Electromagnetic Induction MCQs with Answers – Part 4 (Class 12 Physics)

Explanation: \( \textbf{Self-induced emf magnitude:} \) \[ |\varepsilon|=L\left|\frac{dI}{dt}\right| \] \( \textbf{Same condition:} \) Both coils have the same \(\left|\frac{dI}{dt}\right|\). \( \textbf{Therefore:} \) The emf magnitude is directly proportional to \(L\). \( \textbf{Ratio setup:} \) \[ |\varepsilon_P|:|\varepsilon_Q|=L_P:L_Q \] \( \textbf{Substitution:} \) \[ |\varepsilon_P|:|\varepsilon_Q|=0.20:0.80 \] \( \textbf{Simplification:} \) \[ |\varepsilon_P|:|\varepsilon_Q|=1:4 \] \( \textbf{Final answer:} \) The ratio is \(1:4\). The larger inductance produces a larger opposing emf for the same current-change rate.

Explanation: Self-induced emf depends on the slope of the \(I\)-\(t\) graph. From \(t=0\) to \(t=2.0\,\text{s}\), the current rises linearly, so the slope \(\frac{dI}{dt}\) is non-zero. During that interval, the coil has a self-induced emf. From \(t=2.0\,\text{s}\) to \(t=5.0\,\text{s}\), the current-time graph is horizontal. A horizontal current-time graph has \(\frac{dI}{dt}=0\), so \(\varepsilon=-L\frac{dI}{dt}=0\).

Explanation: Self-inductance can be written as \(L=\frac{N\phi}{I}\), so it measures flux linkage per unit current. The self-induced emf relation \(\varepsilon=-L\frac{dI}{dt}\) shows that \(L\) controls the emf for a given rate of current change. Its SI unit is henry, written as \(\text{H}\). Resistance \(R\) is different because it relates current to voltage drop through \(V=IR\). Inductance is connected with magnetic flux linkage, while resistance is connected with energy dissipation in a conductor.

Explanation: From the self-induced emf relation, \(\varepsilon=L\left|\frac{dI}{dt}\right|\) in magnitude. Rearranging gives \(L=\frac{\varepsilon}{dI/dt}\). The unit of emf is \(\text{V}\), and the unit of \(\frac{dI}{dt}\) is \(\text{A s}^{-1}\). Therefore the unit of \(L\) is \(\frac{\text{V}}{\text{A s}^{-1}}=\text{V s A}^{-1}\). This unit is called henry, written as \(\text{H}\).

Explanation: Self-inductance is defined by \(N\phi=LI\). For a fixed current \(I\), a larger flux linkage \(N\phi\) means a larger value of \(L\). This is why coils with many turns or magnetic cores can have larger inductance. Resistance is a different circuit property and does not define flux linkage per unit current. Inductance measures how strongly a coil links magnetic flux with its own current.

Explanation: A coil with more turns can produce a stronger magnetic effect for the same current, depending on the geometry. More turns also mean more turns are linked with the magnetic flux. Since self-inductance is flux linkage per unit current, increasing turns generally increases \(L\). In a long solenoid, this appears clearly in the relation \(L=\frac{\mu N^2A}{l}\). The square dependence on \(N\) reflects both field production and linkage contribution.

Explanation: A magnetic core of high permeability allows a larger magnetic field and flux for the same current. Since \(L=\frac{N\phi}{I}\), a larger flux linkage for the same current gives a larger self-inductance. Soft iron is commonly used when stronger magnetic coupling is desired. The core does not increase inductance by making current infinite or by removing turns. It increases the magnetic response of the coil.

Explanation: For a long solenoid, the self-inductance is \(L=\frac{\mu N^2A}{l}\). Increasing the number of turns increases \(L\), and increasing permeability \(\mu\) also increases \(L\). A larger cross-sectional area \(A\) increases flux through each turn, so it increases \(L\). If \(N\) and \(A\) are fixed, increasing the solenoid length \(l\) decreases \(L\), not increases it. Row S is therefore opposite to the solenoid relation.

Explanation: \( \textbf{Starting relation:} \) \[ L=\frac{\varepsilon}{dI/dt} \] \( \textbf{Dimension of emf:} \) \[ [\varepsilon]=[M L^2 T^{-3} A^{-1}] \] \( \textbf{Dimension of current-change rate:} \) \[ \left[\frac{dI}{dt}\right]=[A T^{-1}] \] \( \textbf{Substitution into \(L\):} \) \[ [L]=\frac{[M L^2 T^{-3} A^{-1}]}{[A T^{-1}]} \] \( \textbf{Simplification:} \) \[ [L]=[M L^2 T^{-3} A^{-1}]\,[A^{-1}T] \] \[ [L]=[M L^2 T^{-2} A^{-2}] \] \( \textbf{Final answer:} \) The dimensional formula of self-inductance is \([M L^2 T^{-2} A^{-2}]\).

Explanation: For a long solenoid, the magnetic field inside is approximately uniform and is given by \(B=\mu nI\). Here \(n\) is the number of turns per unit length, and \(\mu\) is the permeability of the medium inside the solenoid. The field is directly proportional to current \(I\). It is also directly proportional to turn density \(n\), not to cross-sectional area \(A\) in the field expression itself. The area appears later while finding flux through each turn.

Explanation: A long solenoid has magnetic field \(B=\mu nI\), where \(n=\frac{N}{l}\). The flux through each turn is \(\phi=BA\) if the field is uniform over area \(A\). The total flux linkage is \(N\phi=NBA\). Substituting \(B=\mu \frac{N}{l}I\) gives \(N\phi=N\left(\mu \frac{N}{l}I\right)A\). Comparing with \(N\phi=LI\), the self-inductance becomes \(L=\frac{\mu N^2A}{l}\).

