Magnetic Effects of Current and Magnetism MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Determine the coil's initial magnetic moment:}\) The moment magnitude is \[ m=NIA \] Substituting, \[ m=(20)(2.0)(0.010) \] \[ m=0.40\,\text{A m}^2 \] Its direction is positive \(z\): \[ \vec m=0.40\hat k\,\text{A m}^2 \] \(\textbf{Calculate the complete magnetic-torque vector:}\) \[ \vec\tau=\vec m\times\vec B \] \[ \vec\tau=(0.40\hat k)\times(0.30\hat i+0.40\hat j) \] Using \[ \hat k\times\hat i=\hat j,\qquad \hat k\times\hat j=-\hat i \] we obtain \[ \vec\tau=0.12\hat j-0.16\hat i \] \[ \vec\tau=(-0.16\hat i+0.12\hat j)\,\text{N m} \] \(\textbf{Trial P — rotation permitted only about }x\textbf{:}\) Only the torque component parallel to the permitted rotation axis causes angular acceleration about that axis. Thus, \[ \tau_x=-0.16\,\text{N m} \] Using the rotational equation, \[ \tau_x=J_x\alpha_x \] \[ \alpha_x=\frac{-0.16}{0.020} \] \[ \alpha_x=-8.0\,\text{rad s}^{-2} \] The bearing supplies the reactions needed to prevent rotations about the other axes. \(\textbf{Trial Q — rotation permitted only about }y\textbf{:}\) The relevant torque component is \[ \tau_y=+0.12\,\text{N m} \] Therefore, \[ \alpha_y=\frac{\tau_y}{J_y} \] \[ \alpha_y=\frac{+0.12}{0.020} \] \[ \alpha_y=+6.0\,\text{rad s}^{-2} \] \(\textbf{Final answer:}\) \[ \boxed{\alpha_x=-8.0\,\text{rad s}^{-2},\qquad \alpha_y=+6.0\,\text{rad s}^{-2}} \] The magnetic field produces the same complete torque vector in both initial configurations. However, the mechanical constraints differ. Only the component of torque about the permitted axis causes the corresponding angular acceleration. Using the complete torque magnitude in either trial would therefore be incorrect. The comparison connects magnetic torque with constrained rotational dynamics.

Explanation: \(\textbf{Resolve the magnetic-moment direction:}\) The moment magnitude is \[ m=1.0\,\text{A m}^2 \] Its direction makes an angle of \(60^\circ\) with positive \(x\). Therefore, \[ \vec m=m(\cos60^\circ\hat i+\sin60^\circ\hat j) \] \[ \vec m=\left(\frac12\hat i+\frac{\sqrt3}{2}\hat j\right)\,\text{A m}^2 \] \(\textbf{Determine the magnetic torque:}\) \[ \vec\tau=\vec m\times\vec B \] Substituting, \[ \vec\tau= \left(\frac12\hat i+\frac{\sqrt3}{2}\hat j\right)\times0.60\hat i \] The parallel component gives zero. The remaining term is \[ \vec\tau=(0.60)\frac{\sqrt3}{2}(\hat j\times\hat i) \] Since \[ \hat j\times\hat i=-\hat k \] we obtain \[ \vec\tau=-0.30\sqrt3\hat k\,\text{N m} \] \(\textbf{Determine instantaneous rotational power:}\) The instantaneous mechanical power delivered by a torque is \[ P=\vec\tau\cdot\vec\omega \] The angular velocity is \[ \vec\omega=+2.0\hat k\,\text{rad s}^{-1} \] Therefore, \[ P=(-0.30\sqrt3\hat k)\cdot(2.0\hat k) \] \[ P=-0.60\sqrt3\,\text{W} \] \(\textbf{Interpret the sign of power:}\) The magnetic torque acts along negative \(z\), while the imposed angular velocity points along positive \(z\). The two vectors oppose each other. Thus, the magnetic field performs negative mechanical work at this instant. To maintain the imposed motion without angular acceleration, the motor must supply an opposing positive torque and positive mechanical power. \(\textbf{Final answer:}\) \[ \boxed{\vec\tau=-0.30\sqrt3\hat k\,\text{N m},\qquad P_B=-0.60\sqrt3\,\text{W}} \] The magnitude of a torque does not determine whether it supplies or removes mechanical energy. The sign of its power depends on the torque–angular-velocity scalar product. Here, the loop rotates away from the field direction, so the magnetic torque opposes its motion. The motor maintains that rotation against the magnetic torque.

