Electromagnetic Induction and Alternating Currents MCQs with Answers – Part 6 (NEET Physics)

Explanation: The common core flux is specified as a sinusoidal function of time. Faraday's law therefore determines the induced voltage per turn directly. The measured primary and secondary RMS voltages can be used to recover the winding turn counts. The primary current then follows from lossless real-power transfer. \(\textbf{Flux amplitude and angular frequency:}\) The maximum magnetic flux through each turn is \[ \Phi_m=2.0\times10^{-3}\,\text{Wb} \] The angular frequency is \[ \omega=100\,\text{rad s}^{-1} \] \(\textbf{Induced voltage per turn:}\) For one turn, \[ e_1=-\frac{d\Phi_B}{dt} \] Differentiating, \[ e_1=-0.20\cos(100t)\,\text{V} \] The peak induced-voltage magnitude per turn is \[ E_{1,m}=0.20\,\text{V} \] Therefore, the RMS voltage magnitude per turn is \[ E_{1,\text{rms}}=\frac{0.20}{\sqrt2}\,\text{V} \] \(\textbf{Recover primary turn count:}\) For \(N_p\) turns, \[ V_{p,\text{rms}}=N_pE_{1,\text{rms}} \] Therefore, \[ N_p=\frac{V_{p,\text{rms}}}{E_{1,\text{rms}}} \] Substituting, \[ N_p=\frac{100\sqrt2}{0.20/\sqrt2} \] \[ N_p=1000 \] \(\textbf{Recover secondary turn count:}\) Similarly, \[ N_s=\frac{V_{s,\text{rms}}}{E_{1,\text{rms}}} \] \[ N_s=\frac{20\sqrt2}{0.20/\sqrt2} \] \[ N_s=200 \] \(\textbf{Secondary load current:}\) The secondary resistor is \[ R=20\,\Omega \] Therefore, \[ I_s=\frac{V_{s,\text{rms}}}{R} \] \[ I_s=\frac{20\sqrt2}{20} \] \[ I_s=\sqrt2\,\text{A} \] \(\textbf{Secondary real output power:}\) \[ P_s=I_s^2R \] \[ P_s=(\sqrt2)^2(20) \] \[ P_s=40\,\text{W} \] \(\textbf{Recover primary RMS current:}\) For an ideal transformer with a resistive load, \[ P_p=P_s \] Therefore, \[ V_{p,\text{rms}}I_p=40\,\text{W} \] Thus, \[ I_p=\frac{40}{100\sqrt2} \] \[ I_p=\frac{\sqrt2}{5}\,\text{A} \] \(\textbf{Turns-ratio consistency check:}\) \[ \frac{N_s}{N_p}=\frac{200}{1000} \] \[ \frac{N_s}{N_p}=\frac15 \] The measured voltage ratio is \[ \frac{V_s}{V_p}=\frac{20\sqrt2}{100\sqrt2} \] \[ \frac{V_s}{V_p}=\frac15 \] Both ratios agree. \(\textbf{Physical interpretation:}\) The common magnetic flux establishes the same induced emf per turn in both windings. The complete winding voltages differ because their turn counts differ. The primary and secondary voltage measurements therefore reveal their respective numbers of turns. The secondary resistor determines the delivered real power. The ideal transformer must draw the corresponding primary current required by energy conservation. The calculation connects a magnetic-flux waveform with winding design and electrical power transfer. \(\textbf{Final answer:}\) \[ \boxed{N_p=1000,\qquad N_s=200,\qquad I_p=\frac{\sqrt2}{5}\,\text{A}} \]

Explanation: Doubling the generator's rotational speed increases both the frequency and voltage amplitude of its sinusoidal output. Changing the transformer secondary tapping alters the fraction of primary voltage appearing across the load. These two adjustments can compensate one another at the load, even though the generator and primary-side currents change. \(\textbf{Initial generator RMS voltage:}\) The initial peak emf is \[ E_{m1}=100\,\text{V} \] Therefore, \[ V_{p1}=\frac{100}{\sqrt2} \] \[ V_{p1}=50\sqrt2\,\text{V} \] \(\textbf{Initial transformer secondary voltage:}\) \[ V_{s1}=V_{p1}\frac{N_{s1}}{N_p} \] Substituting, \[ V_{s1}=50\sqrt2\left(\frac{400}{200}\right) \] \[ V_{s1}=100\sqrt2\,\text{V} \] \(\textbf{Initial resistor-heating power:}\) \[ P_1=\frac{V_{s1}^2}{R} \] \[ P_1=\frac{(100\sqrt2)^2}{50} \] \[ P_1=400\,\text{W} \] \(\textbf{Initial primary RMS current:}\) The ideal transformer transfers the same real power from primary to secondary. Therefore, \[ I_{p1}=\frac{P_1}{V_{p1}} \] \[ I_{p1}=\frac{400}{50\sqrt2} \] \[ I_{p1}=4\sqrt2\,\text{A} \] \(\textbf{Generator output after speed doubling:}\) The generated peak emf is proportional to angular speed: \[ E_m=NBA\omega \] Therefore, \[ E_{m2}=2E_{m1} \] \[ E_{m2}=200\,\text{V} \] The new primary RMS voltage is \[ V_{p2}=\frac{200}{\sqrt2} \] \[ V_{p2}=100\sqrt2\,\text{V} \] \(\textbf{Secondary voltage after changing the tap:}\) The new secondary turn count is \[ N_{s2}=200 \] Therefore, \[ V_{s2}=V_{p2}\frac{N_{s2}}{N_p} \] \[ V_{s2}=100\sqrt2\left(\frac{200}{200}\right) \] \[ V_{s2}=100\sqrt2\,\text{V} \] The secondary RMS voltage remains unchanged. \(\textbf{Final load power:}\) \[ P_2=\frac{V_{s2}^2}{R} \] \[ P_2=\frac{(100\sqrt2)^2}{50} \] \[ P_2=400\,\text{W} \] Therefore, \[ \frac{P_2}{P_1}=1 \] \(\textbf{Final primary RMS current:}\) \[ I_{p2}=\frac{P_2}{V_{p2}} \] \[ I_{p2}=\frac{400}{100\sqrt2} \] \[ I_{p2}=2\sqrt2\,\text{A} \] Hence, \[ \frac{I_{p2}}{I_{p1}}=\frac{2\sqrt2}{4\sqrt2} \] \[ \frac{I_{p2}}{I_{p1}}=\frac12 \] \(\textbf{Compare generator driving torques:}\) The generator's average mechanical power equals the fixed load power: \[ P=\tau_{\text{avg}}\omega \] Initially, \[ \tau_1=\frac{400}{100} \] \[ \tau_1=4.0\,\text{N m} \] Finally, \[ \tau_2=\frac{400}{200} \] \[ \tau_2=2.0\,\text{N m} \] Therefore, \[ \frac{\tau_2}{\tau_1}=\frac12 \] \(\textbf{Physical interpretation:}\) The generator's doubled speed raises its source voltage and frequency. The transformer secondary turn count is simultaneously halved relative to its initial value. The two changes preserve the RMS voltage applied to the external resistor. Consequently, the resistor consumes the same real power. The generator supplies that unchanged power at twice the shaft angular speed and therefore requires half the average driving torque. Its primary-side electrical current also halves because the primary RMS voltage doubles while real-power demand remains fixed. This illustrates coordinated generator and transformer control that preserves useful output while changing mechanical and electrical operating conditions. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{p2}}{I_{p1}}=\frac12,\qquad \frac{\tau_2}{\tau_1}=\frac12,\qquad \frac{P_2}{P_1}=1} \]

