Electromagnetic Induction and Alternating Currents MCQs with Answers – Part 1 (NEET Physics)

Explanation: Magnetic flux measures the magnetic field passing through an oriented surface. For a plane surface placed in a uniform magnetic field, the magnetic flux is \[ \Phi_B=BA\cos\theta \] Here, \(\theta\) is the angle between the magnetic field and the area vector, not the angle between the magnetic field and the plane. The area vector is perpendicular to the plane of the coil. Since the magnetic field makes an angle of \(30^\circ\) with the plane, its angle with the area vector is \(60^\circ\). Therefore, \[ |\Phi_B|=BA\cos60^\circ \] Using the complementary-angle identity, \[ |\Phi_B|=BA\sin30^\circ=\frac{BA}{2} \] \(\textbf{Final answer:}\) \[ \boxed{|\Phi_B|=BA\sin30^\circ} \] A useful limiting check is that a field parallel to the plane gives zero flux, while a field perpendicular to the plane gives maximum flux magnitude. The SI unit of magnetic flux is weber, with \(1\,\text{Wb}=1\,\text{T m}^2\).

Explanation: \(\textbf{Given:}\) \[ B=0.30\,\text{T},\qquad A=0.20\,\text{m}^2 \] The loop initially has its chosen area vector parallel to the magnetic field. \(\textbf{Initial magnetic flux:}\) \[ \Phi_i=BA\cos0^\circ \] Substituting, \[ \Phi_i=(0.30)(0.20)(1) \] \[ \Phi_i=+0.060\,\text{Wb} \] The positive sign indicates that the field component is along the chosen area vector. \(\textbf{Flux after rotation:}\) After rotation, the angle between the chosen area vector and the magnetic field becomes \(120^\circ\). \[ \Phi_f=BA\cos120^\circ \] \[ \Phi_f=(0.30)(0.20)\left(-\frac12\right) \] \[ \Phi_f=-0.030\,\text{Wb} \] The negative sign means that the magnetic field component normal to the loop is now opposite to the chosen area-vector direction. The field magnitude and area have not changed; only the orientation has changed. Magnetic flux is a signed scalar associated with an oriented surface, so a negative value is physically meaningful. \(\textbf{Final answer:}\) \[ \boxed{(\Phi_i,\Phi_f)=(+0.060,-0.030)\,\text{Wb}} \]

Explanation: The Assertion is true because electromagnetic induction does not necessarily require mechanical motion of a conductor. Faraday's law relates induced emf to the time rate of change of magnetic flux linkage. For a stationary plane loop in a spatially uniform magnetic field, \[ \Phi_B=BA\cos\theta \] Even when \(A\) and \(\theta\) remain constant, a time-dependent magnetic field can change the flux through the loop. Consequently, the induced emf can be nonzero. For a coil having \(N\) identical turns, Faraday's law is \[ \varepsilon=-N\frac{d\Phi_B}{dt} \] The Reason is therefore also true. It directly identifies the physical condition responsible for induction in the stationary loop. The essential requirement is a changing magnetic flux linkage, not motion by itself. A stationary loop in a constant magnetic field has no induced emf merely because a nonzero magnetic flux exists. Conversely, a stationary loop exposed to an appropriately changing magnetic field can develop an induced emf without any displacement. Thus, both statements are correct, and the Reason provides the correct explanation of the Assertion.

Explanation: \(\textbf{Given:}\) \[ N=50,\quad A=0.040\,\text{m}^2,\quad \theta=60^\circ \] \[ B_i=0.20\,\text{T},\quad B_f=0.50\,\text{T},\quad \Delta t=0.10\,\text{s} \] The coil remains stationary, so its area and orientation are constant. \(\textbf{Magnetic flux change per turn:}\) \[ \Phi_B=BA\cos\theta \] Therefore, \[ \Delta\Phi_B=A(B_f-B_i)\cos\theta \] Substituting, \[ \Delta\Phi_B=(0.040)(0.50-0.20)\cos60^\circ \] \[ \Delta\Phi_B=(0.040)(0.30)\left(\frac12\right) \] \[ \Delta\Phi_B=0.0060\,\text{Wb} \] \(\textbf{Apply Faraday's law:}\) For \(N\) identical turns, \[ |\varepsilon_{\text{avg}}|=N\frac{|\Delta\Phi_B|}{\Delta t} \] \[ |\varepsilon_{\text{avg}}|=\frac{50(0.0060)}{0.10} \] \[ |\varepsilon_{\text{avg}}|=\frac{0.30}{0.10}=3.0\,\text{V} \] The factor \(N\) is essential because the induced emf depends on total flux linkage. The factor \(\cos60^\circ\) is equally important because only the field component normal to the coil contributes to magnetic flux. Since the field changes uniformly, the induced emf is constant during this interval under the stated geometry. \(\textbf{Final answer:}\) \[ \boxed{|\varepsilon_{\text{avg}}|=3.0\,\text{V}} \]