Explanation: The magnetic torque on a plane current coil is \[ \tau=NIAB\sin\theta \] Here, \(\theta\) is the angle between the coil's magnetic moment and the magnetic field. In a moving-coil galvanometer, the shaped pole pieces and soft iron cylinder help produce a strong radial magnetic field. The field is arranged to remain in the plane of the moving coil as it rotates. Consequently, the magnetic moment remains perpendicular to the field. The angle in the torque expression is therefore \(90^\circ\). Statement I is correct. The magnetic torque becomes \[ \tau_B=NIAB \] The restoring spring exerts an opposing torque of magnitude \[ \tau_r=C\phi \] where \(C\) is the torsional constant and \(\phi\) is the angular deflection. At equilibrium, \[ NIAB=C\phi \] Therefore, \[ \phi=\frac{NAB}{C}I \] The deflection is proportional to current, making Statement II correct. In an ordinary uniform field, the angle between the moment and the field changes as the coil rotates. The sine factor generally changes, so the magnetic torque is not necessarily \(NIAB\) at every orientation. Statement III is incorrect. At fixed current and magnetic parameters, increasing \(C\) decreases the equilibrium deflection. Statement IV is correct. The radial field is important because it removes the changing-angle factor from the torque during galvanometer operation. The restoring spring then converts the electromagnetic torque into a reproducible angular reading.

Explanation: \(\textbf{Use the radial-field torque balance:}\) The magnetic torque is \[ \tau_B=NIAB \] At equilibrium, it balances the restoring torque: \[ NIAB=C\phi \] Therefore, \[ C=\frac{NIAB}{\phi} \] \(\textbf{Calculate the original magnetic torque:}\) The given current is \[ I_1=20\,\mu\text{A}=20\times10^{-6}\,\text{A} \] Substitute the original values: \[ \tau_{B1}=(100)(20\times10^{-6})(2.0\times10^{-4})(0.50) \] \[ \tau_{B1}=2.0\times10^{-7}\,\text{N m} \] \(\textbf{Determine the torsional constant:}\) The measured deflection is \[ \phi_1=0.20\,\text{rad} \] Thus, \[ C=\frac{2.0\times10^{-7}}{0.20} \] \[ C=1.0\times10^{-6}\,\text{N m rad}^{-1} \] \(\textbf{Apply the changed field and current:}\) The new magnetic field is \[ B_2=1.25B_1 \] \[ B_2=(1.25)(0.50) \] \[ B_2=0.625\,\text{T} \] The new current is \[ I_2=30\times10^{-6}\,\text{A} \] \(\textbf{Calculate the new equilibrium deflection:}\) \[ \phi_2=\frac{NI_2AB_2}{C} \] Substituting, \[ \phi_2= \frac{(100)(30\times10^{-6})(2.0\times10^{-4})(0.625)} {1.0\times10^{-6}} \] \[ \phi_2=0.375\,\text{rad} \] \(\textbf{Check using proportionality:}\) At unchanged \(N\), \(A\), and \(C\), \[ \frac{\phi_2}{\phi_1} =\frac{I_2}{I_1}\frac{B_2}{B_1} \] \[ \frac{\phi_2}{\phi_1} =\frac{30}{20}(1.25) \] \[ \frac{\phi_2}{\phi_1}=1.875 \] \[ \phi_2=(0.20)(1.875)=0.375\,\text{rad} \] \(\textbf{Final answer:}\) \[ \boxed{C=1.0\times10^{-6}\,\text{N m rad}^{-1},\qquad \phi_2=0.375\,\text{rad}} \] The first reading determines the mechanical restoring constant from electromagnetic torque balance. The second reading combines two independent changes: current and magnetic-field magnitude. The radial-field design keeps the torque proportional to current over the permitted deflection range. The restoring constant remains unchanged, so the new deflection follows directly from the changed magnetic torque.