Explanation: The rotating coil generates a sinusoidal emf whose peak value is fixed by its magnetic field, geometry and angular speed. However, the useful power delivered to the external resistor depends on the division of electrical energy between external resistance and internal winding resistance. The condition for maximum external power is not the same as the condition for maximum generator efficiency. \(\textbf{Generated RMS emf:}\) \[ E_{\text{rms}}=\frac{E_m}{\sqrt2} \] Substituting, \[ E_{\text{rms}}=\frac{100}{\sqrt2} \] \[ E_{\text{rms}}=50\sqrt2\,\text{V} \] \(\textbf{Circuit current for an arbitrary load:}\) The total resistance is \[ R_{\text{total}}=R+10 \] Therefore, \[ I_{\text{rms}}=\frac{50\sqrt2}{R+10} \] \(\textbf{Useful external-load power:}\) \[ P_R=I_{\text{rms}}^2R \] Substituting, \[ P_R=\frac{(50\sqrt2)^2R}{(R+10)^2} \] \[ P_R=\frac{5000R}{(R+10)^2} \] \(\textbf{Identify the maximum-power condition:}\) For positive \(R\), \[ (R-10)^2\geq0 \] Expanding, \[ R^2+100\geq20R \] Adding \(20R\) to both sides, \[ (R+10)^2\geq40R \] Therefore, \[ \frac{R}{(R+10)^2}\leq\frac1{40} \] Equality occurs when \[ R=10\,\Omega \] \(\textbf{Maximum useful power:}\) \[ P_{R,\max}=\frac{5000(10)}{(10+10)^2} \] \[ P_{R,\max}=\frac{50000}{400} \] \[ P_{R,\max}=125\,\text{W} \] \(\textbf{Winding loss at the optimum:}\) The RMS current becomes \[ I_{\text{rms}}=\frac{50\sqrt2}{20} \] \[ I_{\text{rms}}=\frac5{\sqrt2}\,\text{A} \] The winding dissipates \[ P_{\text{winding}}=I_{\text{rms}}^2r \] \[ P_{\text{winding}}=\left(\frac5{\sqrt2}\right)^2(10) \] \[ P_{\text{winding}}=125\,\text{W} \] \(\textbf{Mechanical input and driving torque:}\) With negligible losses other than winding heating, \[ P_{\text{mechanical}}=P_R+P_{\text{winding}} \] \[ P_{\text{mechanical}}=125+125 \] \[ P_{\text{mechanical}}=250\,\text{W} \] At constant angular speed, \[ \tau_{\text{avg}}=\frac{P_{\text{mechanical}}}{\omega} \] \[ \tau_{\text{avg}}=\frac{250}{50} \] \[ \tau_{\text{avg}}=5.0\,\text{N m} \] \(\textbf{Physical interpretation:}\) At maximum external-load power, the load resistance equals the generator's internal resistance. The generator then dissipates as much average power in its own winding as it delivers usefully to the load. Thus, the maximum-power operating condition has an electrical efficiency of only \(50\%\) in this model. Increasing the external resistance beyond the optimum can improve efficiency while reducing useful output power. The calculation connects induced emf, internal resistance, external power transfer and mechanical driving torque. \(\textbf{Final answer:}\) \[ \boxed{R=10\,\Omega,\qquad P_{R,\max}=125\,\text{W},\qquad \tau_{\text{avg}}=5.0\,\text{N m}} \]

Explanation: A transformer core experiences changing magnetic flux during AC operation. If its magnetic material also conducts electricity, changing flux can induce circulating eddy currents within the core. These currents cause resistive heating and contribute to transformer energy loss. A solid conducting core permits comparatively large current paths through its bulk. Thin insulated laminations restrict many of these paths, reducing the associated eddy-current heating. Statement I is correct. The insulation between successive sheets interrupts electrical conduction across the lamination boundaries. This prevents many larger loops from forming across adjacent sheets. Statement II is correct. Hysteresis loss has a different physical origin from eddy-current heating. It is associated with repeated magnetisation and demagnetisation of the magnetic material. Consequently, interrupting conducting paths does not necessarily eliminate hysteresis loss. Statement III is incorrect. Transformer efficiency depends on the useful real output relative to the real input. If useful output and winding losses remain unchanged while eddy-current loss decreases, the required input real power can fall. The ideal voltage-turns relation remains governed by the common changing magnetic flux and winding turns. Statement IV is correct. This experiment separates a loss-reduction design choice from the basic voltage-transformation mechanism. A laminated core can improve efficiency without producing a new output frequency or violating energy conservation. The correct combination is Statements I, II and IV only.

Explanation: The linked magnetic flux changes for two separate reasons. The coil rotates relative to the field, and the magnetic-field magnitude increases independently with time. Faraday's law must account for both contributions. Using only the standard constant-field generator relation would omit the field-variation contribution. \(\textbf{Write the changing flux linkage:}\) \[ \lambda(t)=NA B(t)\cos(20t) \] The product of turn count and coil area is \[ NA=(100)(0.010) \] \[ NA=1.0\,\text{m}^2 \] Therefore, \[ \lambda(t)=(0.50+0.10t)\cos(20t) \] \(\textbf{Differentiate the complete expression:}\) Faraday's law gives \[ e(t)=-\frac{d\lambda}{dt} \] Applying the product rule, \[ \frac{d\lambda}{dt}=0.10\cos(20t)-20(0.50+0.10t)\sin(20t) \] Hence, \[ e(t)=-0.10\cos(20t)+20(0.50+0.10t)\sin(20t) \] \(\textbf{Emf at the initial orientation:}\) At \(t=0\), \[ \cos0=1,\qquad \sin0=0 \] Therefore, \[ e(0)=-0.10(1)+20(0.50)(0) \] \[ e(0)=-0.10\,\text{V} \] \(\textbf{Emf after one-quarter revolution:}\) At \[ t_1=\frac{\pi}{40}\,\text{s} \] the rotational phase is \[ 20t_1=\frac{\pi}{2} \] Thus, \[ e(t_1)=20\left(0.50+0.10\frac{\pi}{40}\right) \] \[ e(t_1)=10+\frac{\pi}{20}\,\text{V} \] \(\textbf{Initial induced current:}\) \[ i(0)=\frac{e(0)}R \] \[ i(0)=\frac{-0.10}{10} \] \[ i(0)=-0.010\,\text{A} \] \(\textbf{Initial resistor-heating power:}\) \[ p_R(0)=i^2(0)R \] \[ p_R(0)=(-0.010)^2(10) \] \[ p_R(0)=1.0\times10^{-3}\,\text{W} \] \(\textbf{Physical interpretation:}\) At the initial instant, the coil has maximum geometrical flux projection because its area vector is parallel to the field. Nevertheless, the flux is increasing because the field magnitude is increasing. The induced emf is therefore nonzero. After one-quarter revolution, the area projection is zero, but its rotational rate of change is maximum. The changing field also affects the field magnitude at that later instant. This problem demonstrates that the familiar zero-emf condition at maximum geometric flux requires the usual constant-field assumption. \(\textbf{Final answer:}\) \[ \boxed{e(0)=-0.10\,\text{V},\qquad e\left(\frac{\pi}{40}\right)=10+\frac{\pi}{20}\,\text{V},\qquad p_R(0)=1.0\times10^{-3}\,\text{W}} \]