Explanation: The induced emf is determined by the slope of the magnetic-flux-versus-time graph, not by the flux value itself. Faraday's law gives \[ \varepsilon=-N\frac{d\Phi_B}{dt} \] \(\textbf{Statement I:}\) During the first interval, flux decreases uniformly from a positive value to an equal-magnitude negative value. It therefore passes through zero at the midpoint, \(t=0.05\,\text{s}\). However, the graph has a constant nonzero negative slope throughout this interval. Consequently, the induced emf remains nonzero when the flux crosses zero. Statement I is true. \(\textbf{Statement II:}\) During the second interval, the graph is horizontal. A horizontal flux-time graph has zero slope. Therefore, the induced emf is zero throughout this interval even though the magnetic flux has the nonzero value \(-0.020\,\text{Wb}\). Statement II is also true. The important distinction is between magnetic flux and the rate of change of magnetic flux. Zero magnetic flux does not necessarily imply zero induced emf. Similarly, a nonzero but constant magnetic flux does not produce an induced emf. Thus, both statements are correct.

Explanation: The changing magnetic field changes the flux linkage through each stationary coil. According to Faraday's law, this produces an induced emf in both coils. The existence of induced emf does not require a closed conducting circuit. Coil P has an effectively open conducting path because the ideal voltmeter draws negligible current. It can therefore indicate an induced voltage while appreciable conduction current remains absent. Coil Q, however, provides a closed conducting path through the galvanometer and resistance. The induced emf can drive a current through this path, resulting in galvanometer deflection. The magnitude of that current depends on the induced emf and the circuit resistance. Since the coils have identical flux-linkage histories, their induced-emf magnitudes are identical under the stated conditions. When the magnetic field becomes constant, the flux linkage also becomes constant. The rate of change of flux linkage then becomes zero, so both induced emfs vanish. Consequently, the induced current in the purely resistive circuit Q also becomes zero. A constant magnetic field may maintain nonzero magnetic flux, but it cannot maintain induction merely because that flux exists. This distinction between induced emf and induced current is fundamental to interpreting electromagnetic-induction experiments.

Explanation: Both processes produce induction, but they change magnetic flux by different physical mechanisms. \(\textbf{Process P — Changing field strength:}\) The coil orientation remains fixed, with \(\theta=0^\circ\). \[ \Delta\Phi_P=A(B_f-B_i)\cos0^\circ \] \[ \Delta\Phi_P=(0.10)(0.50-0.20)(1) \] \[ \Delta\Phi_P=+0.030\,\text{Wb per turn} \] Using Faraday's law, \[ \varepsilon_{P,\text{avg}}=-N\frac{\Delta\Phi_P}{\Delta t_P} \] \[ \varepsilon_{P,\text{avg}}=-\frac{40(0.030)}{0.10} \] \[ \varepsilon_{P,\text{avg}}=-12\,\text{V} \] \(\textbf{Process Q — Changing orientation:}\) The field strength and coil area remain constant. \[ \Delta\Phi_Q=BA(\cos60^\circ-\cos0^\circ) \] \[ \Delta\Phi_Q=(0.40)(0.10)\left(\frac12-1\right) \] \[ \Delta\Phi_Q=-0.020\,\text{Wb per turn} \] Therefore, \[ \varepsilon_{Q,\text{avg}}=-\frac{40(-0.020)}{0.20} \] \[ \varepsilon_{Q,\text{avg}}=+4\,\text{V} \] \(\textbf{Magnitude comparison:}\) \[ |\varepsilon_P|:|\varepsilon_Q|=12:4=3:1 \] The induced emfs have opposite algebraic signs because the flux increases in P but decreases in Q. Their opposite signs correspond to opposite induced-emf polarities under the specified common orientation convention. The essential reasoning is that field variation and orientation variation can both cause induction, but the signed flux change and elapsed time determine the average emf. \(\textbf{Final answer:}\) The average induced-emf magnitudes are in the ratio \(3:1\), with opposite polarities.