Explanation: For a moving-coil galvanometer in a radial magnetic field, equilibrium requires \[ NIAB=C\phi \] Therefore, \[ \phi=\frac{NAB}{C}I \] The quantities \(A\), \(B\) and \(C\) are identical for the two instruments. In Test P, equal currents pass through both moving coils. The deflection is then directly proportional to turn number. Since \(G_2\) has twice as many turns, it develops twice the magnetic torque at the same current. Its equilibrium deflection is therefore twice that of \(G_1\). In Test Q, the applied voltage is the same across the two moving coils. Let the resistance of \(G_1\) be \(R\). The resistance of \(G_2\) is \(2R\). By Ohm's law, \[ I_1=\frac VR,\qquad I_2=\frac{V}{2R} \] Thus, \(G_2\) carries half the current of \(G_1\) in this test. Its number of turns is twice as large. The products controlling their magnetic torques are \[ N_1I_1 \] and \[ N_2I_2=(2N_1)\frac{I_1}{2}=N_1I_1 \] Therefore, the two deflections are equal under the same applied voltage. Increasing the number of turns improves current sensitivity under the stated conditions. However, the associated increase in coil resistance compensates for that advantage when the instruments are operated at the same voltage. The distinction between equal-current and equal-voltage tests is essential.

Explanation: \(\textbf{Determine the torque in the ordinary uniform field:}\) The magnetic-torque magnitude on a coil is \[ \tau_B=NIAB\sin\theta \] The angle between magnetic moment and field is \[ \theta=90^\circ-\phi \] Therefore, \[ \tau_B=NIAB\cos\phi \] The torsional restoring torque has magnitude \[ \tau_r=C\phi \] At equilibrium, \[ NIAB\cos\phi=C\phi \] Thus, \[ I(\phi)=\frac{C\phi}{NAB\cos\phi} \] \(\textbf{Compare the currents at the two deflections:}\) For the specified angles, \[ \frac{I_{60}}{I_{30}} =\frac{(60^\circ)/\cos60^\circ}{(30^\circ)/\cos30^\circ} \] The ratio of angular deflections is \[ \frac{60^\circ}{30^\circ}=2 \] Using \[ \cos30^\circ=\frac{\sqrt3}{2},\qquad \cos60^\circ=\frac12 \] we obtain \[ \frac{I_{60}}{I_{30}} =2\frac{\cos30^\circ}{\cos60^\circ} \] \[ \frac{I_{60}}{I_{30}}=2\sqrt3 \] \(\textbf{Compare with a radial-field galvanometer:}\) In a radial field, the magnetic moment remains perpendicular to the magnetic field. Consequently, \[ \tau_B=NIAB \] The equilibrium relation is \[ NIAB=C\phi \] Therefore, \[ I\propto\phi \] For the same two angular deflections, \[ \frac{I_{60}}{I_{30}}=\frac{60^\circ}{30^\circ}=2 \] \(\textbf{Final answer:}\) \[ \boxed{\left(\frac{I_{60}}{I_{30}}\right)_{\mathrm{ordinary}}=2\sqrt3,\qquad \left(\frac{I_{60}}{I_{30}}\right)_{\mathrm{radial}}=2} \] The ordinary uniform field introduces a changing cosine factor into the magnetic torque. As the coil deflects, the magnetic torque per unit current decreases. A larger increase in current is therefore needed to reach the second deflection. In a radial field, this angular reduction is avoided, giving a linear current–deflection relation.

Explanation: \(\textbf{Separate the zero offset from the physical deflection:}\) The indicated reading contains a constant additive offset. Therefore, \[ \phi_{\mathrm{ind}}=\phi_0+\phi \] where \(\phi\) is the actual mechanical deflection. The graph intercept gives \[ \phi_0=0.040\,\text{rad} \] The intercept does not represent magnetic deflection produced by zero current. \(\textbf{Find the current sensitivity from the graph slope:}\) The slope is \[ S_I=\frac{\Delta\phi_{\mathrm{ind}}}{\Delta I} \] Substituting the two calibration points, \[ S_I=\frac{0.140-0.040}{(20-0)\times10^{-6}} \] \[ S_I=\frac{0.100}{20\times10^{-6}} \] \[ S_I=5.0\times10^3\,\text{rad A}^{-1} \] Subtracting a constant zero offset does not change this slope. \(\textbf{Determine the torsional constant:}\) For a radial-field galvanometer, \[ \phi=\frac{NAB}{C}I \] Thus, \[ S_I=\frac{NAB}{C} \] Rearranging, \[ C=\frac{NAB}{S_I} \] Substitute the coil parameters: \[ C=\frac{(100)(2.0\times10^{-4})(0.50)} {5.0\times10^3} \] \[ C=\frac{0.010}{5.0\times10^3} \] \[ C=2.0\times10^{-6}\,\text{N m rad}^{-1} \] \(\textbf{Determine the mechanical deflection at }50\,\mu\text{A}\textbf{:}\) The actual mechanical deflection is \[ \phi=S_I I \] \[ \phi=(5.0\times10^3)(50\times10^{-6}) \] \[ \phi=0.250\,\text{rad} \] The scale would indicate \[ \phi_{\mathrm{ind}}=0.040+0.250 \] \[ \phi_{\mathrm{ind}}=0.290\,\text{rad} \] \(\textbf{Final answer:}\) \[ \boxed{C=2.0\times10^{-6}\,\text{N m rad}^{-1},\qquad \phi=0.250\,\text{rad}} \] The graph slope determines current sensitivity, whereas the intercept identifies a fixed reading error. The mechanical restoring constant must be inferred from the current-dependent deflection, not the raw indicated reading. At \(50\,\mu\text{A}\), the instrument indicates \(0.290\,\text{rad}\), but the moving coil has actually turned through \(0.250\,\text{rad}\). Separating calibration offset from physical response prevents an incorrect sensitivity calculation.

