so the same current flows through both.

   Step 3 — Current from Ohm's law for the complete circuit.

   The total resistance in the loop is $R + r$:

   $I = \frac{\varepsilon}{R + r} = \frac{2}{3.9 + 0.1} = \frac{2}{4.0} = 0.5\ \text{A}$

   Step 4 — The internal ("lost") volts.

   $I r = (0.5)(0.1) = 0.05\ \text{V}$

   Step 5 — The terminal voltage.

   $V = \varepsilon - I r = 2 - 0.05 = 1.95\ \text{V}$

   Step 6 — Cross-check via the external resistor.

   $V = I R = (0.5)(3.9) = 1.95\ \text{V} \quad\checkmark$

   The two independent routes agree, confirming the arithmetic.

   Sanity note: because $r \ll R$, the terminal voltage should come out only slightly below the emf of 2 V — which 1.95 V is. Option (C), $2.71\ \text{V}$, is impossible: a discharging cell can never deliver more than its emf. Options (A) and (D) are far too low for such a small internal resistance.

   ✓Final answer

   The correct option is (B) — $1.95\ \text{V}$.

   ANSWER: B

6. KCET 2025Set D-41 markMCQ
   Q.A solenoid is 1m long and 4 cm in diameter. It has five layers of windings of 1000 turns each and carries a current of 7A. The magnetic field at the centre of the solenoid is (A) $0.4396 \times 10^{-5}$ T (B) $4.396 \times 10^{-2}$ T (C) $43.96 \times 10^{-2}$ T (D) $439.6$ T

   ›Reveal solutionSolution

   Add the layers to get the total turns, divide by the length for $n$, and apply $B = \mu_0 n I$ — the diameter is irrelevant.

   Step 1 — The formula and why it applies.

   For a long solenoid (length $\gg$ diameter), Ampère's law applied to a rectangular loop straddling the winding gives a field that is uniform inside and essentially zero outside:

   $\boxed{B = \mu_0 n I}$

   where $n$ is the number of turns per unit length and $I$ the current. Check the geometry: length $= 1\ \text{m}$, diameter $= 4\ \text{cm} = 0.04\ \text{m}$. The ratio is $25:1$, comfortably "long", so the formula is valid at the centre.

   Note what does not appear: the diameter. The interior field of an ideal solenoid is independent of its radius — the $4\ \text{cm}$ is given only as a distractor (and to let you verify the long-solenoid condition).

   Step 2 — Count the total turns.

   The windings are in five layers of 1000 turns each. Every layer wraps the same $1\ \text{m}$ length and every turn of every layer carries the same current in the same sense, so their contributions simply add:

   $N = 5 \times 1000 = 5000 \text{ turns}$

   Step 3 — Turns per unit length.

   $n = \frac{N}{L} = \frac{5000}{1\ \text{m}} = 5000\ \text{turns m}^{-1}$

   Step 4 — Substitute.

   With $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$ and $I = 7\ \text{A}$:

   $B = (4\pi \times 10^{-7})(5000)(7)$

   First combine the numbers:

   $5000 \times 7 = 35\,000 = 3.5 \times 10^{4}$

   $B = 4\pi \times 10^{-7} \times 3.5\times10^{4} = 4 \times 3.1416 \times 3.5 \times 10^{-3}$

   $4 \times 3.1416 = 12.566; \qquad 12.566 \times 3.5 = 43.98$

   $B = 43.98 \times 10^{-3}\ \text{T} = 4.398\times10^{-2}\ \text{T}$

   Step 5 — Match to an option.

   $4.398\times10^{-2}\ \text{T}$ matches option (B), $4.396\times10^{-2}\ \text{T}$, to within rounding of $\pi$.

   Step 6 — Why the others fail.