Explanation: \(\textbf{Graph interpretation:}\) The slope of each straight segment gives the rate of change of magnetic flux per turn. For \(N=20\) turns, \[ \varepsilon=-N\frac{\Delta\Phi_B}{\Delta t} \] \(\textbf{First interval — From }0\textbf{ to }0.10\,\text{s}\textbf{:}\) \[ \Delta\Phi_1=0.010-(-0.020)=+0.030\,\text{Wb} \] \[ \Delta t_1=0.10\,\text{s} \] Therefore, \[ \varepsilon_1=-20\left(\frac{+0.030}{0.10}\right) \] \[ \varepsilon_1=-6\,\text{V} \] \(\textbf{Second interval — From }0.10\textbf{ to }0.30\,\text{s}\textbf{:}\) The magnetic flux remains constant. \[ \Delta\Phi_2=0 \] Hence, \[ \varepsilon_2=0\,\text{V} \] \(\textbf{Third interval — From }0.30\textbf{ to }0.40\,\text{s}\textbf{:}\) \[ \Delta\Phi_3=-0.010-(+0.010)=-0.020\,\text{Wb} \] \[ \Delta t_3=0.10\,\text{s} \] Therefore, \[ \varepsilon_3=-20\left(\frac{-0.020}{0.10}\right) \] \[ \varepsilon_3=+4\,\text{V} \] \(\textbf{Physical interpretation:}\) The first segment has positive flux-time slope and hence negative induced emf. The middle segment has zero slope and hence zero emf. The final segment has negative slope and hence positive induced emf. The flux values themselves can be positive or negative without directly determining the induced-emf sign. The algebraic sign depends on the direction of flux change and the fixed circuit convention. The values describe the open intervals; at an ideal sharp corner, the instantaneous slope is not uniquely defined. \(\textbf{Final answer:}\) \[ \boxed{(\varepsilon_1,\varepsilon_2,\varepsilon_3)=(-6,0,+4)\,\text{V}} \]

Explanation: Faraday's law determines the induced emf from the rate of change of magnetic flux linkage. For either coil, \[ |\varepsilon|=N\left|\frac{d\Phi_B}{dt}\right| \] The two coils have identical numbers of turns and identical flux histories. Therefore, their induced-emf magnitudes are equal. Circuit resistance does not directly appear in Faraday's emf relation. The induced current, however, depends on the total resistance of the conducting path. With negligible self-inductance, Ohm's law gives \[ I=\frac{|\varepsilon|}{R} \] For coil P, \[ I_P=\frac{|\varepsilon_P|}{R_P} \] For coil Q, \[ I_Q=\frac{|\varepsilon_Q|}{R_Q} \] Since \(R_Q=2R_P\) and \(|\varepsilon_Q|=|\varepsilon_P|\), \[ \frac{I_Q}{I_P}=\frac{R_P}{R_Q}=\frac12 \] Thus the second coil carries half the current while having the same induced-emf magnitude. The essential distinction is between the cause of emf and the circuit response to that emf. The prescribed changing flux fixes the emf, whereas resistance determines the resulting current. \(\textbf{Final answer:}\) \[ \boxed{|\varepsilon_Q|=|\varepsilon_P|,\qquad I_Q=\frac{I_P}{2}} \]