Explanation: \(\textbf{Determine the force on an active side:}\) The magnetic-force magnitude on a straight conductor perpendicular to a magnetic field is \[ F=ILB \] For one turn, the active-side length is \[ L=0.050\,\text{m} \] The current is \[ I=50\times10^{-6}\,\text{A} \] The field magnitude is \[ B=0.20\,\text{T} \] Therefore, the force on one active side of one turn is \[ F_{\mathrm{one}}=(50\times10^{-6})(0.050)(0.20) \] \[ F_{\mathrm{one}}=5.0\times10^{-7}\,\text{N} \] \(\textbf{Account for all turns:}\) The galvanometer has \[ N=200 \] identical turns. The total force associated with the corresponding active sides is \[ F_{\mathrm{side}}=NF_{\mathrm{one}} \] \[ F_{\mathrm{side}}=200(5.0\times10^{-7}) \] \[ F_{\mathrm{side}}=1.0\times10^{-4}\,\text{N} \] \(\textbf{Construct the magnetic couple:}\) The forces on the two active sides act in opposite directions and form a couple. Their perpendicular separation is the coil width: \[ w=0.020\,\text{m} \] The torque magnitude is \[ \tau_B=F_{\mathrm{side}}w \] \[ \tau_B=(1.0\times10^{-4})(0.020) \] \[ \tau_B=2.0\times10^{-6}\,\text{N m} \] \(\textbf{Balance magnetic and restoring torques:}\) At equilibrium, \[ \tau_B=C\phi \] Therefore, \[ \phi=\frac{\tau_B}{C} \] Substituting, \[ \phi=\frac{2.0\times10^{-6}}{5.0\times10^{-6}} \] \[ \phi=0.40\,\text{rad} \] \(\textbf{Check using magnetic moment:}\) The area of one rectangular turn is \[ A=(0.050)(0.020)=1.0\times10^{-3}\,\text{m}^2 \] The radial-field torque relation gives \[ \tau_B=NIAB \] \[ \tau_B=(200)(50\times10^{-6})(1.0\times10^{-3})(0.20) \] \[ \tau_B=2.0\times10^{-6}\,\text{N m} \] This confirms the force-couple calculation. \(\textbf{Final answer:}\) \[ \boxed{F_{\mathrm{side}}=1.0\times10^{-4}\,\text{N},\qquad \phi=0.40\,\text{rad}} \] The galvanometer's magnetic torque arises from forces acting on opposite active sides of its rectangular coil. The force on one active side must account for every turn, but the two side forces must not be added as scalar forces to calculate torque. Their separation forms the moment arm of the couple. The result agrees with the compact radial-field torque expression \(NIAB\).

