Electromagnetic Induction and Alternating Currents MCQs with Answers – Part 2 (NEET Physics)

Explanation: The inductors have different inductances but equal initial magnetic energies. Therefore, their initial currents cannot be equal. The initial-current relation must be established before comparing the self-induced emfs. \(\textbf{Equal initial magnetic energies:}\) For a linear inductor, \[ U=\frac12LI^2 \] The given condition is \[ U_P=U_Q \] Therefore, \[ \frac12L_PI_P^2=\frac12L_QI_Q^2 \] Since \[ L_Q=9L_P \] we obtain \[ L_PI_P^2=9L_PI_Q^2 \] Cancelling \(L_P\), \[ I_P^2=9I_Q^2 \] Both initial currents are positive, so \[ I_P=3I_Q \] Hence, \[ I_Q=\frac{I_P}{3} \] \(\textbf{Uniform reduction to zero:}\) The currents are reduced to zero over the same duration \(\Delta t\). For P, \[ \left|\frac{dI_P}{dt}\right|=\frac{I_P}{\Delta t} \] For Q, \[ \left|\frac{dI_Q}{dt}\right|=\frac{I_Q}{\Delta t} \] Thus, \[ \left|\frac{dI_Q}{dt}\right| =\frac13\left|\frac{dI_P}{dt}\right| \] \(\textbf{Compare self-induced emf magnitudes:}\) \[ |\varepsilon|=L\left|\frac{dI}{dt}\right| \] Therefore, \[ \frac{|\varepsilon_Q|}{|\varepsilon_P|} =\frac{L_Q(I_Q/\Delta t)}{L_P(I_P/\Delta t)} \] Substituting, \[ \frac{|\varepsilon_Q|}{|\varepsilon_P|} =\frac{9L_P(I_P/3)}{L_PI_P} \] \[ \frac{|\varepsilon_Q|}{|\varepsilon_P|}=3 \] \(\textbf{Compare released energies:}\) Both inductors finish with zero current. Their final stored energies are therefore zero. Each releases precisely its own initial stored magnetic energy under the stated ideal conditions. Since those initial energies are equal, their released energies are equal. The larger inductance produces a greater emf despite Q having a smaller current-change rate. This is possible because the emf depends on the product of inductance and current-change rate. Equal stored energies do not imply equal self-induced emfs during prescribed current histories. \(\textbf{Final answer:}\) \[ \boxed{\frac{|\varepsilon_Q|}{|\varepsilon_P|}=3,\qquad \frac{\Delta U_Q}{\Delta U_P}=1} \]

Explanation: An ideal inductor stores energy in its magnetic field. When its current rises, additional work must be done against the self-induced emf. The energy transferred depends on the change in \(I^2\), rather than only on the change in current. \(\textbf{Initial magnetic energy:}\) For a linear inductor, \[ U=\frac12LI^2 \] The inductance is \[ L=1.5\,\text{H} \] Initially, \[ I_i=2.0\,\text{A} \] Therefore, \[ U_i=\frac12(1.5)(2.0)^2 \] \[ U_i=0.75(4.0) \] \[ U_i=3.0\,\text{J} \] \(\textbf{Final magnetic energy:}\) The final current is \[ I_f=4.0\,\text{A} \] Hence, \[ U_f=\frac12(1.5)(4.0)^2 \] \[ U_f=0.75(16) \] \[ U_f=12\,\text{J} \] \(\textbf{Increase in stored energy:}\) \[ \Delta U=U_f-U_i \] \[ \Delta U=12-3.0 \] \[ \Delta U=9.0\,\text{J} \] \(\textbf{Average power delivered to the magnetic field:}\) The duration is \[ \Delta t=0.50\,\text{s} \] The average rate of energy transfer is \[ P_{\text{avg}}=\frac{\Delta U}{\Delta t} \] Substituting, \[ P_{\text{avg}}=\frac{9.0}{0.50} \] \[ P_{\text{avg}}=18\,\text{W} \] \(\textbf{Check using the induced voltage:}\) The current-change rate is \[ \frac{dI}{dt}=\frac{4.0-2.0}{0.50} \] \[ \frac{dI}{dt}=4.0\,\text{A s}^{-1} \] The voltage supplied across the ideal inductor in the direction of increasing current is \[ V_L=L\frac{dI}{dt} \] \[ V_L=(1.5)(4.0) \] \[ V_L=6.0\,\text{V} \] The instantaneous power entering the inductor is \[ P(t)=V_LI(t) \] Because current rises uniformly from \(2.0\) to \(4.0\,\text{A}\), its average value over the interval is \[ I_{\text{avg}}=\frac{2.0+4.0}{2}=3.0\,\text{A} \] Therefore, \[ P_{\text{avg}}=(6.0)(3.0) \] \[ P_{\text{avg}}=18\,\text{W} \] Both methods agree. The instantaneous power is not constant because the current increases even though the induced-voltage magnitude remains constant. The result also shows why multiplying the final power by the full time interval would overestimate the energy transferred. \(\textbf{Final answer:}\) \[ \boxed{\Delta U=9.0\,\text{J},\qquad P_{\text{avg}}=18\,\text{W}} \]