Explanation: \(\textbf{Given:}\) \[ N=25,\qquad R=5\,\Omega \] The magnetic flux changes in two successive stages. Since the coil circuit is resistive, \[ I=\frac{\varepsilon}{R} \] The algebraic charge transferred is \[ q=\int I\,dt \] Combining this with Faraday's law for constant resistance gives \[ q=-\frac{N}{R}\Delta\Phi_B \] \(\textbf{Stage 1 — Increasing magnetic flux:}\) \[ \Delta\Phi_1=0.040-0=+0.040\,\text{Wb} \] Therefore, \[ q_1=-\frac{25}{5}(0.040) \] \[ q_1=-0.20\,\text{C} \] \(\textbf{Stage 2 — Decreasing magnetic flux:}\) \[ \Delta\Phi_2=0.010-0.040=-0.030\,\text{Wb} \] Therefore, \[ q_2=-\frac{25}{5}(-0.030) \] \[ q_2=+0.15\,\text{C} \] \(\textbf{Net algebraic charge:}\) The induced-current direction reverses between the two stages because the direction of flux change reverses. Charges crossing the chosen section in opposite directions contribute opposite algebraic signs. Hence, \[ q_{\text{net}}=q_1+q_2 \] \[ q_{\text{net}}=-0.20+0.15 \] \[ q_{\text{net}}=-0.05\,\text{C} \] The magnitude of the total charge passage without regard to direction would be \(0.35\,\text{C}\), but that is not the requested algebraic charge. The durations affect the induced-current magnitudes, but with constant resistance the net signed charge depends only on the initial and final flux values. \(\textbf{Final answer:}\) \[ \boxed{q_{\text{net}}=-0.05\,\text{C}} \]

Explanation: The magnetic flux through the loop depends jointly on field strength and orientation. For a plane loop in a uniform magnetic field, \[ \Phi_B=BA\cos\theta \] \(\textbf{Initial condition:}\) The area vector is parallel to the magnetic field. \[ \Phi_i=BA\cos0^\circ=BA \] \(\textbf{Final condition:}\) The field strength has increased to \(2B\), while the angle has become \(60^\circ\). \[ \Phi_f=(2B)A\cos60^\circ \] \[ \Phi_f=(2B)A\left(\frac12\right)=BA \] Thus, \[ \Delta\Phi_B=\Phi_f-\Phi_i=0 \] Faraday's law gives the average induced emf: \[ \varepsilon_{\text{avg}}=-\frac{\Delta\Phi_B}{\Delta t}=0 \] However, equal initial and final flux values do not establish that the flux stayed constant during the entire interval. The field strength and orientation can vary in such a way that the flux changes temporarily before returning to its original value. In that case, the instantaneous induced emf becomes nonzero during portions of the process. Alternatively, if \(B(t)\cos\theta(t)\) is deliberately maintained constant at every instant, the instantaneous induced emf remains zero throughout. The given information does not specify the detailed time dependence of these simultaneous changes. Therefore, zero average emf is certain, but zero instantaneous emf at every instant is not. This distinction prevents the common error of treating zero net flux change as proof that no induction occurred at any intermediate instant.

Explanation: The average induced emf depends on total flux change divided by elapsed time. The instantaneous induced emf depends on the instantaneous time derivative of the flux. These are different calculations when the flux-time relation is nonlinear. \(\textbf{Given:}\) \[ N=10,\qquad \Phi_B(t)=0.06-0.04t+0.01t^2 \] \(\textbf{Stage 1 — Average induced emf:}\) At \(t=0\), \[ \Phi_B(0)=0.06\,\text{Wb} \] At \(t=2\,\text{s}\), \[ \Phi_B(2)=0.06-0.04(2)+0.01(2)^2 \] \[ \Phi_B(2)=0.06-0.08+0.04=0.02\,\text{Wb} \] Therefore, \[ \Delta\Phi_B=0.02-0.06=-0.04\,\text{Wb} \] Using Faraday's law, \[ \varepsilon_{\text{avg}}=-N\frac{\Delta\Phi_B}{\Delta t} \] \[ \varepsilon_{\text{avg}}=-10\left(\frac{-0.04}{2}\right) \] \[ \varepsilon_{\text{avg}}=+0.20\,\text{V} \] \(\textbf{Stage 2 — Instantaneous induced emf:}\) Differentiate the supplied flux expression: \[ \frac{d\Phi_B}{dt}=-0.04+0.02t \] Hence, \[ \varepsilon(t)=-10(-0.04+0.02t) \] At \(t=2\,\text{s}\), \[ \varepsilon(2)=-10[-0.04+0.02(2)] \] \[ \varepsilon(2)=0\,\text{V} \] \(\textbf{Physical interpretation:}\) The magnetic flux has decreased over the complete two-second interval, giving a positive average induced emf. At \(t=2\,\text{s}\), however, the instantaneous slope of the flux-time curve is zero. The instantaneous induced emf therefore vanishes at that particular instant. There is no contradiction between a nonzero average emf and zero instantaneous emf at the endpoint. \(\textbf{Final answer:}\) \[ \boxed{(\varepsilon_{\text{avg}},\varepsilon_{t=2\,\text{s}})=(+0.20,0)\,\text{V}} \]