Explanation: An ideal induced secondary emf does not necessarily equal the terminal voltage of a practical loaded transformer. Current through the winding resistance produces an internal voltage drop. The first load condition allows that equivalent resistance to be determined. The second condition then requires a fresh circuit calculation because the current changes. \(\textbf{Current in the first experiment:}\) The measured terminal RMS voltage is \[ V_1=100\,\text{V} \] The load resistance is \[ R_1=20\,\Omega \] Therefore, \[ I_1=\frac{V_1}{R_1} \] \[ I_1=\frac{100}{20} \] \[ I_1=5.0\,\text{A} \] \(\textbf{Recover equivalent winding resistance:}\) For the specified resistive model, \[ E_s=I_1(R_1+r) \] Therefore, \[ r=\frac{E_s}{I_1}-R_1 \] Substituting, \[ r=\frac{120}{5.0}-20 \] \[ r=24-20 \] \[ r=4.0\,\Omega \] \(\textbf{Current with the smaller load resistance:}\) The new load resistance is \[ R_2=8.0\,\Omega \] The total equivalent resistance becomes \[ R_{\text{total},2}=R_2+r \] \[ R_{\text{total},2}=8.0+4.0 \] \[ R_{\text{total},2}=12\,\Omega \] Thus, \[ I_2=\frac{E_s}{R_{\text{total},2}} \] \[ I_2=\frac{120}{12} \] \[ I_2=10\,\text{A} \] \(\textbf{New secondary terminal voltage:}\) \[ V_2=I_2R_2 \] \[ V_2=(10)(8.0) \] \[ V_2=80\,\text{V} \] \(\textbf{Winding-heating power:}\) \[ P_{\text{winding},2}=I_2^2r \] \[ P_{\text{winding},2}=(10)^2(4.0) \] \[ P_{\text{winding},2}=400\,\text{W} \] \(\textbf{Real-power consistency check:}\) The external load receives \[ P_{\text{load},2}=I_2^2R_2 \] \[ P_{\text{load},2}=(10)^2(8.0) \] \[ P_{\text{load},2}=800\,\text{W} \] The ideal internal secondary emf supplies \[ P_{\text{internal}}=E_sI_2 \] \[ P_{\text{internal}}=(120)(10) \] \[ P_{\text{internal}}=1200\,\text{W} \] This equals the sum of useful load power and winding heating. \(\textbf{Physical interpretation:}\) Reducing the external load resistance increases the secondary current. The larger current causes a greater voltage drop across the winding resistance. Consequently, the terminal voltage decreases even though the induced internal emf remains unchanged. The winding loss also increases because it depends on current squared. This distinguishes induced emf, terminal voltage, useful output power and internal heating in a practical transformer. \(\textbf{Final answer:}\) \[ \boxed{r=4.0\,\Omega,\qquad V_2=80\,\text{V},\qquad P_{\text{winding},2}=400\,\text{W}} \]

Explanation: The magnetic field affects the generator's emf amplitude. However, the generator is not constrained to maintain its original angular speed. Instead, the driving mechanism supplies a fixed torque. The final rotational speed must therefore be found by balancing mechanical input power against electrical output power. \(\textbf{Generated emf:}\) For the rotating coil, \[ E_m=NBA\omega \] The RMS voltage is \[ E_{\text{rms}}=\frac{NBA\omega}{\sqrt2} \] \(\textbf{Average output power:}\) For the fixed pure resistor, \[ P_R=\frac{E_{\text{rms}}^2}{R} \] Therefore, \[ P_R=\frac{(NBA)^2\omega^2}{2R} \] \(\textbf{Mechanical power supplied:}\) At constant angular speed, \[ P_{\text{mechanical}}=\tau\omega \] With negligible losses, \[ \tau\omega=\frac{(NBA)^2\omega^2}{2R} \] For nonzero rotation, cancel one factor of \(\omega\): \[ \tau=\frac{(NA)^2B^2\omega}{2R} \] \(\textbf{Apply the fixed-torque condition:}\) The driving torque, coil geometry and external resistance remain unchanged. Therefore, \[ B_1^2\omega_1=B_2^2\omega_2 \] With \[ B_2=2B_1 \] we obtain \[ B_1^2\omega_1=4B_1^2\omega_2 \] Hence, \[ \frac{\omega_2}{\omega_1}=\frac14 \] \(\textbf{Compare generated peak emfs:}\) \[ \frac{E_{m2}}{E_{m1}}=\frac{B_2\omega_2}{B_1\omega_1} \] Substituting, \[ \frac{E_{m2}}{E_{m1}}=2\left(\frac14\right) \] \[ \frac{E_{m2}}{E_{m1}}=\frac12 \] \(\textbf{Compare average resistor powers:}\) The mechanical torque is unchanged: \[ P=\tau\omega \] Therefore, \[ \frac{P_2}{P_1}=\frac{\omega_2}{\omega_1} \] \[ \frac{P_2}{P_1}=\frac14 \] The same ratio follows from the square of the generated RMS-voltage ratio. \(\textbf{Physical interpretation:}\) A stronger magnetic field makes the generator produce a larger emf at any fixed angular speed. It also increases the electromagnetic opposition to rotation for a given speed and load. When the external driving torque is held fixed, the system reaches a lower steady speed. The speed decrease is sufficiently large that the final generated emf is actually smaller. Therefore, doubling the magnetic field does not necessarily increase electrical output when the mechanical driving condition is changed. The problem distinguishes fixed-speed and fixed-torque generator operation. \(\textbf{Final answer:}\) \[ \boxed{\frac{\omega_2}{\omega_1}=\frac14,\qquad \frac{E_{m2}}{E_{m1}}=\frac12,\qquad \frac{P_2}{P_1}=\frac14} \]