Explanation: The solenoid undergoes simultaneous changes in length, cross-sectional area and current. Its self-inductance depends on geometry, whereas its magnetic field also depends on current. Stored magnetic energy depends on both inductance and the square of current. These dependences must be applied separately. \(\textbf{Initial self-inductance:}\) For a long air-core solenoid, \[ L_i=\frac{\mu_0N^2A}{l} \] \(\textbf{Changed geometry:}\) The final dimensions are \[ l_f=2l \] \[ A_f=\frac A2 \] The number of turns remains \(N\). Therefore, \[ L_f=\frac{\mu_0N^2(A/2)}{2l} \] \[ L_f=\frac{\mu_0N^2A}{4l} \] Comparing with the initial expression, \[ L_f=\frac14L_i \] \(\textbf{Internal magnetic-field comparison:}\) For an ideal long solenoid, \[ B=\mu_0\frac Nl I \] Initially, \[ B_i=\mu_0\frac Nl I \] Finally, the current is \(2I\) and the length is \(2l\). \[ B_f=\mu_0\frac{N}{2l}(2I) \] Simplifying, \[ B_f=\mu_0\frac Nl I \] Therefore, \[ B_f=B_i \] \(\textbf{Stored magnetic-energy comparison:}\) Initially, \[ U_i=\frac12L_iI^2 \] Finally, \[ U_f=\frac12L_f(2I)^2 \] Substituting \(L_f=L_i/4\), \[ U_f=\frac12\left(\frac{L_i}{4}\right)(4I^2) \] \[ U_f=\frac12L_iI^2 \] Hence, \[ U_f=U_i \] \(\textbf{Consistency with magnetic-energy density:}\) The magnetic field remains unchanged. The solenoid volume also remains unchanged because \[ A_fl_f=\left(\frac A2\right)(2l)=Al \] For an air-core solenoid, magnetic-energy density depends on \(B^2\). Since both the field and the interior volume remain unchanged, the total magnetic energy is unchanged. This independently confirms the result obtained from \(\frac12LI^2\). The simultaneous parameter changes produce an invariant magnetic field and stored energy even though inductance decreases. \(\textbf{Final answer:}\) \[ \boxed{\frac{L_f}{L_i}=\frac14,\qquad \frac{B_f}{B_i}=1,\qquad \frac{U_f}{U_i}=1} \]

Explanation: The magnetic field inside an ideal long solenoid depends on permeability, turn density and current. Changing the magnetic medium therefore changes the current required to establish a given magnetic field. The self-inductance also changes because it depends on permeability. \(\textbf{Initial magnetic field:}\) For a long solenoid, \[ B=\mu\frac Nl I \] Initially, \[ B_0=\mu_0\frac Nl I_0 \] \(\textbf{Final current for unchanged field:}\) The new permeability is \[ \mu_f=4\mu_0 \] The field is required to remain \(B_0\). Therefore, \[ B_0=4\mu_0\frac Nl I_f \] Equating the initial and final field expressions, \[ \mu_0\frac Nl I_0=4\mu_0\frac Nl I_f \] Cancelling common factors, \[ I_0=4I_f \] Thus, \[ I_f=\frac{I_0}{4} \] \(\textbf{Inductance change:}\) For a long solenoid, \[ L=\frac{\mu N^2A}{l} \] The geometry and turn number are unchanged. Therefore, \[ L_f=4L_0 \] \(\textbf{Stored magnetic energy:}\) Initially, \[ U_0=\frac12L_0I_0^2 \] Finally, \[ U_f=\frac12L_fI_f^2 \] Substituting the changed quantities, \[ U_f=\frac12(4L_0)\left(\frac{I_0}{4}\right)^2 \] \[ U_f=\frac12(4L_0)\frac{I_0^2}{16} \] \[ U_f=\frac14\left(\frac12L_0I_0^2\right) \] Hence, \[ U_f=\frac{U_0}{4} \] \(\textbf{Physical check:}\) The field remains unchanged even though current decreases because the permeability increases. For a linear magnetic medium, \[ u_B=\frac{B^2}{2\mu} \] At fixed \(B\), increasing permeability by a factor of four reduces the magnetic-energy density by the same factor. Since the solenoid volume remains unchanged, this agrees with the calculated stored-energy ratio. The magnetic field, current, inductance and stored energy therefore do not all scale in the same way when the magnetic medium changes. \(\textbf{Final answer:}\) \[ \boxed{I_f=\frac{I_0}{4},\qquad U_f=\frac{U_0}{4}} \]

Explanation: The current changes uniformly from a positive value to an equally large negative value. The self-induced emf depends on the signed current-change rate. Magnetic energy, however, depends on the square of current. \(\textbf{Current-change rate:}\) \[ I_i=+2.0\,\text{A} \] \[ I_f=-2.0\,\text{A} \] \[ \Delta t=0.20\,\text{s} \] Therefore, \[ \frac{dI}{dt}=\frac{I_f-I_i}{\Delta t} \] \[ \frac{dI}{dt}=\frac{-2.0-(+2.0)}{0.20} \] \[ \frac{dI}{dt}=\frac{-4.0}{0.20} \] \[ \frac{dI}{dt}=-20\,\text{A s}^{-1} \] \(\textbf{Self-induced emf:}\) For constant inductance, \[ \varepsilon_L=-L\frac{dI}{dt} \] Substituting, \[ \varepsilon_L=-(0.40)(-20) \] \[ \varepsilon_L=+8.0\,\text{V} \] The emf retains this sign throughout the uniform reversal. \(\textbf{Initial magnetic energy:}\) \[ U_i=\frac12LI_i^2 \] \[ U_i=\frac12(0.40)(+2.0)^2 \] \[ U_i=0.80\,\text{J} \] \(\textbf{Final magnetic energy:}\) \[ U_f=\frac12LI_f^2 \] \[ U_f=\frac12(0.40)(-2.0)^2 \] \[ U_f=0.80\,\text{J} \] Hence, \[ \Delta U=U_f-U_i \] \[ \Delta U=0.80-0.80 \] \[ \Delta U=0 \] \(\textbf{Energy transfer during reversal:}\) At the midpoint of the uniform reversal, the current becomes zero. The stored magnetic energy also becomes zero at that instant. Energy is released by the inductor during the first half of the reversal. During the second half, energy is supplied again to establish a magnetic field in the opposite direction. The total initial and final energies are equal even though substantial energy transfer occurs during the process. A zero net change in stored energy therefore does not imply that the self-induced emf vanishes throughout the process. \(\textbf{Final answer:}\) \[ \boxed{\varepsilon_L=+8.0\,\text{V},\qquad \Delta U=0} \]