   • (A) $0.4396\times10^{-5}\ \text{T}$ — smaller by a factor of $\sim10^4$; this is what you get if the five layers are ignored and a power of ten is dropped.
   • (C) $43.96\times10^{-2}\ \text{T}$ — ten times too large (a misplaced decimal: $43.98\times10^{-3}$, not $\times10^{-2}$).
   • (D) $439.6\ \text{T}$ — physically absurd; fields of that size exist only in the most extreme pulsed laboratory magnets, not in a hand-wound solenoid carrying $7\ \text{A}$.

   ✓Final answer

   The correct option is (B) — $4.396\times10^{-2}\ \text{T}$.

   ANSWER: B

7. COMEDK 2025Set 2025-A1 markMCQ
   Q.A circular coil of area $2 \sqrt{2} \mathrm{~cm}^2$ and resistance $2 \boldsymbol{\Omega}$ is arranged vertically in the east -west direction. A uniform magnetic field 0.2 T is set up across the plane in the north to east direction. Now the magnetic field is removed at a steady rate in 0.4 s What is the current developed in the coil? (A) $0.5 \times 10^{-3} A$ (B) $0.5 \times 10^{-4} A$ (C) $1 \times 10^{-4} A$ (D) $0.5 \times 10^{-5} A$

   ›Reveal solutionSolution

   The induced current is found from Faraday’s law: the rate of change of magnetic flux through the coil equals the induced emf, which divided by the resistance gives the current. The result is $0.5 \times 10^{-4} \, \text{A}$, so option (B) is correct.

   Concept & Intuition

   When the magnetic field through a coil changes, an emf is induced. Here the field is uniform initially and then removed steadily, so the flux changes at a constant rate. The key is to find the component of the magnetic field perpendicular to the coil’s plane — only that component contributes to the flux. The coil is vertical and oriented east–west; the field is directed from north to east, so we must resolve it into components relative to the coil’s normal.

   Step-by-step solution

   1. Identify the geometry

      The coil is vertical and its plane faces east–west. That means the normal to the coil’s plane points either north or south (perpendicular to the plane). Let’s take the normal pointing north. The magnetic field of $0.2 \, \text{T}$ is directed from north to east — that is, at $45^\circ$ east of north. So the angle between the field direction and the coil’s normal (north) is $45^\circ$.

   2. Compute the initial magnetic flux

      Magnetic flux $\Phi = B A \cos\theta$, where $\theta$ is the angle between the field and the normal.

      Area $A = 2\sqrt{2} \, \text{cm}^2 = 2\sqrt{2} \times 10^{-4} \, \text{m}^2$.

      So

   $\Phi_i = (0.2) \times (2\sqrt{2} \times 10^{-4}) \times \cos 45^\circ.$

   Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$,

   $\Phi_i = 0.2 \times 2\sqrt{2} \times 10^{-4} \times \frac{1}{\sqrt{2}} = 0.2 \times 2 \times 10^{-4} = 4 \times 10^{-5} \, \text{Wb}.$

   3. Final flux

      The field is removed completely, so final flux $\Phi_f = 0$.

   4. Induced emf

      By Faraday’s law, the magnitude of the induced emf is

   $|\mathcal{E}| = \left|\frac{\Delta \Phi}{\Delta t}\right| = \frac{4 \times 10^{-5} - 0}{0.4} = 1 \times 10^{-4} \, \text{V}.$

   5. Induced current Using Ohm’s law, $I = \mathcal{E} / R$. Resistance $R = 2 \, \Omega$.

   $I = \frac{1 \times 10^{-4}}{2} = 0.5 \times 10^{-4} \, \text{A}.$

   Watch out

   A common mistake is to forget that only the component of the magnetic field perpendicular to the coil’s plane contributes to flux. Using the full field strength without the $\cos 45^\circ$ factor would give double the correct flux and hence double the current.