Explanation: \(\textbf{Given:}\) Both coils have the same number of turns and equal magnitudes of flux change per turn. \[ N=40,\qquad |\Delta\Phi_B|=0.010\,\text{Wb} \] Their flux-change durations and resistances are different. \(\textbf{Coil X — Induced emf:}\) \[ |\varepsilon_X|=N\frac{|\Delta\Phi_B|}{\Delta t_X} \] \[ |\varepsilon_X|=\frac{40(0.010)}{0.20} \] \[ |\varepsilon_X|=2\,\text{V} \] Current in coil X: \[ I_X=\frac{|\varepsilon_X|}{R_X}=\frac{2}{4} \] \[ I_X=0.50\,\text{A} \] \(\textbf{Coil Y — Induced emf:}\) \[ |\varepsilon_Y|=\frac{40(0.010)}{0.10} \] \[ |\varepsilon_Y|=4\,\text{V} \] Current in coil Y: \[ I_Y=\frac{|\varepsilon_Y|}{R_Y}=\frac{4}{8} \] \[ I_Y=0.50\,\text{A} \] \(\textbf{Compare Joule-heating powers:}\) For a resistive circuit, \[ P=I^2R \] The induced-current magnitudes are equal, but the circuit resistances differ. \[ P_X=(0.50)^2(4) \] \[ P_X=1\,\text{W} \] Similarly, \[ P_Y=(0.50)^2(8) \] \[ P_Y=2\,\text{W} \] Therefore, \[ \frac{P_Y}{P_X}=\frac{2}{1}=2 \] The faster flux change in Y doubles its induced emf. Doubling the circuit resistance simultaneously keeps its current equal to that in X. However, the Joule-heating power is twice as large in Y because the same current passes through twice the resistance. The equal currents must not be mistaken for equal heating powers. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_Y}{P_X}=2} \]

Explanation: The induced-emf history provides information about changes in magnetic flux linkage. Faraday's law can therefore be used in reverse to reconstruct the final magnetic field. \(\textbf{Initial magnetic flux per turn:}\) \[ B_i=0.60\,\text{T},\qquad A=0.050\,\text{m}^2 \] Since the field initially points along the chosen area vector, \[ \Phi_i=B_iA \] \[ \Phi_i=(0.60)(0.050)=+0.030\,\text{Wb} \] \(\textbf{First interval — Negative induced emf:}\) For a constant emf over a finite interval, \[ \varepsilon=-N\frac{\Delta\Phi_B}{\Delta t} \] Rearranging, \[ \Delta\Phi_B=-\frac{\varepsilon\Delta t}{N} \] Substituting the first-interval data, \[ \Delta\Phi_1=-\frac{(-2.0)(0.10)}{10} \] \[ \Delta\Phi_1=+0.020\,\text{Wb} \] The flux after the first interval is \[ \Phi_1=0.030+0.020=+0.050\,\text{Wb} \] \(\textbf{Second interval — Positive induced emf:}\) \[ \Delta\Phi_2=-\frac{(+4.0)(0.10)}{10} \] \[ \Delta\Phi_2=-0.040\,\text{Wb} \] The final flux is therefore \[ \Phi_f=\Phi_1+\Delta\Phi_2 \] \[ \Phi_f=0.050-0.040=+0.010\,\text{Wb} \] \(\textbf{Recover the final magnetic field:}\) The field remains perpendicular to the coil, so the signed flux is \(BA\). \[ B_{f,\text{signed}}=\frac{\Phi_f}{A} \] \[ B_{f,\text{signed}}=\frac{+0.010}{0.050} \] \[ B_{f,\text{signed}}=+0.20\,\text{T} \] The positive sign establishes that the final field points along the chosen area vector. The final magnetic field is weaker than the initial field but has not reversed direction. The negative and positive induced-emf intervals correspond to increasing and decreasing signed flux, respectively. Both changes must be included; using only the second interval would give an incorrect result. \(\textbf{Final answer:}\) The final field has magnitude \(0.20\,\text{T}\) and points along the area vector.