Explanation: The stored magnetic energy depends on current magnitude. Its instantaneous rate of change depends on both the current and its rate of change. Therefore, two inductors carrying the same current can have opposite energy-transfer directions. \(\textbf{Magnetic-energy relation:}\) For a linear inductor, \[ U=\frac12LI^2 \] Differentiating with respect to time while \(L\) remains constant, \[ \frac{dU}{dt}=LI\frac{dI}{dt} \] This gives the instantaneous rate at which magnetic energy changes. \(\textbf{Experiment P — Increasing current:}\) \[ L=2.0\,\text{H} \] \[ I=+3.0\,\text{A} \] \[ \frac{dI}{dt}=+0.50\,\text{A s}^{-1} \] Substituting, \[ \left(\frac{dU}{dt}\right)_P=(2.0)(3.0)(+0.50) \] \[ \left(\frac{dU}{dt}\right)_P=+3.0\,\text{W} \] The positive sign means that the magnetic field is gaining energy. \(\textbf{Experiment Q — Decreasing current:}\) The instantaneous current remains \(+3.0\,\text{A}\), but \[ \frac{dI}{dt}=-0.50\,\text{A s}^{-1} \] Therefore, \[ \left(\frac{dU}{dt}\right)_Q=(2.0)(3.0)(-0.50) \] \[ \left(\frac{dU}{dt}\right)_Q=-3.0\,\text{W} \] The negative sign means that the magnetic field is releasing stored energy. \(\textbf{Relation to self-induced emf:}\) The self-induced emf is \[ \varepsilon_L=-L\frac{dI}{dt} \] In P, \[ \varepsilon_P=-(2.0)(+0.50)=-1.0\,\text{V} \] In Q, \[ \varepsilon_Q=-(2.0)(-0.50)=+1.0\,\text{V} \] The equal emf magnitudes reflect equal current-change-rate magnitudes. Their opposite signs reflect the opposite directions of current change. The stored magnetic energy is increasing in P and decreasing in Q. The current magnitude alone is insufficient to determine whether an inductor is storing or releasing energy. \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{dU}{dt}\right)_P=+3.0\,\text{W},\qquad \left(\frac{dU}{dt}\right)_Q=-3.0\,\text{W}} \]

Explanation: A constant current does not always imply constant magnetic flux linkage. If the inductance changes because the magnetic medium or geometry changes, the flux linkage can change even while current remains constant. Faraday's law must be applied to the complete flux-linkage relation. \(\textbf{Changing flux linkage:}\) For the stated linear magnetic system, \[ \Lambda=LI \] The induced emf is \[ \varepsilon=-\frac{d\Lambda}{dt} \] Therefore, \[ \varepsilon=-\frac{d(LI)}{dt} \] Expanding, \[ \varepsilon=-L\frac{dI}{dt}-I\frac{dL}{dt} \] \(\textbf{Constant-current condition:}\) The current controller maintains \[ I=2.0\,\text{A} \] Therefore, \[ \frac{dI}{dt}=0 \] The emf reduces to \[ \varepsilon=-I\frac{dL}{dt} \] \(\textbf{Rate of inductance change:}\) \[ \Delta L=0.50-0.20 \] \[ \Delta L=0.30\,\text{H} \] The interval is \[ \Delta t=0.10\,\text{s} \] Hence, \[ \frac{dL}{dt}=\frac{0.30}{0.10} \] \[ \frac{dL}{dt}=3.0\,\text{H s}^{-1} \] Thus, \[ \varepsilon=-(2.0)(3.0) \] \[ \varepsilon=-6.0\,\text{V} \] The induced-emf magnitude is \(6.0\,\text{V}\). Its polarity opposes the increase in magnetic flux linkage caused by insertion of the core. \(\textbf{Initial stored magnetic energy:}\) \[ U_i=\frac12L_iI^2 \] \[ U_i=\frac12(0.20)(2.0)^2 \] \[ U_i=0.40\,\text{J} \] \(\textbf{Final stored magnetic energy:}\) \[ U_f=\frac12L_fI^2 \] \[ U_f=\frac12(0.50)(2.0)^2 \] \[ U_f=1.0\,\text{J} \] \(\textbf{Increase in energy:}\) \[ \Delta U=U_f-U_i \] \[ \Delta U=1.0-0.40 \] \[ \Delta U=0.60\,\text{J} \] The standard expression \(\varepsilon=-L\,dI/dt\) assumes constant inductance. It cannot be used alone when the inductance changes during the process. The increased magnetic energy involves the external current controller and the mechanical process of moving the core. The electrical energy transferred by the controller need not equal the magnetic-energy increase by itself because mechanical work can also enter the energy balance. \(\textbf{Final answer:}\) \[ \boxed{|\varepsilon|=6.0\,\text{V},\qquad \Delta U=0.60\,\text{J}} \]

