Electromagnetic Induction MCQs with Answers – Part 2 (Class 12 Physics)

Explanation: Faraday's law does not use final flux alone. It uses the rate of change of flux, written as \(\frac{d\phi_B}{dt}\) for instantaneous emf or \(\frac{\Delta\phi_B}{\Delta t}\) for average emf. Two situations can have the same final flux but different induced emf if their initial fluxes or time intervals are different. A steady final flux after the change is over gives no continuing emf. The time history of flux change is what matters for induction.

Explanation: For one turn, \(\varepsilon=-\frac{d\phi_B}{dt}\). A straight-line \(\phi_B\)-\(t\) graph has a constant slope. If the slope is negative, then \(\frac{d\phi_B}{dt}\) is negative. The negative sign in Faraday's law makes the induced emf positive. Since the slope is constant, the induced emf is also constant during that interval.

Explanation: In an \(N\)-turn coil, the relevant quantity is flux linkage. If each turn links the same flux \(\phi_B\), the total flux linkage is \(N\phi_B\). Faraday's law says that the induced emf equals the negative rate of change of flux linkage. Therefore \(\varepsilon=-\frac{d(N\phi_B)}{dt}=-N\frac{d\phi_B}{dt}\) when \(N\) is constant. The factor \(N\) multiplies the emf because the turns contribute in series.

Explanation: \( \textbf{Number of turns:} \) \(N=250\). \( \textbf{Initial flux per turn:} \) \(\phi_1=1.2\times10^{-3}\,Wb\). \( \textbf{Final flux per turn:} \) \(\phi_2=0.20\times10^{-3}\,Wb\). \( \textbf{Time interval:} \) \(\Delta t=0.050\,s\). \( \textbf{Magnitude of flux change per turn:} \) \[ |\Delta\phi_B|=|0.20\times10^{-3}-1.2\times10^{-3}| \] \[ |\Delta\phi_B|=1.0\times10^{-3}\,Wb \] \( \textbf{Average emf magnitude for \(N\) turns:} \) \[ |\varepsilon_{avg}|=N\frac{|\Delta\phi_B|}{\Delta t} \] \( \textbf{Substitution:} \) \[ |\varepsilon_{avg}|=250\left(\frac{1.0\times10^{-3}}{0.050}\right) \] \( \textbf{Intermediate value:} \) \[ \frac{1.0\times10^{-3}}{0.050}=2.0\times10^{-2}\,V \] \( \textbf{Final calculation:} \) \[ |\varepsilon_{avg}|=250(2.0\times10^{-2})=5.0\,V \] \( \textbf{Final answer:} \) The induced emf magnitude is \(5.0\,V\).

Explanation: \( \textbf{Faraday's law for a coil:} \) \[ |\varepsilon|=N\left|\frac{\Delta\phi_B}{\Delta t}\right| \] \( \textbf{Same condition:} \) Both coils have the same \(\Delta\phi_B\) per turn and the same \(\Delta t\). \( \textbf{Therefore:} \) The induced emf magnitude is directly proportional to \(N\). \( \textbf{For the first coil:} \) \[ |\varepsilon_1|\propto100 \] \( \textbf{For the second coil:} \) \[ |\varepsilon_2|\propto400 \] \( \textbf{Ratio:} \) \[ |\varepsilon_1|:|\varepsilon_2|=100:400 \] \( \textbf{Simplification:} \) \[ |\varepsilon_1|:|\varepsilon_2|=1:4 \] \( \textbf{Final answer:} \) The emf ratio is \(1:4\). The turn factor multiplies the flux-change rate when all turns link the same changing flux.

Explanation: In the common form \(\varepsilon=-N\frac{d\phi_B}{dt}\), \(\phi_B\) represents the magnetic flux linked with one turn. The factor \(N\) accounts for the total number of identical turns. If one already uses total flux linkage, then the law is written as \(\varepsilon=-\frac{d(N\phi_B)}{dt}\). Mixing up \(\phi_B\) with \(N\phi_B\) can lead to counting the turn factor twice. The formula is simplest when the flux through each turn is the same.