   Tip

   Notice that the $\sqrt{2}$ in the area cancels neatly with the $\cos 45^\circ = 1/\sqrt{2}$, simplifying the arithmetic. Always look for such cancellations in problems with $45^\circ$ angles.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. COMEDK 2025Set 2025-E1 markMCQ
   Q.A rectangular coil of 250 turns has an average area of $20 \mathrm{~cm} \times 15 \mathrm{~cm}$. The coil rotates with a speed of 60 cycles per second in a uniform magnetic field of $2 \times 10^{-2} \mathrm{~T}$ about an axis perpendicular to the field. The peak value of the induced emf is: (A) $36 \pi$ volt (B) $30 \pi$ volt (C) $18 \pi$ volt (D) $12 \pi$ volt

   ›Reveal solutionSolution

   Peak emf $\varepsilon_0 = NBA\omega$ with $\omega=2\pi f$: $250\times(2\times10^{-2})\times(0.03)\times120\pi = 18\pi$ volt.

   Given: $N=250$ turns, area $A = 20\text{ cm}\times15\text{ cm} = 300\text{ cm}^2 = 300\times10^{-4}\,\text{m}^2 = 0.03\,\text{m}^2$, frequency $f = 60\,\text{Hz}$, $B = 2\times10^{-2}\,\text{T}$.

   Angular frequency:

   $\omega = 2\pi f = 2\pi(60) = 120\pi\ \text{rad/s}$

   Peak (maximum) induced emf of a rotating coil:

   $\varepsilon_0 = N B A \omega$

   $\varepsilon_0 = 250 \times (2\times10^{-2}) \times 0.03 \times 120\pi$

   Step by step:

   $250 \times 0.02 = 5,\qquad 5 \times 0.03 = 0.15,\qquad 0.15 \times 120 = 18$

   $\varepsilon_0 = 18\pi\ \text{volt}$

   ✓Final answer

   Peak induced emf $= 18\pi$ volt — option (C).

9. COMEDK 2025Set 2025-M1 markMCQ
   Q.The magnetic flux $\phi$ through a stationary loop of wire having a resistance R varies with time as $\phi=4 t^2+3 t$. The average emf and total charge flowing in the loop in the time interval $t=0$ to $t=\tau$ respectively are (A) $\tau+3, \quad \frac{4 \tau^2+3 \tau}{R}$ (B) $3 \tau+4, \quad \frac{4 \tau^2+3 \tau}{R}$ (C) $4 \tau+3, \quad \frac{4 r+3}{R}$ (D) $4 \tau+3, \quad \frac{4 \tau^2+3 \tau}{R}$

   ›Reveal solutionSolution

   The average emf is the total change in flux divided by the time interval, giving $4\tau + 3$; the total charge is the total change in flux divided by resistance, giving $\frac{4\tau^2 + 3\tau}{R}$. The correct option is (D).

   The key idea here is that average emf is not the same as instantaneous emf. For a stationary loop, Faraday’s law gives the instantaneous induced emf as $\mathcal{E} = -\frac{d\phi}{dt}$. But the average emf over a time interval is simply the total change in magnetic flux divided by that time interval — no calculus needed for the average. Similarly, the total charge that flows is linked to the total change in flux via $Q = \frac{\Delta \phi}{R}$, because the induced current is $I = \mathcal{E}/R$ and charge is the integral of current over time.

   Let’s work through it step by step.

   1. Find the total change in flux from $t = 0$ to $t = \tau$. At $t = 0$: $\phi(0) = 4(0)^2 + 3(0) = 0$. At $t = \tau$: $\phi(\tau) = 4\tau^2 + 3\tau$. So the total change is

   $\Delta \phi = \phi(\tau) - \phi(0) = 4\tau^2 + 3\tau.$

   2. Average emf is defined as the total change in flux divided by the total time (since emf is the rate of change of flux, its average over an interval is just the net change per unit time).

   $\text{Average emf} = \frac{\Delta \phi}{\Delta t} = \frac{4\tau^2 + 3\tau}{\tau} = 4\tau + 3.$

   Notice this is not the same as evaluating $-\frac{d\phi}{dt}$ at some midpoint — it’s a pure average.