Explanation: A steady current in a coil establishes a magnetic field and stores energy in that field. For a linear inductor, the stored energy is \[ U=\frac12LI^2 \] When the switch opens, the circuit attempts to reduce the current rapidly. However, the changing current changes the magnetic flux linked with the coil. This produces a self-induced emf according to Faraday's law. For constant inductance, \[ \varepsilon_L=-L\frac{dI}{dt} \] The magnitude of the emf increases when the magnitude of the current-change rate increases. Therefore, an abrupt attempted interruption can produce a large momentary induced voltage. The polarity of this voltage tends to maintain the current that was flowing before the switch opened. In a real circuit, the voltage may become sufficient to produce electrical breakdown of the air gap between the contacts. A brief spark can then provide a temporary path for current. The energy sustaining this transient effect comes at least partly from the magnetic field previously established around the coil. The effect does not mean that the inductor creates energy. It transfers previously stored magnetic energy into other forms, including heating and light during the spark. The current through an ideal inductor cannot undergo an instantaneous finite jump when the applied voltage remains finite. Real switching arrangements therefore experience a transient current-decay process rather than an ideal instantaneous disappearance of current. The Reason correctly connects stored magnetic energy, rapid current change and large induced emf. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: The primary coil produces a magnetic field when current flows through it. A portion of the magnetic flux produced by this field links the turns of the nearby secondary coil. Mutual induction occurs when the magnetic flux linkage of the secondary changes because of a change in the primary current. Direct electrical contact between the coils is not required. \(\textbf{Switch-closing interval:}\) When the primary switch is closed, the current in the primary rises from zero toward its steady value. The magnetic field produced by the primary therefore increases. The magnetic flux linked with the secondary also changes. By Faraday's law, an emf is induced in the secondary. Because the secondary circuit contains a galvanometer and forms a closed conducting path, a transient current flows. This produces a galvanometer pulse. \(\textbf{Steady-primary-current interval:}\) Once the primary current becomes constant, its magnetic field becomes steady. The flux linked with the stationary secondary coil then remains constant. There is no continuing mutually induced emf. Therefore, the secondary galvanometer returns to zero after the transient has ended. A nonzero steady magnetic field does not by itself produce a sustained induced current. \(\textbf{Switch-opening interval:}\) Opening the primary switch causes its current to decrease. The associated magnetic flux through the secondary changes in the opposite sense to the change during switch closing. Consequently, the induced-emf polarity in the secondary reverses. The galvanometer therefore gives a transient pulse in the opposite direction. The magnitudes and durations of the two pulses need not be identical because the primary-current rise and decay may occur at different rates. The distinguishing pattern is a pulse, followed by zero steady response, followed by an oppositely directed pulse. This is a direct demonstration of mutual induction and the dependence of induced emf on changing flux linkage.

Explanation: Mutual induction depends on the magnetic flux of one coil linking the turns of another coil. The flux linkage can change either because the primary current changes or because the magnetic coupling between the coils changes. In this experiment, the primary current remains constant, but the mutual inductance decreases as the coils move apart. \(\textbf{Secondary flux linkage:}\) For the stated linear coupling, \[ \Lambda_2=MI_1 \] where \(M\) is mutual inductance and \(I_1\) is primary current. Faraday's law gives \[ \varepsilon_2=-\frac{d\Lambda_2}{dt} \] Therefore, \[ \varepsilon_2=-\frac{d(MI_1)}{dt} \] Expanding, \[ \varepsilon_2=-M\frac{dI_1}{dt}-I_1\frac{dM}{dt} \] \(\textbf{Constant-primary-current condition:}\) The primary current is fixed at \[ I_1=2.0\,\text{A} \] Thus, \[ \frac{dI_1}{dt}=0 \] The induced emf becomes \[ \varepsilon_2=-I_1\frac{dM}{dt} \] \(\textbf{Rate of change of mutual inductance:}\) \[ M_i=0.50\,\text{H} \] \[ M_f=0.20\,\text{H} \] The mutual-inductance change is \[ \Delta M=0.20-0.50 \] \[ \Delta M=-0.30\,\text{H} \] Over the stated interval, \[ \frac{dM}{dt}=\frac{-0.30}{0.15} \] \[ \frac{dM}{dt}=-2.0\,\text{H s}^{-1} \] \(\textbf{Mutually induced emf:}\) \[ \varepsilon_2=-(2.0)(-2.0) \] \[ \varepsilon_2=+4.0\,\text{V} \] Its magnitude is \(4.0\,\text{V}\). The polarity tends to oppose the decrease in secondary flux linkage caused by the separation. \(\textbf{Secondary current:}\) The secondary resistance is \[ R_2=2.0\,\Omega \] Using Ohm's law, \[ I_2=\frac{|\varepsilon_2|}{R_2} \] \[ I_2=\frac{4.0}{2.0} \] \[ I_2=2.0\,\text{A} \] \(\textbf{Transferred charge:}\) Because the mutual inductance changes uniformly while primary current remains constant, the induced emf and secondary current are constant during the specified interval. Therefore, \[ q=I_2\Delta t \] \[ q=(2.0)(0.15) \] \[ q=0.30\,\text{C} \] \(\textbf{Physical interpretation:}\) The secondary current is nonzero even though the primary current is constant. This does not contradict the usual mutual-induction relation \(\varepsilon_2=-M\,dI_1/dt\), because that shorter expression assumes constant mutual inductance. Here, the changing relative position changes magnetic coupling. The secondary flux linkage consequently decreases even without a change in primary current. The induced current and its magnetic effects oppose the flux change associated with the motion. The external controller and the mechanical agent moving the secondary participate in the energy balance. Thus, mutual induction is governed by changing flux linkage, not exclusively by changing primary current. \(\textbf{Final answer:}\) \[ \boxed{|\varepsilon_2|=4.0\,\text{V},\qquad I_2=2.0\,\text{A},\qquad q=0.30\,\text{C}} \]