Explanation: \(\textbf{Use the static equilibrium reading:}\) The radial-field magnetic torque is constant for a fixed current: \[ \tau_B=NIAB \] The restoring torque is proportional to deflection: \[ \tau_r=C\phi \] At static equilibrium, \[ \tau_B=C\phi_{\mathrm{eq}} \] The measured equilibrium deflection is \[ \phi_{\mathrm{eq}}=0.20\,\text{rad} \] Therefore, \[ \tau_B=0.20C \] \(\textbf{Identify the energy changes during undamped motion:}\) The coil starts from rest at zero deflection. As it rotates, the constant magnetic torque does positive rotational work: \[ W_B=\tau_B\phi \] The torsion spring stores potential energy: \[ U_s=\frac12C\phi^2 \] Because damping is negligible, mechanical energy is conserved after accounting for the work done by the maintained-current magnetic torque. \(\textbf{Apply the first-turning-point condition:}\) At the first maximum deflection, the coil momentarily has zero angular velocity. Its rotational kinetic energy is therefore again zero. The magnetic work supplied since release equals the spring energy: \[ \tau_B\phi_{\max}=\frac12C\phi_{\max}^2 \] The nonzero turning-point solution satisfies \[ \phi_{\max}=\frac{2\tau_B}{C} \] Using the static equilibrium condition, \[ \phi_{\max}=2\phi_{\mathrm{eq}} \] \[ \phi_{\max}=2(0.20) \] \[ \phi_{\max}=0.40\,\text{rad} \] \(\textbf{Determine the net torque at the turning point:}\) The signed net torque in the direction of increasing deflection is \[ \tau_{\mathrm{net}}=\tau_B-C\phi \] At \(\phi=0.40\,\text{rad}\), \[ \tau_{\mathrm{net}}=0.20C-0.40C \] \[ \tau_{\mathrm{net}}=-0.20C \] The negative sign indicates a restoring net torque towards decreasing deflection. \(\textbf{Final answer:}\) \[ \boxed{\phi_{\max}=0.40\,\text{rad},\qquad \tau_{\mathrm{net}}=-0.20C} \] The static equilibrium deflection is not the same as the first turning point of an undamped coil released from rest. At the equilibrium angle, the moving coil can still possess angular speed and pass through that position. It stops momentarily at twice the equilibrium deflection because magnetic work has been stored in the torsion spring. The net torque there points back towards equilibrium. Practical galvanometers use damping to suppress such sustained oscillations and permit stable readings.

Explanation: \(\textbf{Current sensitivity of the original galvanometer:}\) In a radial magnetic field, the magnetic torque is \[ \tau_m=NIAB \] The restoring torque at equilibrium is \[ \tau_r=\kappa\phi \] Therefore, \[ NIAB=\kappa\phi \] The current sensitivity is \[ S_I=\frac{\phi}{I}=\frac{NAB}{\kappa} \] Substituting the original parameters, \[ S_I= \frac{(200)(2.0\times10^{-3})(0.25)} {5.0\times10^{-4}} \] \[ S_I=200\,\text{rad A}^{-1} \] \(\textbf{Original voltage sensitivity:}\) The voltage sensitivity of the bare galvanometer is \[ S_V=\frac{\phi}{V}=\frac{S_I}{G} \] Thus, \[ S_V=\frac{200}{50} \] \[ S_V=4.0\,\text{rad V}^{-1} \] At an applied voltage of \(0.060\,\text{V}\), the original equilibrium deflection is \[ \phi=S_VV=4.0(0.060) \] \[ \phi=0.240\,\text{rad} \] \(\textbf{Redesigned current sensitivity:}\) The magnetic field doubles, while the restoring constant triples. The number of turns and coil area are unchanged. Therefore, \[ \frac{S'_I}{S_I}=\frac23 \] \[ S'_I=\frac23(200) \] \[ S'_I=\frac{400}{3}\,\text{rad A}^{-1} \] \(\textbf{Redesigned voltage sensitivity:}\) The winding resistance becomes \(100\,\Omega\). Hence, \[ S'_V=\frac{S'_I}{G'} \] \[ S'_V=\frac{400/3}{100} \] \[ S'_V=\frac43\,\text{rad V}^{-1} \] Consequently, \[ \frac{S'_V}{S_V}=\frac{4/3}{4}=\frac13 \] \(\textbf{New equilibrium deflection:}\) At the same applied voltage, \[ \phi'=S'_VV \] \[ \phi'=\frac43(0.060) \] \[ \phi'=0.080\,\text{rad} \] Although the stronger magnetic field increases magnetic torque per unit current, the larger restoring constant reduces current sensitivity. The increased winding resistance further reduces the current produced by a fixed applied voltage. Therefore, current sensitivity and voltage sensitivity must be compared separately. The final voltage sensitivity is one-third of the original, and the equilibrium deflection is \(0.080\,\text{rad}\).