   3. Total charge flowing through the loop: The induced current at any instant is $I(t) = \frac{\mathcal{E}(t)}{R} = -\frac{1}{R}\frac{d\phi}{dt}$. The total charge is the integral of current over time:

   $Q = \int_0^\tau I(t)\,dt = \int_0^\tau -\frac{1}{R}\frac{d\phi}{dt}\,dt = -\frac{1}{R} \int_{\phi(0)}^{\phi(\tau)} d\phi = -\frac{1}{R} \bigl(\phi(\tau) - \phi(0)\bigr).$

   The minus sign just indicates direction; the magnitude of charge is

   $Q = \frac{|\Delta \phi|}{R} = \frac{4\tau^2 + 3\tau}{R}.$

   Tip

   A neat shortcut: For any induced emf, the total charge that flows depends only on the total change in flux and the resistance, not on how fast the flux changes. So you never need to differentiate — just compute $\Delta \phi / R$.

   4. Match with the options: Average emf = $4\tau + 3$ and total charge = $\frac{4\tau^2 + 3\tau}{R}$. This matches option (D) exactly.

   Watch out

   A common mistake is to compute the instantaneous emf at $t = \tau$ (which is $8\tau + 3$) and mistake that for the average. Always remember: average emf = total flux change / total time, not the derivative at the endpoint.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

10. COMEDK 2024Set 2024-A1 markMCQ
    Q.A wire ' 1 ' $\mathrm{cm}$ long bent into a circular loop is placed perpendicular to the magnetic field of flux density '$B^{\prime} W b \mathrm{~m}^{-2}$. Within $0.1 \mathrm{sec}$, the loop is changed into a square of side '$a$' $\mathrm{cm}$ and flux density is doubled. The value of e.m.f. induced is (A) $\frac{-5 B\left(8 л \mathrm{a}^2-\mathrm{l}^2\right)}{2}$ (B) $\frac{5 B\left(8 л a^2-l^2\right)}{2 л}$ (C) $\frac{-5 B\left(8 л a^2-l^2\right)}{2 л}$ (D) $\frac{-5 B\left(8 a^2-l^2\right)}{2 \pi}$

    ›Reveal solutionSolution

    The induced emf is found from Faraday’s law by computing the change in magnetic flux through the loop as its shape changes from a circle to a square and the field doubles. The correct expression is option (C).

    The key idea is that the induced emf depends only on the rate of change of magnetic flux through the loop. Here, two things change simultaneously: the shape (which alters the area) and the magnetic field strength. The wire length is fixed, so the perimeter of the circle equals the perimeter of the square, giving a relation between the original length $l$ and the square’s side $a$. We then compute the flux before and after, take the difference, divide by the time interval, and include the negative sign from Lenz’s law.

    1. Relate the wire length to the square’s side. The wire is $l$ cm long. Bent into a circle, its circumference is $l$. Bent into a square of side $a$ cm, the perimeter is $4a$. Since the same wire is used,

    $l = 4a \quad \Rightarrow \quad a = \frac{l}{4}.$

    This is crucial: the problem gives both $l$ and $a$ as symbols, but they are not independent — the square’s side is determined by the wire length. However, the answer choices treat $l$ and $a$ as separate variables, so we must keep both in the final expression, using the relation only to simplify if needed.

    2. Find the initial area (circular loop). Circumference $= l = 2\pi r$, so radius $r = \frac{l}{2\pi}$. Area of circle:

    $A_\text{circle} = \pi r^2 = \pi \left(\frac{l}{2\pi}\right)^2 = \frac{l^2}{4\pi}.$

    3. Find the final area (square loop). Side $a$ is given, so area:

    $A_\text{square} = a^2.$

    (Using the relation $a = l/4$ would give $a^2 = l^2/16$, but we keep $a$ as is for the answer.)