Explanation: The magnetic flux linkage of coil P contains two contributions. One arises from its own current, while the other arises from the current in neighbouring coil Q. Because both currents change, both contributions must be included when calculating the total induced emf in P. \(\textbf{Flux-linkage relation for P:}\) \[ \Lambda_P=L_PI_P+MI_Q \] Faraday's law gives \[ \varepsilon_P=-\frac{d\Lambda_P}{dt} \] Since the inductances remain constant, \[ \varepsilon_P=-L_P\frac{dI_P}{dt}-M\frac{dI_Q}{dt} \] \(\textbf{Self-induced contribution in P:}\) \[ \varepsilon_{P,\text{self}}=-(0.50)(+4.0) \] \[ \varepsilon_{P,\text{self}}=-2.0\,\text{V} \] This contribution opposes the increasing positive current in P. \(\textbf{Mutually induced contribution in P:}\) \[ \varepsilon_{P,\text{mutual}}=-(0.20)(-10) \] \[ \varepsilon_{P,\text{mutual}}=+2.0\,\text{V} \] The two induced-emf contributions therefore oppose each other. \(\textbf{Total induced emf in P:}\) \[ \varepsilon_P=-2.0+2.0 \] \[ \varepsilon_P=0 \] The total flux linkage of P has zero instantaneous rate of change even though both currents are changing. \(\textbf{Mutual contribution in Q:}\) The emf induced in Q by the changing current in P is \[ \varepsilon_{Q,\text{mutual}}=-M\frac{dI_P}{dt} \] Substituting, \[ \varepsilon_{Q,\text{mutual}}=-(0.20)(+4.0) \] \[ \varepsilon_{Q,\text{mutual}}=-0.80\,\text{V} \] This is the mutual contribution alone; the total induced emf in Q would additionally involve its self-inductance. The cancellation in P therefore does not imply that every induced-emf contribution in the coupled system vanishes. The result depends on the algebraic rates of change of both currents and their respective inductance coefficients. \(\textbf{Final answer:}\) \[ \boxed{\varepsilon_P=0,\qquad \varepsilon_{Q,\text{mutual}}=-0.80\,\text{V}} \]

Explanation: The magnetic field, inductance and stored energy of a solenoid depend differently on its turns, dimensions and current. The two solenoids have the same length, but Q has twice the number of turns and twice the cross-sectional area. Its current is reduced to half. \(\textbf{Compare magnetic fields:}\) For a long air-core solenoid, \[ B=\mu_0\frac{N}{l}I \] For P, \[ B_P=\mu_0\frac{N}{l}I \] For Q, \[ B_Q=\mu_0\frac{2N}{l}\left(\frac I2\right) \] Simplifying, \[ B_Q=\mu_0\frac{N}{l}I \] Therefore, \[ \frac{B_Q}{B_P}=1 \] \(\textbf{Compare self-inductances:}\) The self-inductance of a long solenoid is \[ L=\frac{\mu_0N^2A}{l} \] Hence, \[ L_P=\frac{\mu_0N^2A}{l} \] For Q, \[ L_Q=\frac{\mu_0(2N)^2(2A)}{l} \] \[ L_Q=\frac{8\mu_0N^2A}{l} \] Consequently, \[ \frac{L_Q}{L_P}=8 \] The number of turns enters quadratically, whereas cross-sectional area enters linearly. \(\textbf{Compare stored magnetic energies:}\) For an ideal linear inductor, \[ U=\frac12LI^2 \] Therefore, \[ U_P=\frac12L_PI^2 \] For Q, \[ U_Q=\frac12(8L_P)\left(\frac I2\right)^2 \] \[ U_Q=\frac12(8L_P)\frac{I^2}{4} \] \[ U_Q=2U_P \] Thus, \[ \frac{U_Q}{U_P}=2 \] \(\textbf{Independent physical check:}\) Both solenoids have the same magnetic-field magnitude. The magnetic-energy density in an air-core solenoid is \[ u_B=\frac{B^2}{2\mu_0} \] Their energy densities are therefore equal. However, Q has twice the interior volume because its cross-sectional area is doubled while length remains unchanged. Its total stored energy must consequently be twice that of P. This agrees with the result obtained using inductance and current. \(\textbf{Final answer:}\) \[ \boxed{\frac{B_Q}{B_P}=1,\qquad \frac{L_Q}{L_P}=8,\qquad \frac{U_Q}{U_P}=2} \]