Explanation: The magnetic flux in a voltage-driven transformer is constrained by Faraday's law. With a sinusoidal primary voltage and negligible winding voltage drop, \[ V_{p,\text{rms}}=\frac{N_p\omega\Phi_m}{\sqrt2} \] The primary voltage, frequency and turn count remain unchanged. Therefore, the maximum common magnetic flux \(\Phi_m\) remains unchanged. Statement I is correct. Magnetic flux density is flux per unit area: \[ B_m=\frac{\Phi_m}{A} \] The same maximum flux passes through twice the cross-sectional area in experiment Q. Consequently, maximum flux density becomes one-half its original value. Statement II is correct. For a linear magnetic circuit with unchanged magnetic-path length and permeability, the magnetising inductance varies directly with core cross-sectional area. Thus, doubling area doubles magnetising inductance. At fixed RMS voltage and frequency, the corresponding ideal magnetising current is inversely proportional to magnetising inductance. Its magnitude therefore becomes one-half. Statement III is correct. The secondary voltage is determined by the same changing magnetic flux and the secondary turn count: \[ V_{s,\text{rms}}=\frac{N_s\omega\Phi_m}{\sqrt2} \] Both quantities remain unchanged in this expression. Therefore, secondary induced voltage does not double. Statement IV is incorrect. The experiment illustrates that core flux, magnetic flux density, magnetising current and secondary voltage are different physical quantities. Increasing core cross-sectional area can reduce the magnetic field intensity needed for operation without automatically increasing induced voltage. The correct statements are I, II and III only.

Explanation: Both experiments have identical secondary RMS voltage and current. Therefore, their secondary apparent powers are equal. However, the loads have different power factors and consequently absorb different real powers. The transformer losses remain fixed by the explicit assumptions, so the efficiency changes. \(\textbf{Secondary apparent power:}\) \[ S_s=V_sI_s \] Substituting, \[ S_s=(100)(5.0) \] \[ S_s=500\,\text{VA} \] \(\textbf{Total transformer loss:}\) \[ P_{\text{loss}}=P_{\text{core}}+P_{\text{winding}} \] \[ P_{\text{loss}}=20+30 \] \[ P_{\text{loss}}=50\,\text{W} \] This value is unchanged between P and Q. \(\textbf{Real output power in P:}\) \[ P_{\text{out},P}=S_s\cos\phi_P \] \[ P_{\text{out},P}=(500)(0.60) \] \[ P_{\text{out},P}=300\,\text{W} \] \(\textbf{Real input power in P:}\) \[ P_{\text{in},P}=P_{\text{out},P}+P_{\text{loss}} \] \[ P_{\text{in},P}=300+50 \] \[ P_{\text{in},P}=350\,\text{W} \] \(\textbf{Efficiency in P:}\) \[ \eta_P=\frac{P_{\text{out},P}}{P_{\text{in},P}} \] \[ \eta_P=\frac{300}{350} \] \[ \eta_P=\frac67 \] \(\textbf{Real output power in Q:}\) The new power factor is unity: \[ P_{\text{out},Q}=(500)(1) \] \[ P_{\text{out},Q}=500\,\text{W} \] \(\textbf{Real input power in Q:}\) \[ P_{\text{in},Q}=500+50 \] \[ P_{\text{in},Q}=550\,\text{W} \] \(\textbf{Efficiency in Q:}\) \[ \eta_Q=\frac{500}{550} \] \[ \eta_Q=\frac{10}{11} \] \(\textbf{Compare internal losses:}\) \[ \frac{P_{\text{loss},Q}}{P_{\text{loss},P}}=\frac{50}{50} \] \[ \frac{P_{\text{loss},Q}}{P_{\text{loss},P}}=1 \] \(\textbf{Physical interpretation:}\) The higher power factor allows the transformer to deliver more useful real power at the same secondary RMS voltage and current. The internal loss remains \(50\,\text{W}\) in both experiments. This fixed loss forms a smaller fraction of the useful output in Q. The transformer therefore becomes more efficient even though its absolute internal heating is unchanged. The result distinguishes efficiency improvement from loss reduction. \(\textbf{Final answer:}\) \[ \boxed{\eta_P=\frac67,\qquad \eta_Q=\frac{10}{11},\qquad \frac{P_{\text{loss},Q}}{P_{\text{loss},P}}=1} \]

Explanation: A secondary short circuit removes the external load resistance but does not remove the internal induced emf. Current is then limited principally by the effective internal winding impedance. In the stated simplified model, this impedance is purely resistive. The short-circuit current can therefore be much larger than the ordinary loaded current. \(\textbf{Current in normal operation:}\) The total resistance in P is \[ R_P=18+2.0 \] \[ R_P=20\,\Omega \] Therefore, \[ I_P=\frac{E_s}{R_P} \] \[ I_P=\frac{100}{20} \] \[ I_P=5.0\,\text{A} \] \(\textbf{Short-circuit resistance:}\) In Q, external resistance is negligible. Therefore, \[ R_Q=2.0\,\Omega \] \(\textbf{Short-circuit current:}\) \[ I_Q=\frac{100}{2.0} \] \[ I_Q=50\,\text{A} \] \(\textbf{Compare current magnitudes:}\) \[ \frac{I_Q}{I_P}=\frac{50}{5.0} \] \[ \frac{I_Q}{I_P}=10 \] \(\textbf{Winding heating during the short circuit:}\) \[ P_{\text{winding},Q}=I_Q^2r \] Substituting, \[ P_{\text{winding},Q}=(50)^2(2.0) \] \[ P_{\text{winding},Q}=5000\,\text{W} \] \(\textbf{Power delivered to an external load:}\) A short circuit has negligible external resistance. Therefore, \[ P_{\text{external},Q}=I_Q^2R_{\text{external}} \] \[ P_{\text{external},Q}=0 \] \(\textbf{Internal energy conversion:}\) The internally generated secondary emf supplies \[ P_{\text{internal}}=E_sI_Q \] \[ P_{\text{internal}}=(100)(50) \] \[ P_{\text{internal}}=5000\,\text{W} \] This entire power is dissipated in the effective winding resistance in the stated model. \(\textbf{Physical interpretation:}\) A nearly zero secondary terminal voltage does not imply that the transformer has stopped producing induced emf. During a short circuit, the internal emf drives current through the remaining winding resistance. The resulting current can become dangerously large. Its heating effect rises with current squared. The experiment explains why a shorted transformer secondary is fundamentally different from an open-circuited secondary, even though neither condition supplies useful power to an ordinary external load. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_Q}{I_P}=10,\qquad P_{\text{winding},Q}=5000\,\text{W},\qquad P_{\text{external},Q}=0} \]