Explanation: \(\textbf{Distinguish current and voltage sensitivity:}\) Current sensitivity is \[ S_I=\frac{\phi}{I} \] For a bare galvanometer of resistance \(G\), \[ V=IG \] Therefore, voltage sensitivity is \[ S_V=\frac{\phi}{V} \] \[ S_V=\frac{S_I}{G} \] \(\textbf{Compare the two voltage sensitivities:}\) For instrument P, \[ S_{V,P}=\frac{6000}{60} \] \[ S_{V,P}=100\,\text{rad V}^{-1} \] For instrument Q, \[ S_{V,Q}=\frac{6000}{180} \] \[ S_{V,Q}=\frac{100}{3}\,\text{rad V}^{-1} \] Hence, \[ \frac{S_{V,P}}{S_{V,Q}}=3 \] \(\textbf{Determine their full-scale currents:}\) At the maximum permissible angular deflection, \[ I_g=\frac{\phi_{\max}}{S_I} \] For P, \[ I_{g,P}=\frac{0.24}{6000} \] \[ I_{g,P}=40\,\mu\text{A} \] For Q, \[ I_{g,Q}=\frac{0.36}{6000} \] \[ I_{g,Q}=60\,\mu\text{A} \] \(\textbf{Convert these limits into terminal voltages:}\) For P, \[ V_{g,P}=I_{g,P}G_P \] \[ V_{g,P}=(40\times10^{-6})(60) \] \[ V_{g,P}=2.4\,\text{mV} \] For Q, \[ V_{g,Q}=(60\times10^{-6})(180) \] \[ V_{g,Q}=10.8\,\text{mV} \] \(\textbf{Assess the applied test voltage:}\) The applied \(3.0\,\text{mV}\) exceeds P's permissible \(2.4\,\text{mV}\). It remains below Q's permissible \(10.8\,\text{mV}\). Thus, only P is driven beyond its full-scale limit. \(\textbf{Final answer:}\) \[ \boxed{\frac{S_{V,P}}{S_{V,Q}}=3,\quad V_{g,P}=2.4\,\text{mV},\quad V_{g,Q}=10.8\,\text{mV}} \] Equal current sensitivities do not imply equal voltage sensitivities. The internal resistance determines how much current a specified terminal voltage produces. The permissible angular deflection independently determines the full-scale current. A galvanometer can therefore have greater voltage sensitivity but a smaller safe voltage range.

Explanation: \(\textbf{Express the directly connected calibration slope:}\) The galvanometer deflection is \[ \phi=S_I I \] When voltage \(V\) is applied directly across its resistance \(G\), \[ I=\frac VG \] Therefore, \[ \phi=\frac{S_I}{G}V \] The measured slope of graph P is \[ s_P=\frac{S_I}{G}=60\,\text{rad V}^{-1} \] \(\textbf{Include the known series resistance:}\) In arrangement Q, the galvanometer and \(100\,\Omega\) resistor carry the same current. Therefore, \[ I_Q=\frac{V}{G+100} \] The corresponding deflection is \[ \phi_Q=\frac{S_I}{G+100}V \] Thus, graph Q has slope \[ s_Q=\frac{S_I}{G+100}=20\,\text{rad V}^{-1} \] \(\textbf{Infer the internal resistance from the slope ratio:}\) \[ \frac{s_P}{s_Q}=\frac{G+100}{G} \] Substituting, \[ \frac{60}{20}=\frac{G+100}{G} \] \[ 3G=G+100 \] \[ 2G=100 \] \[ G=50\,\Omega \] \(\textbf{Recover the current sensitivity:}\) \[ S_I=s_PG \] \[ S_I=(60)(50) \] \[ S_I=3000\,\text{rad A}^{-1} \] \(\textbf{Predict the deflection at the specified voltage:}\) The voltage in arrangement Q is \[ V=20\,\text{mV}=0.020\,\text{V} \] Using its measured slope, \[ \phi_Q=s_QV \] \[ \phi_Q=(20)(0.020) \] \[ \phi_Q=0.40\,\text{rad} \] \(\textbf{Final answer:}\) \[ \boxed{G=50\,\Omega,\quad S_I=3000\,\text{rad A}^{-1},\quad \phi_Q=0.40\,\text{rad}} \] The two calibration slopes separate the galvanometer's magnetic sensitivity from its electrical resistance. Adding the series resistor reduces the current generated by a prescribed applied voltage. It does not alter the intrinsic current sensitivity of the galvanometer. The comparison therefore permits both unknown instrument parameters to be reconstructed from voltage–deflection measurements.