    4. Compute the initial and final magnetic flux. Flux $\Phi = B A$ (since field is perpendicular). Initial flux:

    $\Phi_i = B \cdot \frac{l^2}{4\pi}.$

    Final flux: field doubles to $2B$, area becomes $a^2$:

    $\Phi_f = (2B) \cdot a^2 = 2B a^2.$

    5. Find the change in flux.

    $\Delta \Phi = \Phi_f - \Phi_i = 2B a^2 - \frac{B l^2}{4\pi}.$

    Factor $B$:

    $\Delta \Phi = B \left( 2a^2 - \frac{l^2}{4\pi} \right).$

    6. Apply Faraday’s law. Induced emf $\mathcal{E} = -\frac{\Delta \Phi}{\Delta t}$. Time interval $\Delta t = 0.1$ s.

    $\mathcal{E} = -\frac{B}{0.1} \left( 2a^2 - \frac{l^2}{4\pi} \right) = -10B \left( 2a^2 - \frac{l^2}{4\pi} \right).$

    7. Simplify to match the answer choices. Write as a single fraction:

    $\mathcal{E} = -10B \left( \frac{8\pi a^2 - l^2}{4\pi} \right) = -\frac{10B}{4\pi} (8\pi a^2 - l^2) = -\frac{5B}{2\pi} (8\pi a^2 - l^2).$

    This is exactly option (C).

    Watch out

    A common mistake is to forget that the wire length is fixed, so $l = 4a$. But here the answer choices treat $l$ and $a$ as independent symbols — you must not substitute $a = l/4$ unless the final expression simplifies to one of the options. In this case, substituting would give a different form not listed, so keep both variables.

    Tip

    Notice that the factor $0.1$ s in the denominator becomes $10$ when moved to the numerator. Always check the time interval: here it’s $0.1$ s, so the rate factor is $10$.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

11. COMEDK 2024Set 2024-A1 markMCQ
    Q.In an inductor of self-inductance $2 \mathrm{~mH}$, current changes with time (in sec) according to the relation, $I=(3 t^2-3 t+8) A$. The emf becomes zero at (A) 2 sec (B) 6 sec (C) 0.5 sec (D) 5 sec

    ›Reveal solutionSolution

    The induced emf in an inductor is given by $\mathcal{E} = -L \frac{dI}{dt}$. Setting this to zero means $\frac{dI}{dt}=0$. Differentiating $I=3t^2-3t+8$ gives $6t-3=0$, so $t=0.5$ s. The correct option is (C).

    The key idea is that the emf across an inductor depends on the rate of change of current, not the current itself. When the current is at a turning point (maximum or minimum), its derivative is zero, so the induced emf vanishes. This is a classic application of Faraday's law in the context of a single inductor.

    1. Recall the fundamental relation for an inductor.

       The self-induced emf in an inductor is $\mathcal{E} = -L \frac{dI}{dt}$, where $L$ is the inductance and $\frac{dI}{dt}$ is the instantaneous rate of change of current. The negative sign indicates Lenz’s law, but for finding when the emf is zero, we only care about the magnitude being zero.

    2. Set the condition for zero emf.

       Since $L$ is a constant (here $2 \text{ mH} = 2 \times 10^{-3} \text{ H}$), $\mathcal{E}=0$ exactly when $\frac{dI}{dt}=0$. So we need to find the time $t$ at which the derivative of the given current function is zero.

    3. Differentiate the current function.

       Given $I(t) = 3t^2 - 3t + 8$, differentiate term by term:

    $\frac{dI}{dt} = \frac{d}{dt}(3t^2) - \frac{d}{dt}(3t) + \frac{d}{dt}(8) = 6t - 3.$

    4. Solve for $t$ when the derivative is zero.

       Set $6t - 3 = 0$, so $6t = 3$, giving $t = 0.5$ seconds.

    5. Interpret the result.

       At $t=0.5$ s, the current is at a stationary point (a minimum, since the parabola opens upward). At that instant, the current is not changing, so no emf is induced across the inductor.