Explanation: The generated emf depends on both the number of turns and the area enclosed by each turn. Increasing the number of turns does not necessarily increase total flux linkage when the total available wire length is fixed. Making more turns also makes each square turn smaller. \(\textbf{Geometry of design P:}\) The total perimeter is \[ 4a_P=8.0 \] Therefore, \[ a_P=2.0\,\text{m} \] Its area is \[ A_P=a_P^2 \] \[ A_P=4.0\,\text{m}^2 \] There is one turn: \[ N_P=1 \] Thus, \[ N_PA_P=4.0\,\text{m}^2 \] \(\textbf{Geometry of design Q:}\) The same total wire length is divided equally between two square turns. The perimeter of each turn is \[ \ell_Q=\frac{8.0}{2} \] \[ \ell_Q=4.0\,\text{m} \] Therefore, the side of each square is \[ a_Q=\frac{4.0}{4} \] \[ a_Q=1.0\,\text{m} \] The enclosed area per turn is \[ A_Q=1.0\,\text{m}^2 \] Since there are two turns, \[ N_QA_Q=(2)(1.0) \] \[ N_QA_Q=2.0\,\text{m}^2 \] \(\textbf{Compare generated peak emfs:}\) For a uniformly rotating coil, \[ E_m=NBA\omega \] Magnetic field and angular speed are unchanged. Therefore, \[ \frac{E_{m,Q}}{E_{m,P}}=\frac{N_QA_Q}{N_PA_P} \] \[ \frac{E_{m,Q}}{E_{m,P}}=\frac{2.0}{4.0} \] \[ \frac{E_{m,Q}}{E_{m,P}}=\frac12 \] \(\textbf{Compare total circuit resistances:}\) The same wire length, material and cross-sectional area are used. Consequently, the winding resistance is the same in both designs. The external load resistance is also unchanged. Thus, both complete electrical circuits have the same total resistance. \(\textbf{Compare external heating powers:}\) The RMS generated voltage is proportional to peak emf. With unchanged total resistance, \[ \frac{I_Q}{I_P}=\frac{E_{m,Q}}{E_{m,P}} \] \[ \frac{I_Q}{I_P}=\frac12 \] The same external resistor is used in both experiments: \[ P_R=I_{\text{rms}}^2R \] Therefore, \[ \frac{P_{R,Q}}{P_{R,P}}=\left(\frac12\right)^2 \] \[ \frac{P_{R,Q}}{P_{R,P}}=\frac14 \] \(\textbf{Compare generated frequencies:}\) Both coils rotate at the same angular speed. Therefore, their electrical frequencies are equal: \[ \frac{f_Q}{f_P}=1 \] \(\textbf{Physical interpretation:}\) Doubling the number of turns appears favourable only if the area per turn can be maintained. With a fixed wire-length budget, more turns require smaller individual loops. In this comparison, the decrease in area per turn more than offsets the increased number of turns. The generated emf becomes smaller even though the turn count increases. Because the total wire resistance and external resistor are unchanged, the useful heating power falls to one-quarter. This demonstrates why generator design cannot be judged from turn count alone. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_{m,Q}}{E_{m,P}}=\frac12,\qquad \frac{P_{R,Q}}{P_{R,P}}=\frac14,\qquad \frac{f_Q}{f_P}=1} \]

Explanation: The transformer operates through a common alternating magnetic flux linking its windings. When the secondary circuit is open, there is no load current in the secondary. The primary draws only the small magnetising current required by the idealised model. Connecting the resistor allows the induced secondary emf to drive a load current. That current produces its own magnetic effect. According to Lenz's law, the induced secondary current acts to oppose the change in magnetic flux responsible for its production. If this opposing effect were uncompensated, it would tend to alter the core flux. However, the primary is connected to a maintained sinusoidal voltage source. The source supplies additional primary current whose magnetic effect compensates for the opposing secondary effect. The common flux can therefore remain approximately determined by the primary applied voltage, frequency and turn count. At the same time, primary current increases to provide the real electrical power dissipated in the secondary resistor. The ideal transformer does not create this additional power. It draws it from the primary source. The Assertion correctly describes the simultaneous increase in primary current and approximately unchanged core flux. The Reason explains this behaviour through Lenz's law and the magnetic coupling between the windings. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: The transformer is driven by a fixed RMS primary voltage rather than a fixed primary current. Therefore, the applied voltage determines the required rate of change of magnetic flux linkage. Reducing source frequency while maintaining the same RMS voltage increases the magnetic-flux amplitude needed to produce the same induced emf. \(\textbf{Primary induced-emf relation:}\) For sinusoidal core flux, \[ \Phi_B(t)=\Phi_m\sin\omega t \] The primary induced-emf peak magnitude is \[ E_{p,m}=N_p\omega\Phi_m \] Therefore, the RMS magnitude is \[ V_{p,\text{rms}}=\frac{N_p\omega\Phi_m}{\sqrt2} \] \(\textbf{Recover the frequency dependence of flux:}\) Rearranging, \[ \Phi_m=\frac{\sqrt2V_{p,\text{rms}}}{N_p\omega} \] The applied RMS voltage and primary turn count remain unchanged. Thus, \[ \Phi_m\propto\frac1\omega \] \(\textbf{Compare maximum core fluxes:}\) The frequency is halved: \[ \frac{\omega_2}{\omega_1}=\frac{50}{100} \] \[ \frac{\omega_2}{\omega_1}=\frac12 \] Therefore, \[ \frac{\Phi_{m2}}{\Phi_{m1}}=\frac{\omega_1}{\omega_2} \] \[ \frac{\Phi_{m2}}{\Phi_{m1}}=2 \] \(\textbf{Secondary voltage relation:}\) For the ideal transformer, \[ \frac{V_s}{V_p}=\frac{N_s}{N_p} \] The primary RMS voltage and turns ratio are unchanged. Therefore, \[ V_{s2}=V_{s1} \] and \[ \frac{V_{s2}}{V_{s1}}=1 \] \(\textbf{Compare resistive load powers:}\) The same pure resistor remains connected across the secondary: \[ P=\frac{V_{s,\text{rms}}^2}{R} \] Since secondary RMS voltage is unchanged, \[ \frac{P_2}{P_1}=1 \] \(\textbf{Core magnetic-flux-density consequence:}\) The magnetic-core cross-sectional area remains constant: \[ B_m=\frac{\Phi_m}{A} \] Therefore, \[ \frac{B_{m2}}{B_{m1}}=2 \] \(\textbf{Physical interpretation:}\) Halving frequency reduces the rate at which a sinusoidal flux of a given amplitude changes. To maintain the same induced primary voltage, the transformer requires twice the magnetic-flux amplitude. The ideal secondary RMS voltage nevertheless remains unchanged because the reduced frequency is compensated by the increased flux amplitude. The useful resistive load power also remains unchanged. In a real magnetic core, the increased flux density may create a risk of saturation. The no-saturation condition is therefore important for the validity of this comparison. \(\textbf{Final answer:}\) \[ \boxed{\frac{\Phi_{m2}}{\Phi_{m1}}=2,\qquad \frac{V_{s2}}{V_{s1}}=1,\qquad \frac{P_2}{P_1}=1} \]