Explanation: The current is negative relative to the chosen reference direction. It is also becoming more negative. Therefore, both the instantaneous current and its rate of change are negative. The signs of these quantities must be retained when calculating the induced emf and energy-transfer rate. \(\textbf{Current-change rate:}\) \[ I=-4.0\,\text{A} \] \[ \frac{dI}{dt}=-2.0\,\text{A s}^{-1} \] The inductance is \[ L=0.50\,\text{H} \] \(\textbf{Self-induced emf:}\) For constant inductance, \[ \varepsilon_L=-L\frac{dI}{dt} \] Substituting, \[ \varepsilon_L=-(0.50)(-2.0) \] \[ \varepsilon_L=+1.0\,\text{V} \] The positive emf opposes the increasing magnitude of current in the negative direction. \(\textbf{Magnetic-energy relation:}\) The stored energy is \[ U=\frac12LI^2 \] Differentiating with respect to time, \[ \frac{dU}{dt}=LI\frac{dI}{dt} \] Substituting the signed values, \[ \frac{dU}{dt}=(0.50)(-4.0)(-2.0) \] \[ \frac{dU}{dt}=+4.0\,\text{W} \] The positive result means that magnetic energy is increasing. \(\textbf{Physical interpretation:}\) Although the signed current is decreasing, its magnitude is increasing. A larger current magnitude produces greater stored magnetic energy. The sign of \(dU/dt\) therefore cannot be inferred from the sign of \(dI/dt\) alone. It depends on the product of current and current-change rate. The result also demonstrates that positive self-induced emf need not imply decreasing stored energy. Current reference directions and physical energy transfer must be treated consistently. \(\textbf{Final answer:}\) \[ \boxed{\varepsilon_L=+1.0\,\text{V},\qquad \frac{dU}{dt}=+4.0\,\text{W}} \]

Explanation: The given magnetic flux is the flux through one turn of Q due to current in P. To determine mutual inductance, the total flux linkage of Q must be used. The second experiment reverses the roles of the driving and receiving coils. For fixed linear magnetic coupling, the mutual-inductance coefficient is the same in either direction. \(\textbf{Flux linkage of Q due to P:}\) \[ N_Q=150 \] \[ \Phi_{QP}=2.0\times10^{-3}\,\text{Wb} \] Therefore, \[ \Lambda_{QP}=N_Q\Phi_{QP} \] \[ \Lambda_{QP}=150(2.0\times10^{-3}) \] \[ \Lambda_{QP}=0.300\,\text{Wb} \] \(\textbf{Mutual-inductance coefficient:}\) For linear coupling, \[ \Lambda_{QP}=MI_P \] Hence, \[ M=\frac{\Lambda_{QP}}{I_P} \] Substituting, \[ M=\frac{0.300}{4.0} \] \[ M=0.075\,\text{H} \] \(\textbf{Interchange the driving and receiving coils:}\) The mutual-inductance reciprocity relation gives \[ M_{PQ}=M_{QP}=M \] The current in Q now changes from \(+1.0\,\text{A}\) to \(-1.0\,\text{A}\). \[ \frac{dI_Q}{dt}=\frac{-1.0-(+1.0)}{0.10} \] \[ \frac{dI_Q}{dt}=-20\,\text{A s}^{-1} \] \(\textbf{Mutually induced emf in P:}\) \[ \varepsilon_P=-M\frac{dI_Q}{dt} \] Substituting, \[ \varepsilon_P=-(0.075)(-20) \] \[ \varepsilon_P=+1.5\,\text{V} \] The positive sign follows from the specified consistent winding conventions. The equality of the two mutual-inductance coefficients does not mean that the coils have equal self-inductances. It also does not mean that the induced emf must have the same magnitude in experiments with different current-change rates. Here, reciprocity allows the coefficient measured using P as the source to predict induction when Q becomes the source. \(\textbf{Final answer:}\) \[ \boxed{M=0.075\,\text{H},\qquad \varepsilon_P=+1.5\,\text{V}} \]

Explanation: For two long coaxial solenoids, mutual inductance depends on their turn densities, common length and the effective area linked by the magnetic field. The magnetic field produced by the inner long solenoid is approximately confined to its interior. Therefore, the relevant common flux area is the inner solenoid's cross-sectional area. \(\textbf{Mutual-inductance relation:}\) For the stated coaxial arrangement, \[ M=\mu_0n_1n_2Al \] The parameters are \[ n_1=1000\,\text{m}^{-1} \] \[ n_2=500\,\text{m}^{-1} \] \[ A=1.0\times10^{-3}\,\text{m}^2 \] \[ l=0.50\,\text{m} \] \(\textbf{Calculate mutual inductance:}\) \[ M=(4\pi\times10^{-7})(1000)(500) (1.0\times10^{-3})(0.50) \] Combining the numerical factors, \[ (1000)(500)(1.0\times10^{-3})(0.50)=250 \] Therefore, \[ M=(4\pi\times10^{-7})(250) \] \[ M=\pi\times10^{-4}\,\text{H} \] Using \(\pi=3.14\), \[ M=3.14\times10^{-4}\,\text{H} \] \[ M=0.314\,\text{mH} \] \(\textbf{Mutually induced emf:}\) For constant mutual inductance, \[ |\varepsilon_2|=M\left|\frac{dI_1}{dt}\right| \] Substituting, \[ |\varepsilon_2|=(3.14\times10^{-4})(10) \] \[ |\varepsilon_2|=3.14\times10^{-3}\,\text{V} \] \[ |\varepsilon_2|=3.14\,\text{mV} \] \(\textbf{Physical check:}\) The larger radius of the outer solenoid does not make its entire cross-sectional area the relevant flux area. In the ideal long-solenoid approximation, the changing magnetic field produced by the inner solenoid is concentrated within the inner cross-section. Only this common flux effectively links the outer winding. Increasing the inner solenoid's area would change mutual inductance, whereas increasing the outer radius alone would not increase it under the stated ideal approximation. \(\textbf{Final answer:}\) \[ \boxed{M=0.314\,\text{mH},\qquad |\varepsilon_2|=3.14\,\text{mV}} \]