Explanation: \( \textbf{Number of turns:} \) \(N=80\). \( \textbf{Flux per turn:} \) \[ \phi_B=5.0\times10^{-4}t\,Wb \] \( \textbf{Flux-change rate per turn:} \) \[ \frac{d\phi_B}{dt}=5.0\times10^{-4}\,Wb\,s^{-1} \] \( \textbf{Faraday's law magnitude for \(N\) turns:} \) \[ |\varepsilon|=N\left|\frac{d\phi_B}{dt}\right| \] \( \textbf{Substitution:} \) \[ |\varepsilon|=80(5.0\times10^{-4}) \] \( \textbf{Calculation:} \) \[ |\varepsilon|=400\times10^{-4}\,V \] \[ |\varepsilon|=4.0\times10^{-2}\,V \] \( \textbf{Final answer:} \) The induced emf magnitude is \(4.0\times10^{-2}\,V\). The emf is constant here because the flux per turn changes linearly with time.

Explanation: Average induced emf is based on the change in flux linkage divided by the time taken. If each turn has flux changing from \(\phi_1\) to \(\phi_2\), the change in flux per turn is \(\phi_2-\phi_1\). For \(N\) identical turns, the corresponding change in flux linkage is \(N(\phi_2-\phi_1)\). Faraday's law adds the negative sign to show the Lenz-law direction. The time interval remains in the denominator because emf depends on rate of change.

Explanation: \( \textbf{Number of turns:} \) \(N=60\). \( \textbf{Flux change per turn:} \) \(|\Delta\phi_B|=2.0\times10^{-4}\,Wb\). \( \textbf{Time interval:} \) \(\Delta t=0.010\,s\). \( \textbf{Average emf magnitude:} \) \[ |\varepsilon_{avg}|=N\frac{|\Delta\phi_B|}{\Delta t} \] \( \textbf{Substitution:} \) \[ |\varepsilon_{avg}|=60\left(\frac{2.0\times10^{-4}}{0.010}\right) \] \( \textbf{Rate of flux change per turn:} \) \[ \frac{2.0\times10^{-4}}{0.010}=2.0\times10^{-2}\,Wb\,s^{-1} \] \( \textbf{Multiply by turns:} \) \[ |\varepsilon_{avg}|=60(2.0\times10^{-2}) \] \( \textbf{Calculation:} \) \[ |\varepsilon_{avg}|=1.2\,V \] \( \textbf{Final answer:} \) The magnitude of average induced emf is \(1.2\,V\).

Explanation: Statement I is true because the induced emf of a coil depends on the rate of change of total flux linkage. If every turn links the same changing flux, the linkage is \(N\phi_B\), so increasing \(N\) increases the emf magnitude. Statement II is true because the negative sign in Faraday's law represents Lenz's law. Statement III is false because resistance does not appear in the expression for induced emf due to flux change. Resistance becomes important when finding induced current using a circuit relation such as \(I=\frac{\varepsilon}{R}\).

Explanation: A changing magnetic flux linked with a circuit produces induced emf. Current, however, needs a complete conducting path so that charges can circulate. If the circuit is open, an emf may be induced across the break, but continuous current cannot flow. If the flux is constant, the basic cause of induced emf is absent. Both conditions must be separated: flux change produces emf, while circuit closure permits current.

Explanation: Once an induced emf is produced in a closed circuit, the current is found using the circuit relation \(I=\frac{\varepsilon}{R}\). The resistance \(R\) controls how much current flows for a given emf. A larger resistance gives a smaller current, while a smaller resistance gives a larger current. The induced emf itself is produced by flux change, not by the resistance. The relation applies to the current response of the closed circuit after the emf has appeared.

Explanation: \( \textbf{Given:} \) Induced emf magnitude \(|\varepsilon|=0.20\,V\), and resistance \(R=5.0\,\Omega\). \( \textbf{Required:} \) Magnitude of induced current \(I\). \( \textbf{Circuit relation:} \) \[ I=\frac{|\varepsilon|}{R} \] \( \textbf{Why this relation applies:} \) The coil is closed, so the induced emf can drive current through its resistance. \( \textbf{Substitution:} \) \[ I=\frac{0.20}{5.0} \] \( \textbf{Calculation:} \) \[ I=0.040\,A \] \( \textbf{Final answer:} \) The magnitude of induced current is \(0.040\,A\). The flux change decides the emf first; the resistance then decides how large the current becomes.