Explanation: Lenz's law states that the direction of induced current is such that the magnetic field produced by it opposes the change in magnetic flux responsible for induction. This opposition concerns the change in magnetic flux, not necessarily the original magnetic field itself. \(\textbf{Situation I:}\) The external magnetic field is directed into the page and is increasing. The loop therefore experiences increasing magnetic flux into the page. To oppose this increase, the induced current must produce a magnetic field directed out of the page. By the right-hand thumb rule, an anticlockwise current produces a magnetic field out of the page when viewed from the front. Hence, the induced current is anticlockwise. \(\textbf{Situation II:}\) The external magnetic field is directed out of the page but is decreasing in magnitude. The outward magnetic flux through the loop is therefore decreasing. To oppose this decrease, the induced current must produce a magnetic field directed out of the page. Again, the right-hand thumb rule requires an anticlockwise current. Thus, the two physically different changes produce induced currents in the same direction. A common mistake is to assume that reversing the original magnetic-field direction must always reverse the induced current. The direction of change must also be considered. An increasing inward field and a decreasing outward field both require an induced field directed outward. Therefore, both induced currents are anticlockwise.

Explanation: The moving rod changes the area enclosed by the conducting circuit. Since the magnetic field is perpendicular to the circuit plane, the changing area produces an induced emf. \(\textbf{Motional-emf relation:}\) For the stated perpendicular geometry, \[ |\varepsilon|=Blv \] where \(l\) is the separation between the rails. \(\textbf{Initial condition:}\) \[ B=0.50\,\text{T},\quad l=0.40\,\text{m},\quad v_1=5.0\,\text{m s}^{-1} \] \[ |\varepsilon_1|=(0.50)(0.40)(5.0) \] \[ |\varepsilon_1|=1.0\,\text{V} \] The initial circuit resistance is \(R_1=2.0\,\Omega\). \[ I_1=\frac{|\varepsilon_1|}{R_1}=\frac{1.0}{2.0} \] \[ I_1=0.50\,\text{A} \] \(\textbf{Second condition — Doubled speed and resistance:}\) \[ v_2=10.0\,\text{m s}^{-1},\qquad R_2=4.0\,\Omega \] The new induced emf is \[ |\varepsilon_2|=Blv_2 \] \[ |\varepsilon_2|=(0.50)(0.40)(10.0) \] \[ |\varepsilon_2|=2.0\,\text{V} \] The new induced current is \[ I_2=\frac{|\varepsilon_2|}{R_2}=\frac{2.0}{4.0} \] \[ I_2=0.50\,\text{A} \] Thus, doubling both speed and resistance leaves the induced-current magnitude unchanged. \(\textbf{Magnetic force on the rod:}\) For a current-carrying rod perpendicular to the magnetic field, \[ F_B=BI_2l \] \[ F_B=(0.50)(0.50)(0.40) \] \[ F_B=0.10\,\text{N} \] By Lenz's law, this magnetic force opposes the rod's rightward motion. To maintain constant speed, the external agent must apply an equal rightward force. \(\textbf{Mechanical power required:}\) \[ P_{\text{mech}}=F_{\text{ext}}v_2 \] \[ P_{\text{mech}}=(0.10)(10.0) \] \[ P_{\text{mech}}=1.0\,\text{W} \] \(\textbf{Energy consistency check:}\) The supplied mechanical power should equal the rate of electrical energy dissipation in the resistor for the stated ideal arrangement. \[ P_{\text{electrical}}=I_2^2R_2 \] \[ P_{\text{electrical}}=(0.50)^2(4.0) \] \[ P_{\text{electrical}}=1.0\,\text{W} \] The mechanical and electrical powers agree. The magnetic force remains unchanged because the current remains unchanged, but the mechanical power doubles because the rod moves twice as fast. This is consistent with conservation of energy and Lenz's law. \(\textbf{Final answer:}\) \[ \boxed{F_B=0.10\,\text{N},\qquad P_{\text{mech}}=1.0\,\text{W}} \]