Explanation: The measured secondary flux linkage has two contributions. One is constant, while the other depends on the primary current. Only the changing contribution produces the mutually induced emf. The measurements can be used to determine the mutual-inductance coefficient from the slope of flux linkage against primary current. \(\textbf{Flux-linkage model:}\) \[ \Lambda_2=\Lambda_0+MI_1 \] Here, \(\Lambda_0\) is the constant contribution. Therefore, \[ M=\frac{\Delta\Lambda_2}{\Delta I_1} \] \(\textbf{Determine mutual inductance:}\) Using the measurements at \(I_1=0\) and \(I_1=+3.0\,\text{A}\), \[ M=\frac{0.450-0.150}{3.0-0} \] \[ M=\frac{0.300}{3.0} \] \[ M=0.100\,\text{H} \] The measurement at \(-1.0\,\text{A}\) independently confirms this result: \[ \Lambda_2=0.150+(0.100)(-1.0) \] \[ \Lambda_2=0.050\,\text{Wb} \] \(\textbf{Primary-current variation:}\) The primary current decreases at the specified rate: \[ \frac{dI_1}{dt}=-20\,\text{A s}^{-1} \] Since \(\Lambda_0\) is constant, \[ \frac{d\Lambda_2}{dt}=M\frac{dI_1}{dt} \] \(\textbf{Induced emf:}\) Faraday's law gives \[ \varepsilon_2=-\frac{d\Lambda_2}{dt} \] Therefore, \[ \varepsilon_2=-M\frac{dI_1}{dt} \] Substituting, \[ \varepsilon_2=-(0.100)(-20) \] \[ \varepsilon_2=+2.0\,\text{V} \] \(\textbf{Secondary current:}\) \[ I_2=\frac{\varepsilon_2}{R} \] \[ I_2=\frac{+2.0}{4.0} \] \[ I_2=+0.50\,\text{A} \] The permanent-field contribution makes the total flux linkage nonzero even when the primary current is zero. However, it does not contribute to the induced emf because it remains constant. The relevant coefficient is the slope of the flux-linkage graph, not the ratio of total flux linkage to primary current. Using the latter would incorrectly include the constant magnetic-flux contribution. \(\textbf{Final answer:}\) \[ \boxed{\varepsilon_2=+2.0\,\text{V},\qquad I_2=+0.50\,\text{A}} \]

Explanation: The magnetic flux linked with the secondary depends on the component of primary magnetic field normal to its turns. Changing the relative orientation of the coils therefore changes the effectiveness of their magnetic coupling. \(\textbf{Flux linkage due to the primary:}\) For a fixed secondary orientation, \[ \Lambda_2=N_2BA\cos\theta \] The primary field is proportional to its current: \[ B=kI_1 \] where \[ k=0.50\,\text{T A}^{-1} \] Therefore, \[ \Lambda_2=N_2kI_1A\cos\theta \] \(\textbf{Mutual-inductance coefficient:}\) Comparing with \[ \Lambda_2=MI_1 \] gives \[ M=N_2kA\cos\theta \] The signed coefficient depends on the chosen orientation. \(\textbf{Arrangement P:}\) \[ \theta_P=0^\circ \] Therefore, \[ M_P=(40)(0.50)(0.010)\cos0^\circ \] \[ M_P=0.20\,\text{H} \] The induced-emf magnitude is \[ |\varepsilon_P|=M_P\left|\frac{dI_1}{dt}\right| \] \[ |\varepsilon_P|=(0.20)(4.0) \] \[ |\varepsilon_P|=0.80\,\text{V} \] \(\textbf{Arrangement Q:}\) \[ \theta_Q=60^\circ \] The effective normal field component is halved: \[ \cos60^\circ=\frac12 \] Hence, \[ M_Q=(0.20)\left(\frac12\right) \] \[ M_Q=0.10\,\text{H} \] The induced-emf magnitude becomes \[ |\varepsilon_Q|=(0.10)(4.0) \] \[ |\varepsilon_Q|=0.40\,\text{V} \] \(\textbf{Physical interpretation:}\) The primary current-change rate and field strength per ampere remain unchanged. Only the orientation of the receiving coil changes. The reduction in induced emf therefore arises from reduced magnetic flux linkage per unit primary current. Mutual inductance describes magnetic coupling between circuits, not merely the existence of a magnetic field near them. The use of the cosine factor is valid because the field is assumed uniform over the entire secondary in both arrangements. \(\textbf{Final answer:}\) \[ \boxed{|\varepsilon_P|=0.80\,\text{V},\qquad |\varepsilon_Q|=0.40\,\text{V}} \]