Explanation: The changing primary current produces changing magnetic flux linkage in the secondary. The strength of that linkage is represented by mutual inductance. Changing the relative position of the coils changes the induced emf amplitude without necessarily changing its frequency. \(\textbf{Primary current-change rate:}\) \[ i_p(t)=2\sqrt2\sin(100t) \] Differentiating, \[ \frac{di_p}{dt}=200\sqrt2\cos(100t) \] \(\textbf{Mutually induced secondary emf:}\) For fixed coupling, \[ e_s=-M\frac{di_p}{dt} \] Therefore, \[ e_s(t)=-200\sqrt2 M\cos(100t) \] \(\textbf{Arrangement P — Stronger coupling:}\) With \[ M_P=0.20\,\text{H} \] the induced emf is \[ e_{sP}(t)=-(200\sqrt2)(0.20)\cos(100t) \] \[ e_{sP}(t)=-40\sqrt2\cos(100t)\,\text{V} \] Its RMS magnitude is \[ E_{sP,\text{rms}}=40\,\text{V} \] \(\textbf{Secondary current and heating in P:}\) \[ I_{sP}=\frac{40}{20} \] \[ I_{sP}=2.0\,\text{A} \] Therefore, \[ P_P=I_{sP}^2R \] \[ P_P=(2.0)^2(20) \] \[ P_P=80\,\text{W} \] \(\textbf{Arrangement Q — Weaker coupling:}\) The mutual inductance becomes \[ M_Q=0.10\,\text{H} \] Therefore, \[ e_{sQ}(t)=-(200\sqrt2)(0.10)\cos(100t) \] \[ e_{sQ}(t)=-20\sqrt2\cos(100t)\,\text{V} \] Thus, \[ E_{sQ,\text{rms}}=20\,\text{V} \] \(\textbf{Compare induced emfs:}\) \[ \frac{E_{sQ,\text{rms}}}{E_{sP,\text{rms}}}=\frac{20}{40} \] \[ \frac{E_{sQ,\text{rms}}}{E_{sP,\text{rms}}}=\frac12 \] \(\textbf{New secondary current and power:}\) \[ I_{sQ}=\frac{20}{20} \] \[ I_{sQ}=1.0\,\text{A} \] Therefore, \[ P_Q=(1.0)^2(20) \] \[ P_Q=20\,\text{W} \] Hence, \[ \frac{P_Q}{P_P}=\frac{20}{80} \] \[ \frac{P_Q}{P_P}=\frac14 \] \(\textbf{Frequency comparison:}\) Both induced-emf expressions contain \[ \cos(100t) \] Therefore, their angular frequencies are identical: \[ \frac{f_Q}{f_P}=1 \] \(\textbf{Physical interpretation:}\) Reducing mutual inductance weakens the magnetic coupling between the primary and secondary. The externally maintained primary current still oscillates at the same frequency. Therefore, the induced secondary emf has the same frequency but smaller amplitude. With an unchanged resistive secondary circuit, current amplitude decreases proportionally to induced emf. Heating decreases according to the square of the current ratio. The experiment distinguishes the effect of coupling strength from the effect of the frequency of primary excitation. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_{sQ}}{E_{sP}}=\frac12,\qquad \frac{P_Q}{P_P}=\frac14,\qquad \frac{f_Q}{f_P}=1} \]

Explanation: The generator's changing magnetic flux produces induced emf whenever its coil rotates. However, induced emf does not automatically imply a substantial circulating current. The electrical connection determines whether the induced emf drives current through the winding. That current is responsible for electromagnetic braking and resistive heating. \(\textbf{Open-circuit experiment:}\) The terminals are disconnected. Therefore, neglecting leakage, \[ I_P=0 \] The winding still develops the specified induced emf. However, the average electromagnetic braking power associated with the external electrical circuit is zero: \[ P_{\text{electromagnetic},P}=0 \] At nonzero constant angular speed, \[ \tau_P=\frac{P_{\text{electromagnetic},P}}{\omega} \] \[ \tau_P=0 \] \(\textbf{Short-circuit experiment:}\) The external conductor has negligible resistance. The remaining circuit resistance is the winding resistance: \[ R_Q=3.0\,\Omega \] \(\textbf{Short-circuit RMS current:}\) \[ I_Q=\frac{E_{\text{rms}}}{R_Q} \] \[ I_Q=\frac{60}{3.0} \] \[ I_Q=20\,\text{A} \] \(\textbf{Winding-heating power:}\) \[ P_{\text{winding},Q}=I_Q^2R_Q \] Substituting, \[ P_{\text{winding},Q}=(20)^2(3.0) \] \[ P_{\text{winding},Q}=1200\,\text{W} \] \(\textbf{Average electromagnetic braking torque:}\) The mechanical drive must supply the average power dissipated in the winding: \[ P_{\text{mechanical},Q}=\tau_Q\omega \] Therefore, \[ \tau_Q=\frac{1200}{100} \] \[ \tau_Q=12\,\text{N m} \] \(\textbf{Useful external-load power:}\) The external connection has negligible resistance: \[ R_{\text{external}}=0 \] Thus, \[ P_{\text{useful},Q}=I_Q^2R_{\text{external}} \] \[ P_{\text{useful},Q}=0 \] \(\textbf{Physical interpretation:}\) An open-circuited generator can produce a measurable emf without driving a sustained external current. It consequently experiences negligible electromagnetic braking in this idealised model. A short-circuited generator, however, can carry a very large current because only its winding resistance limits the flow. The mechanical drive must overcome the resulting braking torque. The supplied mechanical energy becomes winding heat rather than useful external output. Thus, neither an open circuit nor a short circuit delivers useful average power to an external resistor, but their braking and heating behaviour are dramatically different. \(\textbf{Final answer:}\) \[ \boxed{\tau_P=0,\qquad \tau_Q=12\,\text{N m},\qquad P_{\text{useful},Q}=0} \]