Explanation: The induced emf in a moving conductor depends on the magnetic force acting on its mobile charges. The magnetic force on a positive charge is directed along \(\vec v\times\vec B\). Only the component of this force parallel to the rod contributes to the potential difference between its ends. \(\textbf{Resolve the velocity:}\) The rod lies along the \(y\)-axis, while its velocity has components along both coordinate axes. For a magnetic field perpendicular to the \(xy\)-plane, the velocity component along the \(x\)-axis produces magnetic force along the rod. Therefore, \[ |\varepsilon|=Blv_x \] Since the velocity makes an angle \(\alpha\) with the positive \(x\)-axis, \[ v_x=v\cos\alpha \] Hence, \[ |\varepsilon|=Blv\cos\alpha \] \(\textbf{Experiment P:}\) \[ |\varepsilon_P|=Blv\cos30^\circ \] \[ |\varepsilon_P|=\frac{\sqrt3}{2}Blv \] The magnetic field points into the plane. By the right-hand rule, \(\vec v\times\vec B\) has an upward component along the rod. Positive charges therefore accumulate at the upper end. \(\textbf{Experiment Q:}\) \[ |\varepsilon_Q|=Blv\cos60^\circ \] \[ |\varepsilon_Q|=\frac12Blv \] The magnetic field now points out of the plane. The component of magnetic force along the rod consequently reverses, making the upper end negative. \(\textbf{Magnitude comparison:}\) \[ \frac{|\varepsilon_P|}{|\varepsilon_Q|} =\frac{\sqrt3/2}{1/2} =\sqrt3 \] Reversing the magnetic field changes the polarity of the induced emf but does not change its magnitude when the field strength remains unchanged. The rod's full speed cannot be substituted directly into \(Blv\) without checking its direction relative to the rod and magnetic field. \(\textbf{Final answer:}\) The emf magnitudes are in the ratio \(\sqrt3:1\), with the upper end positive in P and negative in Q.

Explanation: A rotating conducting rod differs from a rod translating uniformly because different points of the rotating rod have different linear speeds. At a distance \(r\) from the rotation axis, the linear speed is \[ v=\omega r \] The magnetic force separates charges along the rotating rod. For a small radial element \(dr\), the magnitude of the induced emf is \[ d\varepsilon=Bv\,dr \] Substituting the rotational speed relation, \[ d\varepsilon=B\omega r\,dr \] \(\textbf{Initial measurement — Complete rod:}\) For the potential difference between the fixed end and the free end, \[ |\varepsilon_1|=\int_0^L B\omega r\,dr \] \[ |\varepsilon_1|=\frac12B\omega L^2 \] \(\textbf{Changed condition — Outer half of the rod:}\) The angular speed becomes \(2\omega\). The voltmeter now measures the emf only between \(r=L/2\) and \(r=L\). Therefore, \[ |\varepsilon_2|=\int_{L/2}^{L}B(2\omega)r\,dr \] \[ |\varepsilon_2|=B\omega\left[L^2-\left(\frac L2\right)^2\right] \] \[ |\varepsilon_2|=B\omega\left(L^2-\frac{L^2}{4}\right) \] \[ |\varepsilon_2|=\frac34B\omega L^2 \] \(\textbf{Compare the measurements:}\) \[ \frac{|\varepsilon_2|}{|\varepsilon_1|} =\frac{(3/4)B\omega L^2}{(1/2)B\omega L^2} \] \[ \frac{|\varepsilon_2|}{|\varepsilon_1|}=\frac32 \] The outer half of a rotating rod contributes more than half of the total induced emf because its points have larger linear speeds. Doubling the angular speed further increases the induced emf in the selected segment. Using \(Blv\) with one common linear speed for the entire rotating rod would give an incorrect result. \(\textbf{Final answer:}\) \[ \boxed{\frac{|\varepsilon_2|}{|\varepsilon_1|}=\frac32} \]