Explanation: The secondary is one continuous winding, but not all its turns link the changing magnetic field. The total induced emf is determined by the sum of the flux-change contributions from the individual turns. Turns situated outside the appreciable primary magnetic field make negligible contributions. \(\textbf{Flux change through linked turns:}\) The number of effectively linked turns is \[ N_{\text{linked}}=60 \] Each linked turn encloses area \[ A=0.010\,\text{m}^2 \] The magnetic-field-change rate is \[ \left|\frac{dB}{dt}\right|=2.0\,\text{T s}^{-1} \] For a linked turn, \[ \left|\frac{d\Phi_B}{dt}\right| =A\left|\frac{dB}{dt}\right| \] Substituting, \[ \left|\frac{d\Phi_B}{dt}\right| =(0.010)(2.0) \] \[ \left|\frac{d\Phi_B}{dt}\right| =0.020\,\text{Wb s}^{-1} \] \(\textbf{Total induced emf:}\) The \(60\) linked turns have aiding emf contributions. Therefore, \[ |\varepsilon|=N_{\text{linked}} \left|\frac{d\Phi_B}{dt}\right| \] \[ |\varepsilon|=(60)(0.020) \] \[ |\varepsilon|=1.2\,\text{V} \] The remaining \(40\) turns make negligible contributions because their linked flux is approximately zero. \(\textbf{Secondary current:}\) The complete secondary circuit has resistance \[ R=3.0\,\Omega \] Using Ohm's law, \[ I=\frac{|\varepsilon|}{R} \] \[ I=\frac{1.2}{3.0} \] \[ I=0.40\,\text{A} \] \(\textbf{Physical significance:}\) All \(100\) turns belong to the same series winding and carry the same induced current. However, they need not all contribute equally to its induced emf. The coefficient of mutual induction depends on the magnetic flux actually linking the receiving turns. Using all \(100\) turns in \(NA\,dB/dt\) would overestimate the induced emf because it would assume that every turn encloses the changing field. The example demonstrates the distinction between total winding turns and effectively magnetically linked turns. Improving magnetic coupling can increase induced emf without changing the electrical connection of the secondary winding. \(\textbf{Final answer:}\) \[ \boxed{|\varepsilon|=1.2\,\text{V},\qquad I=0.40\,\text{A}} \]

Explanation: Mutual induction depends on the rate of change of secondary flux linkage. The flux linkage can vary because the primary current changes, because the mutual inductance changes, or because both change simultaneously. Here, the primary current increases while the magnetic coupling decreases. \(\textbf{Secondary flux linkage:}\) \[ \Lambda_2=MI_1 \] Differentiating, \[ \frac{d\Lambda_2}{dt} =M\frac{dI_1}{dt}+I_1\frac{dM}{dt} \] Faraday's law gives \[ \varepsilon_2=-\frac{d\Lambda_2}{dt} \] Therefore, \[ \varepsilon_2=-M\frac{dI_1}{dt} -I_1\frac{dM}{dt} \] \(\textbf{Contribution from changing primary current:}\) \[ M=0.40\,\text{H} \] \[ \frac{dI_1}{dt}=+5.0\,\text{A s}^{-1} \] Hence, \[ \left(\frac{d\Lambda_2}{dt}\right)_{\text{current}} =(0.40)(+5.0) \] \[ \left(\frac{d\Lambda_2}{dt}\right)_{\text{current}} =+2.0\,\text{Wb s}^{-1} \] This contribution tends to increase the linked flux. \(\textbf{Contribution from changing coupling:}\) \[ I_1=+2.0\,\text{A} \] \[ \frac{dM}{dt}=-1.0\,\text{H s}^{-1} \] Therefore, \[ \left(\frac{d\Lambda_2}{dt}\right)_{\text{coupling}} =(+2.0)(-1.0) \] \[ \left(\frac{d\Lambda_2}{dt}\right)_{\text{coupling}} =-2.0\,\text{Wb s}^{-1} \] This contribution tends to decrease the linked flux. \(\textbf{Net flux-linkage change:}\) \[ \frac{d\Lambda_2}{dt}=+2.0-2.0 \] \[ \frac{d\Lambda_2}{dt}=0 \] Hence, \[ \varepsilon_2=0 \] The secondary therefore has zero instantaneous mutually induced emf despite the increasing primary current. The commonly used expression \(\varepsilon_2=-M\,dI_1/dt\) assumes constant mutual inductance. When the coils move relative to one another, that assumption is not satisfied. Both the current-change and coupling-change terms must be considered. Their exact instantaneous cancellation means that the total linked flux is stationary at that instant. It does not require either the primary current or mutual inductance to be constant. The cancellation may cease at another instant if the two rates no longer satisfy the same relation. The correct conclusion is zero instantaneous induced emf due to opposing flux-linkage changes.

Explanation: The mutual inductance of two coils depends on the magnetic flux produced by one coil that links the turns of the other. The total geometrical area of the larger coil is not by itself the determining quantity. For two concentric circular coils with a much smaller inner radius, the magnetic field produced by the outer coil can be treated as nearly uniform across the inner coil. The magnetic field at the centre of a circular coil carrying current \(I\) is \[ B=\frac{\mu_0NI}{2R} \] where \(R\) is the radius of the source coil and \(N\) its turn count. If the outer coil's radius increases while its current and turn count are fixed, its central magnetic field decreases. The small inner coil therefore links less magnetic flux per unit current in the outer coil. Since mutual inductance is the ratio of linked flux linkage to source current, mutual inductance decreases. The approximate dependence for fixed inner-coil geometry is \[ M\propto\frac{1}{R_{\text{outer}}} \] The inner coil remains coaxial with the outer coil, so their magnetic coupling does not vanish merely because the radii differ. Increasing the area of the outer current loop does not guarantee an increase in the field passing through the inner coil. The field produced per unit current must be considered. The reasoning also remains consistent with the reciprocity of mutual inductance. Whether the inner or outer coil is chosen as the source, the coefficient describing the same fixed coil arrangement is equal. The correct conclusion is that increasing the outer radius weakens the central field per ampere and reduces mutual inductance.