    Tip

    A common mistake is to set the current itself to zero ($I=0$) instead of its derivative. But the emf depends on how fast the current changes, not on its value. Here, $I(0.5) = 3(0.25)-1.5+8 = 7.25$ A — not zero at all, yet the emf is zero.

    Watch out

    Don’t forget that the inductance $L$ is in mH; but since we only set $\frac{dI}{dt}=0$, the value of $L$ doesn’t affect the answer. However, if you were computing the actual emf, you’d need to convert to henries.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

12. COMEDK 2024Set 2024-E1 markMCQ
    Q.A conducting circular loop is placed in a uniform magnetic field $\mathrm{B}=0.125 \mathrm{~T}$ with its plane perpendicular to the loop. If the radius of the loop is made to shrink at a constant rate of $2 \mathrm{~mm} \mathrm{~s}^{-1}$, then the induced emf when the radius is $4 \mathrm{~cm}$ is (A) $0.52 \pi \mu V$ (B) $20 \pi \mu V$ (C) $\frac{2}{3}\mu V$ (D) $\frac{3 \pi}{2} \mu V$

    ›Reveal solutionSolution

    The induced emf is found from Faraday’s law: the rate of change of magnetic flux through the loop as its radius shrinks. The result is $20\pi\ \mu\text{V}$, which corresponds to option (B).

    The key idea is that a changing area in a constant magnetic field produces an induced emf. Here, the loop’s radius decreases at a steady rate, so the area shrinks linearly with time. Faraday’s law tells us that the induced emf equals the negative rate of change of magnetic flux. Since the field is uniform and perpendicular to the loop, the flux is simply $B \times \text{area}$. The problem gives the rate of change of radius, so we can find the rate of change of area and thus the emf.

    1. Set up the magnetic flux. The loop’s plane is perpendicular to $\mathbf{B}$, so the flux through one turn is

    $\Phi = B \cdot (\text{area}) = B \cdot \pi r^2.$

    Here $B = 0.125\ \text{T}$ is constant.

    2. Apply Faraday’s law. The induced emf is

    $\mathcal{E} = -\frac{d\Phi}{dt} = -B \pi \cdot \frac{d}{dt}(r^2).$

    Using the chain rule,

    $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}.$

    So

    $\mathcal{E} = -B \pi \cdot 2r \frac{dr}{dt}.$

    The negative sign indicates direction (Lenz’s law); we care about magnitude.

    3. Plug in the numbers at the instant $r = 4\ \text{cm}$.

       • $B = 0.125\ \text{T}$
       • $r = 4\ \text{cm} = 0.04\ \text{m}$
       • $\frac{dr}{dt} = -2\ \text{mm/s} = -0.002\ \text{m/s}$ (negative because radius is shrinking)

       Magnitude of emf:

    $|\mathcal{E}| = B \pi \cdot 2r \left|\frac{dr}{dt}\right| = 0.125 \times \pi \times 2 \times 0.04 \times 0.002.$

    4. Calculate step by step.

    $0.125 \times 2 = 0.25, \quad 0.25 \times 0.04 = 0.01, \quad 0.01 \times 0.002 = 2 \times 10^{-5}.$

    So

    $|\mathcal{E}| = \pi \times 2 \times 10^{-5} = 2\pi \times 10^{-5}\ \text{V}.$

    5. Convert to microvolts. $1\ \mu\text{V} = 10^{-6}\ \text{V}$, so

    $2\pi \times 10^{-5}\ \text{V} = 20\pi \times 10^{-6}\ \text{V} = 20\pi\ \mu\text{V}.$

    Tip

    A common mistake is forgetting to convert cm to m and mm/s to m/s. Always work in SI units (metres, seconds) before plugging into formulas.

    Watch out

    Some might mistakenly use the formula for motional emf ($B\ell v$) on a straight wire, but here the entire loop’s area changes — Faraday’s law for a changing area is the correct approach.

    ✓Final answer

    The correct option is (B).

    ANSWER: B