Explanation: In a long solenoid, \(n\) represents the number of turns per unit length. Therefore \(n=\frac{N}{l}\). Substituting \(N=nl\) into \(L=\frac{\mu N^2A}{l}\) gives \(L=\frac{\mu (nl)^2A}{l}\). This simplifies to \(L=\mu n^2Al\). Both formulas describe the same solenoid when the symbols are used consistently.

Explanation: \( \textbf{Given:} \) \(N=500\), \(A=4.0\times10^{-4}\,\text{m}^2\), \(l=0.25\,\text{m}\), and \(\mu=4\pi\times10^{-7}\,\text{H m}^{-1}\). \( \textbf{Solenoid inductance:} \) \[ L=\frac{\mu N^2A}{l} \] \( \textbf{Square of turns:} \) \[ N^2=(500)^2=2.5\times10^5 \] \( \textbf{Substitution:} \) \[ L=\frac{(4\pi\times10^{-7})(2.5\times10^5)(4.0\times10^{-4})}{0.25} \] \( \textbf{Combine powers and numbers:} \) \[ (2.5\times10^5)(4.0\times10^{-4})=100 \] \[ L=\frac{(4\pi\times10^{-7})(100)}{0.25} \] \( \textbf{Simplification:} \) \[ L=1600\pi\times10^{-7}\,\text{H} \] \[ L\approx5.0\times10^{-4}\,\text{H} \] \( \textbf{Final answer:} \) The self-inductance is closest to \(5.0\times10^{-4}\,\text{H}\).

Explanation: For a long solenoid, \(L=\frac{\mu N^2A}{l}\). With \(A\), \(l\), and \(\mu\) fixed, the inductance is proportional to \(N^2\). If \(N\) is doubled, \(N^2\) becomes \((2N)^2=4N^2\). Therefore the inductance becomes four times its original value. The square dependence appears because more turns both strengthen the field and increase flux linkage.

Explanation: \( \textbf{Solenoid relation:} \) \[ L=\frac{\mu N^2A}{l} \] \( \textbf{Same quantities:} \) Both solenoids have the same \(N\), \(A\), and \(\mu\). \( \textbf{For solenoid P:} \) \[ L_P=\frac{\mu N^2A}{l} \] \( \textbf{For solenoid Q:} \) \[ L_Q=\frac{\mu N^2A}{2l} \] \( \textbf{Comparison:} \) \[ L_Q=\frac{L_P}{2} \] \( \textbf{Ratio:} \) \[ L_P:L_Q=2:1 \] \( \textbf{Final answer:} \) The ratio is \(2:1\). A longer solenoid has smaller turn density for the same total turns, so its inductance is lower in this comparison.

Explanation: The self-inductance of a long solenoid is \(L=\frac{\mu N^2A}{l}\). If \(N\), \(A\), and \(l\) remain unchanged, \(L\) is directly proportional to permeability \(\mu\). A material with permeability \(200\) times that of air produces \(200\) times the inductance. The factor is not squared because \(\mu\) appears only once in the formula. The core changes magnetic flux linkage for the same current.

Explanation: The magnetic field relation \(B=\mu nI\) is used for a long solenoid. If the field is approximately uniform and along the solenoid axis, the flux through one turn is \(\phi=BA\). The flux linkage is then \(N\phi=NBA\). The turn density \(n\) is the number of turns per unit length, so \(n=\frac{N}{l}\), not \(\frac{l}{N}\). Row S is the faulty step because it reverses the definition of turn density.

Explanation: \( \textbf{Given:} \) \(N=1000\), \(A=2.0\times10^{-4}\,\text{m}^2\), \(l=0.50\,\text{m}\), \(\mu=1.0\times10^{-6}\,\text{H m}^{-1}\), and \(\left|\frac{dI}{dt}\right|=30\,\text{A s}^{-1}\). \( \textbf{First find solenoid inductance:} \) \[ L=\frac{\mu N^2A}{l} \] \( \textbf{Substitution:} \) \[ L=\frac{(1.0\times10^{-6})(1000)^2(2.0\times10^{-4})}{0.50} \] \( \textbf{Simplify \(N^2\):} \) \[ (1000)^2=1.0\times10^6 \] \( \textbf{Inductance:} \) \[ L=\frac{(1.0)(2.0\times10^{-4})}{0.50}=4.0\times10^{-4}\,\text{H} \] \( \textbf{Self-induced emf magnitude:} \) \[ |\varepsilon|=L\left|\frac{dI}{dt}\right| \] \( \textbf{Final calculation:} \) \[ |\varepsilon|=(4.0\times10^{-4})(30)=1.2\times10^{-2}\,\text{V} \] \( \textbf{Final answer:} \) The self-induced emf magnitude is \(1.2\times10^{-2}\,\text{V}\).

Explanation: For a long solenoid, \(L=\frac{\mu N^2A}{l}\). If \(N\), \(l\), and \(\mu\) are fixed, \(L\) is directly proportional to area \(A\). Tripling the cross-sectional area triples the flux through each turn for the same magnetic field. Therefore the total flux linkage per unit current also triples. The area factor is linear, while the turn factor is squared.

Explanation: The self-inductance of a long solenoid is \(L=\frac{\mu N^2A}{l}\). Statement I is true because \(L\) is directly proportional to \(\mu\). Statement II is true because a larger area gives larger flux linkage. Statement III is false in the stated condition because \(L\) is inversely proportional to \(l\) when \(N\), \(A\), and \(\mu\) are fixed. The length dependence changes only if turn density or other quantities are held fixed instead of total turns.