Explanation: An AC transformer's apparent-power rating involves the product of RMS voltage and RMS current. It is not automatically the same as the real power delivered to a load. The load power factor determines the fraction of apparent power converted into useful average real power. \(\textbf{Secondary RMS voltage:}\) For the ideal transformer, \[ V_s=V_p\frac{N_s}{N_p} \] Substituting, \[ V_s=100\left(\frac{800}{400}\right) \] \[ V_s=200\,\text{V} \] \(\textbf{Maximum rated secondary current:}\) The apparent-power rating is \[ S_{\max}=1000\,\text{VA} \] Therefore, \[ I_{s,\max}=\frac{S_{\max}}{V_s} \] \[ I_{s,\max}=\frac{1000}{200} \] \[ I_{s,\max}=5.0\,\text{A} \] \(\textbf{Maximum real power for load P:}\) \[ P_P=V_sI_{s,\max}\cos\phi_P \] Substituting, \[ P_P=(200)(5.0)(0.60) \] \[ P_P=600\,\text{W} \] \(\textbf{Maximum real power for load Q:}\) The power factor is unity: \[ P_Q=(200)(5.0)(1) \] \[ P_Q=1000\,\text{W} \] \(\textbf{Primary RMS current at rated operation:}\) For the ideal transformer, \[ I_p=I_s\frac{N_s}{N_p} \] Substituting, \[ I_p=(5.0)\left(\frac{800}{400}\right) \] \[ I_p=10\,\text{A} \] The magnitude is the same for both loads. Therefore, \[ \frac{I_{p,Q}}{I_{p,P}}=1 \] \(\textbf{Primary real-power check:}\) For P, \[ P_{\text{in},P}=(100)(10)(0.60) \] \[ P_{\text{in},P}=600\,\text{W} \] For Q, \[ P_{\text{in},Q}=(100)(10)(1) \] \[ P_{\text{in},Q}=1000\,\text{W} \] Each primary input equals the corresponding secondary real output. \(\textbf{Physical interpretation:}\) Both loads draw the same rated RMS secondary current. The primary winding therefore carries the same RMS current magnitude in both experiments. However, the real powers differ because their power factors differ. At unity power factor, the full rated apparent power corresponds to useful average real power. At a power factor of \(0.60\), only \(60\%\) of the apparent-power rating becomes real output. The rating must therefore be interpreted as an RMS voltage-current limit rather than an unconditional wattage rating. \(\textbf{Final answer:}\) \[ \boxed{P_P=600\,\text{W},\qquad P_Q=1000\,\text{W},\qquad \frac{I_{p,Q}}{I_{p,P}}=1} \]

Explanation: The generator's peak emf increases with angular speed. However, the connected inductor's reactance also increases with electrical angular frequency. These two effects must be considered together. The generator is voltage-producing, but the external circuit is purely inductive rather than resistive. \(\textbf{Generated peak emf:}\) For a rotating coil, \[ E_m=NBA\omega \] Initially, \[ E_{m1}=NBA\omega \] After doubling speed, \[ E_{m2}=NBA(2\omega) \] Therefore, \[ \frac{E_{m2}}{E_{m1}}=2 \] \(\textbf{Inductive reactance:}\) For the ideal external inductor, \[ X_L=\omega L \] Initially, \[ X_{L1}=\omega L \] Finally, \[ X_{L2}=2\omega L \] Thus, \[ \frac{X_{L2}}{X_{L1}}=2 \] \(\textbf{Peak inductor current:}\) For sinusoidal steady operation, \[ I_m=\frac{E_m}{X_L} \] Substituting the generator and inductor relations, \[ I_m=\frac{NBA\omega}{\omega L} \] Cancelling angular frequency, \[ I_m=\frac{NBA}{L} \] The peak current is independent of angular speed under the stated assumptions. Therefore, \[ \frac{I_{m2}}{I_{m1}}=1 \] \(\textbf{Compare maximum stored magnetic energies:}\) The inductor stores \[ U_L=\frac12Li^2 \] Its maximum stored energy is \[ U_{\max}=\frac12LI_m^2 \] Because \(L\) and \(I_m\) remain unchanged, \[ \frac{U_{\max,2}}{U_{\max,1}}=1 \] \(\textbf{Average real electrical power:}\) The current through an ideal inductor lags its voltage by \(90^\circ\). Therefore, \[ P_{\text{avg}}=E_{\text{rms}}I_{\text{rms}}\cos90^\circ \] \[ P_{\text{avg}}=0 \] This holds in both experiments. \(\textbf{Average mechanical-power requirement:}\) With negligible mechanical and electrical losses, average mechanical power transferred to the electrical circuit equals average real electrical power. Therefore, \[ P_{\text{mechanical,avg}}=0 \] A driving mechanism may still need to exchange energy with the generator during individual parts of the cycle to maintain exactly constant angular speed. However, the net electrical work over a complete cycle is zero. \(\textbf{Physical interpretation:}\) Doubling generator speed doubles generated voltage amplitude and frequency. The inductor's reactance increases by exactly the same factor as voltage amplitude. Consequently, its current amplitude remains unchanged. The maximum magnetic energy stored in the external inductor also remains unchanged. The inductor alternately receives and returns energy, producing wattless current. This provides a useful contrast with a resistive generator load, where increased voltage amplitude generally increases average real-power consumption. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{m2}}{I_{m1}}=1,\qquad \frac{U_{\max,2}}{U_{\max,1}}=1,\qquad P_{\text{mechanical,avg}}=0} \]

Explanation: The rod moves perpendicular to its length and to the applied magnetic field. Magnetic force on its mobile charges produces charge separation along the rod. The motional emf depends on magnetic field, rod length and speed, not on whether the external circuit is open or closed. However, current flows only when a conducting return path is available. \(\textbf{Motional emf of the rod:}\) \[ \varepsilon=B\ell v \] Substituting, \[ \varepsilon=(0.80)(0.50)(5.0) \] \[ \varepsilon=2.0\,\text{V} \] \(\textbf{Open-switch observation:}\) With the switch open, no steady conduction current flows: \[ I_{\text{open}}=0 \] There is no resistive drop within the rod. Consequently, the ideal voltmeter reads the full motional emf: \[ V_{\text{open}}=2.0\,\text{V} \] \(\textbf{Current after closing the switch:}\) The rod and fixed resistor are now in series. The total circuit resistance is \[ R_{\text{total}}=1.0+3.0 \] \[ R_{\text{total}}=4.0\,\Omega \] Therefore, \[ I=\frac{\varepsilon}{R_{\text{total}}} \] \[ I=\frac{2.0}{4.0} \] \[ I=0.50\,\text{A} \] \(\textbf{Loaded terminal voltage:}\) The voltmeter measures the potential difference across the external resistor: \[ V_{\text{closed}}=IR_{\text{external}} \] \[ V_{\text{closed}}=(0.50)(3.0) \] \[ V_{\text{closed}}=1.5\,\text{V} \] The remaining \(0.50\,\text{V}\) is the internal resistive drop along the rod. \(\textbf{Magnetic force opposing motion:}\) A current-carrying rod in the perpendicular magnetic field experiences force \[ F=B I\ell \] Thus, \[ F=(0.80)(0.50)(0.50) \] \[ F=0.20\,\text{N} \] By Lenz's law, this magnetic force opposes the rod's motion. An equal external force must act to the right to maintain constant speed. \(\textbf{Mechanical-energy check:}\) \[ P_{\text{external}}=Fv \] \[ P_{\text{external}}=(0.20)(5.0) \] \[ P_{\text{external}}=1.0\,\text{W} \] The total Joule-heating power is \[ P_{\text{Joule}}=I^2R_{\text{total}} \] \[ P_{\text{Joule}}=(0.50)^2(4.0) \] \[ P_{\text{Joule}}=1.0\,\text{W} \] The powers agree. \(\textbf{Physical interpretation:}\) Opening the circuit removes the steady induced current but does not eliminate the motional emf. Closing the circuit allows current to flow and causes an internal voltage drop. It also produces a magnetic force opposing motion. The external agent must then perform mechanical work that becomes Joule heat. The voltmeter reading and the emf are equal only in the open-circuit condition or when internal resistance is negligible. \(\textbf{Final answer:}\) \[ \boxed{V_{\text{open}}=2.0\,\text{V},\qquad V_{\text{closed}}=1.5\,\text{V},\qquad F_{\text{external}}=0.20\,\text{N}} \]