Explanation: The same changing magnetic flux can induce emf in both coils. Faraday's law concerns the production of emf due to flux change and does not require the circuit to be closed. A closed circuit is needed for continuous current. Coil P has a complete path, so the induced emf can drive current through its resistance. Coil Q may have induced emf across its open ends, but charges cannot circulate around a broken path.

Explanation: \( \textbf{Same flux-change condition:} \) The induced emf magnitude is the same in both trials. \( \textbf{Current relation:} \) \[ I=\frac{\varepsilon}{R} \] \( \textbf{Trial P resistance:} \) \[ R_P=4\,\Omega \] \( \textbf{Trial Q resistance:} \) \[ R_Q=8\,\Omega \] \( \textbf{Current in Trial P:} \) \[ I_P=\frac{\varepsilon}{4} \] \( \textbf{Current in Trial Q:} \) \[ I_Q=\frac{\varepsilon}{8} \] \( \textbf{Comparison:} \) \[ I_Q=\frac{1}{2}I_P \] \( \textbf{Final answer:} \) The current in Trial Q is half the current in Trial P. Changing resistance changes the current response, not the flux-change-produced emf in this comparison.

Explanation: The induced emf is set by the rate of change of magnetic flux. If the flux-change rate is kept the same, Faraday's law gives the same induced emf. The induced current in a closed circuit is \(I=\frac{\varepsilon}{R}\). Increasing \(R\) reduces \(I\) for the same \(\varepsilon\). Resistance controls the size of the current, but it does not create or remove the flux change that produced the emf.

Explanation: \( \textbf{Number of turns:} \) \(N=20\). \( \textbf{Flux-change rate per turn:} \) \[ \left|\frac{d\phi_B}{dt}\right|=3.0\times10^{-3}\,Wb\,s^{-1} \] \( \textbf{Resistance:} \) \[ R=6.0\,\Omega \] \( \textbf{Induced emf magnitude:} \) \[ |\varepsilon|=N\left|\frac{d\phi_B}{dt}\right| \] \( \textbf{Substitution for emf:} \) \[ |\varepsilon|=20(3.0\times10^{-3}) \] \( \textbf{Emf value:} \) \[ |\varepsilon|=6.0\times10^{-2}\,V \] \( \textbf{Current relation:} \) \[ I=\frac{|\varepsilon|}{R} \] \( \textbf{Substitution for current:} \) \[ I=\frac{6.0\times10^{-2}}{6.0} \] \( \textbf{Final answer:} \) The induced current is \(1.0\times10^{-2}\,A\).

Explanation: A closed circuit with changing flux can have induced emf and current. An open circuit with changing flux can still have induced emf, but continuous current cannot circulate through the broken path. A closed circuit with constant flux has no induced emf due to flux change. Row S is inconsistent because an open circuit with constant flux lacks both necessary features for continuous induced current. It has no flux change to produce emf and no closed path to support current.

Explanation: Statement I is true because Faraday's law allows induced emf due to changing magnetic flux even when the circuit is open. Statement II is true because current in a closed circuit is found from \(I=\frac{\varepsilon}{R}\). Statement III is false because for the same emf, increasing \(R\) decreases current. Resistance affects current after the emf is produced. The distinction between emf production and current flow is essential in induction problems.

Explanation: \( \textbf{Induced emf:} \) \(\varepsilon=1.5\,V\). \( \textbf{New resistance:} \) \(R=9.0\,\Omega\). \( \textbf{Flux-change condition:} \) The flux-change rate is unchanged, so the induced emf remains \(1.5\,V\). \( \textbf{Current relation:} \) \[ I=\frac{\varepsilon}{R} \] \( \textbf{Substitution:} \) \[ I=\frac{1.5}{9.0} \] \( \textbf{Calculation:} \) \[ I=0.166\ldots\,A \] \( \textbf{Rounded value:} \) \[ I\approx0.17\,A \] \( \textbf{Final answer:} \) The new current is \(0.17\,A\). Using the old resistance would give the old current, not the current after the circuit resistance changes.