Explanation: A uniformly rotating coil experiences a continuously changing angle between its area vector and the magnetic field. Under the stated initial condition, the magnetic flux linkage is \[ \Phi(t)=\Phi_0\cos\omega t \] Faraday's law gives \[ \varepsilon(t)=-\frac{d\Phi}{dt} \] Therefore, \[ \varepsilon(t)=\omega\Phi_0\sin\omega t \] Since \(\varepsilon_0=\omega\Phi_0\), \[ \varepsilon(t)=\varepsilon_0\sin\omega t \] \(\textbf{Statement I:}\) The magnitude of flux linkage is maximum when \(|\cos\omega t|=1\). At these instants, \(\sin\omega t=0\). Thus, the induced emf is zero whenever the flux-linkage magnitude is maximum. Statement I is correct. \(\textbf{Statement II:}\) The given positive flux linkage requires \[ \cos\omega t=\frac12 \] The additional information that the flux is decreasing establishes \[ \frac{d\Phi}{dt}\lt 0 \] Consequently, the induced emf is positive under the stated convention. The corresponding sine value is \[ \sin\omega t=+\sqrt{1-\left(\frac12\right)^2} =\frac{\sqrt3}{2} \] Hence, \[ \varepsilon=+\frac{\sqrt3}{2}\varepsilon_0 \] Statement II is correct. \(\textbf{Statement III:}\) The frequency of the induced emf is determined by the angular speed: \[ f=\frac{\omega}{2\pi} \] Doubling the number of turns increases the maximum flux linkage and emf amplitude but does not change the angular frequency. Statement III is incorrect. The phase relationship between flux linkage and induced emf is therefore central to interpreting rotating-coil induction. Both quantities vary sinusoidally, but their maxima do not occur at the same instant. The correct combination is Statements I and II only.

Explanation: The instantaneous magnetic flux depends on both magnetic field strength and loop area. For a circular loop whose plane is perpendicular to the magnetic field, \[ \Phi_B=BA=B\pi r^2 \] Both \(B\) and \(r\) vary, so both contributions to the flux-change rate must be included. \(\textbf{Rate of flux change:}\) Differentiating the product gives \[ \frac{d\Phi_B}{dt} =\pi r^2\frac{dB}{dt}+2\pi Br\frac{dr}{dt} \] The first term represents the effect of changing field strength. The second term represents the effect of the expanding area. Because the magnetic field decreases while the loops expand, these two contributions have opposite signs. \(\textbf{Loop P:}\) \[ r_P=0.10\,\text{m} \] Substituting, \[ \frac{d\Phi_P}{dt} =\pi(0.10)^2(-0.20)+2\pi(0.50)(0.10)(0.020) \] \[ \frac{d\Phi_P}{dt}=-0.002\pi+0.002\pi \] \[ \frac{d\Phi_P}{dt}=0 \] Therefore, \[ \varepsilon_P=0 \] The two effects cancel exactly at this instant. \(\textbf{Loop Q:}\) \[ r_Q=0.20\,\text{m} \] \[ \frac{d\Phi_Q}{dt} =\pi(0.20)^2(-0.20)+2\pi(0.50)(0.20)(0.020) \] \[ \frac{d\Phi_Q}{dt}=-0.008\pi+0.004\pi \] \[ \frac{d\Phi_Q}{dt}=-0.004\pi\,\text{Wb s}^{-1} \] Using Faraday's law, \[ \varepsilon_Q=-\frac{d\Phi_Q}{dt} \] \[ \varepsilon_Q=+0.004\pi\,\text{V} \] The outward magnetic flux is decreasing in Q. By Lenz's law, the induced current must produce an outward magnetic field to oppose this decrease. An anticlockwise current, viewed from the front, produces the required outward field. The cancellation in P does not extend automatically to Q because the field-change contribution varies as \(r^2\), whereas the area-expansion contribution varies as \(r\) for fixed radial speed. Thus, equal rates of field variation and radial expansion do not guarantee equal induction effects for loops of different radii. \(\textbf{Final answer:}\) Loop P has zero induced emf, while loop Q has magnitude \(0.004\pi\,\text{V}\) and anticlockwise induced-current direction.