Explanation: Both rods move through the same magnetic field, but they are the opposite boundaries of one closed circuit. The circuit area changes according to their relative velocity. The two rods also experience magnetic forces in opposite horizontal directions because current flows oppositely along them. \(\textbf{Rate of increase of enclosed area:}\) Let \(x_Q-x_P\) denote the separation of the rods. Then, \[ A=\ell(x_Q-x_P) \] Differentiating, \[ \frac{dA}{dt}=\ell(v_Q-v_P) \] Substituting, \[ \frac{dA}{dt}=(1.0)(5.0-2.0) \] \[ \frac{dA}{dt}=3.0\,\text{m}^2\text{s}^{-1} \] \(\textbf{Induced emf magnitude:}\) \[ |\varepsilon|=B\frac{dA}{dt} \] \[ |\varepsilon|=(0.50)(3.0) \] \[ |\varepsilon|=1.50\,\text{V} \] \(\textbf{Induced current:}\) \[ I=\frac{|\varepsilon|}{R} \] \[ I=\frac{1.50}{3.0} \] \[ I=0.50\,\text{A} \] \(\textbf{Determine current circulation:}\) The enclosed area increases while the magnetic field points out of the page. Outward magnetic flux therefore increases. By Lenz's law, the induced current must produce a field into the page. Viewed from the front, this requires clockwise circulation. \(\textbf{Magnetic force on each rod:}\) The force magnitude on either rod is \[ F=BI\ell \] Substituting, \[ F=(0.50)(0.50)(1.0) \] \[ F=0.25\,\text{N} \] Clockwise circulation carries current downward through the right rod Q and upward through left rod P. Using the magnetic-force direction rule, Q experiences force to the left and P experiences force to the right. \(\textbf{Power supplied to rod Q:}\) An external rightward force of \(0.25\,\text{N}\) is required to maintain Q's motion: \[ P_Q=Fv_Q \] \[ P_Q=(0.25)(5.0) \] \[ P_Q=+1.25\,\text{W} \] \(\textbf{Power supplied to rod P:}\) The magnetic force on P already acts in the direction of its motion. Its external agent must therefore exert a leftward force: \[ P_P=-Fv_P \] \[ P_P=-(0.25)(2.0) \] \[ P_P=-0.50\,\text{W} \] \(\textbf{Net mechanical input:}\) \[ P_{\text{net}}=P_Q+P_P \] \[ P_{\text{net}}=1.25-0.50 \] \[ P_{\text{net}}=0.75\,\text{W} \] \(\textbf{Electrical heating check:}\) \[ P_{\text{Joule}}=I^2R \] \[ P_{\text{Joule}}=(0.50)^2(3.0) \] \[ P_{\text{Joule}}=0.75\,\text{W} \] The net mechanical input exactly accounts for the circuit heating. \(\textbf{Physical interpretation:}\) The current is determined by the rate at which the distance between the rods increases, not by either rod's speed considered alone. Both rods experience magnetic forces of equal magnitude but opposite direction. The faster rod's external agent supplies energy, while the slower rod's external agent absorbs mechanical energy. Only the difference between these mechanical-power transfers becomes circuit heat. This is consistent with Faraday's law, Lenz's law and conservation of energy. \(\textbf{Final answer:}\) \[ \boxed{I=0.50\,\text{A}\ \text{(clockwise)},\qquad P_{\text{net}}=0.75\,\text{W}} \]

Explanation: The rod moves through a stationary but spatially nonuniform magnetic field. Its local motional emf depends on the field at its instantaneous position. Because the field becomes stronger towards the right, the induced current increases with time despite constant rod speed. The total transferred charge must account for the entire flux change. \(\textbf{Position of the rod:}\) The rod starts at \(x=0\) and moves at constant speed: \[ x=vt \] With \(v=1.0\,\text{m s}^{-1}\), \[ x=t \] where \(x\) is measured in metres when \(t\) is in seconds. \(\textbf{Magnetic field at the moving rod:}\) \[ B(t)=0.20+0.40t \] \(\textbf{Instantaneous motional emf:}\) \[ \varepsilon=B(t)\ell v \] Substituting, \[ \varepsilon(t)=(0.20+0.40t)(0.50)(1.0) \] \[ \varepsilon(t)=0.10+0.20t\,\text{V} \] \(\textbf{Induced current waveform:}\) \[ I(t)=\frac{\varepsilon(t)}R \] \[ I(t)=\frac{0.10+0.20t}{0.50} \] \[ I(t)=0.20+0.40t\,\text{A} \] \(\textbf{Charge transferred in the first second:}\) The transferred charge is the area under the current-time graph: \[ q=\int_0^1I(t)\,dt \] \[ q=\int_0^1(0.20+0.40t)\,dt \] \[ q=\left[0.20t+0.20t^2\right]_0^1 \] \[ q=0.20+0.20 \] \[ q=0.40\,\text{C} \] \(\textbf{Current at the end of the interval:}\) \[ I(1)=0.20+0.40(1) \] \[ I(1)=0.60\,\text{A} \] \(\textbf{Magnetic field at the final rod position:}\) At \(x=1.0\,\text{m}\), \[ B_f=0.20+0.40(1.0) \] \[ B_f=0.60\,\text{T} \] \(\textbf{Magnetic braking force:}\) \[ F=B_fI_f\ell \] Substituting, \[ F=(0.60)(0.60)(0.50) \] \[ F=0.18\,\text{N} \] The force acts opposite to the rightward motion. \(\textbf{Energy-rate consistency:}\) The final electrical power is \[ P_{\text{Joule}}=I_f^2R \] \[ P_{\text{Joule}}=(0.60)^2(0.50) \] \[ P_{\text{Joule}}=0.18\,\text{W} \] The final mechanical-power requirement is \[ P_{\text{mechanical}}=Fv \] \[ P_{\text{mechanical}}=(0.18)(1.0) \] \[ P_{\text{mechanical}}=0.18\,\text{W} \] The two power values agree. \(\textbf{Physical interpretation:}\) The rod speed is fixed, but the magnetic field experienced by the moving rod becomes stronger. Consequently, emf and current increase rather than remaining constant. The total charge is determined by the accumulated current throughout the interval, not by multiplying final current by total time. The final magnetic braking force depends on both the final field and the final current. The result integrates motional emf, spatially varying flux, charge transfer and mechanical opposition. \(\textbf{Final answer:}\) \[ \boxed{|q|=0.40\,\text{C},\qquad I_f=0.60\,\text{A},\qquad F_f=0.18\,\text{N}} \]