Explanation: The current sensitivity measures angular deflection per unit current through the galvanometer coil. For radial-field operation, \[ S_I=\frac{NAB}{C} \] The coil's electrical resistance does not appear in this relation. Voltage sensitivity measures angular deflection per unit potential difference applied directly across the bare galvanometer. Therefore, \[ S_V=\frac{S_I}{G} \] In modification P, the number of turns doubles while the torsional constant, area and field remain unchanged. Current sensitivity consequently doubles. However, the coil resistance also doubles, so the ratio \(S_I/G\) remains unchanged. Thus, P corresponds to \((2,1)\), which is entry 1. In modification Q, only the torsional constant doubles. Magnetic torque for a given current is unchanged, but the restoring torque per unit angle doubles. Current sensitivity is halved. Because resistance is unchanged, voltage sensitivity is also halved. Thus, Q corresponds to \((1/2,1/2)\), which is entry 2. In modification R, doubling the radial field doubles current sensitivity. Halving the resistance simultaneously doubles the current produced by a given terminal voltage. Consequently, voltage sensitivity becomes four times its original value. Thus, R corresponds to \((2,4)\), which is entry 3. In modification S, the magnetic and mechanical parameters are unchanged. Current sensitivity remains unchanged. Doubling resistance halves voltage sensitivity. Thus, S corresponds to \((1,1/2)\), which is entry 4. The comparison shows that current sensitivity and voltage sensitivity need not change by the same factor. The magnetic and mechanical design controls the deflection per unit current, while electrical resistance additionally controls the deflection per unit voltage.

Explanation: \(\textbf{Identify the unaffected intrinsic sensitivity:}\) For a radial-field galvanometer, \[ S_I=\frac{NAB}{C} \] The stated warming changes only electrical resistance. Therefore, the intrinsic current sensitivity remains \[ S_I=5000\,\text{rad A}^{-1} \] The new coil resistance is \[ G'=(1.25)(40) \] \[ G'=50\,\Omega \] \(\textbf{Experiment P — prescribed current:}\) An ideal current source maintains \[ I_P=20\times10^{-6}\,\text{A} \] The coil deflection is \[ \phi_P=S_I I_P \] \[ \phi_P=(5000)(20\times10^{-6}) \] \[ \phi_P=0.10\,\text{rad} \] Because both current and current sensitivity remain unchanged, warming does not change this deflection. \(\textbf{Experiment Q — prescribed terminal voltage:}\) For the original resistance, \[ I_{Q,1}=\frac{V}{G} \] \[ I_{Q,1}=\frac{2.0\times10^{-3}}{40} \] \[ I_{Q,1}=50\,\mu\text{A} \] Therefore, \[ \phi_{Q,1}=(5000)(50\times10^{-6}) \] \[ \phi_{Q,1}=0.25\,\text{rad} \] After warming, \[ I_{Q,2}=\frac{2.0\times10^{-3}}{50} \] \[ I_{Q,2}=40\,\mu\text{A} \] The new deflection is \[ \phi_{Q,2}=(5000)(40\times10^{-6}) \] \[ \phi_{Q,2}=0.20\,\text{rad} \] \(\textbf{Final answer:}\) \[ \boxed{\phi_P=0.10\,\text{rad}\ \text{in both cases};\qquad \phi_{Q,1}=0.25\,\text{rad},\quad \phi_{Q,2}=0.20\,\text{rad}} \] An increase in coil resistance does not directly alter the magnetic torque produced by a specified current. It does reduce the current supplied by a fixed terminal voltage. The same physical change therefore leaves current-controlled deflection unchanged while decreasing voltage-controlled deflection. Identifying what the source maintains constant is essential.

Explanation: \(\textbf{Calculate the original full-scale current:}\) Current sensitivity is \[ S_I=\frac{\phi}{I} \] At the maximum permissible angle, \[ I_g=\frac{\phi_{\max}}{S_I} \] Substituting, \[ I_g=\frac{0.40}{4000} \] \[ I_g=1.0\times10^{-4}\,\text{A} \] \[ I_g=100\,\mu\text{A} \] \(\textbf{Determine the original full-scale terminal voltage:}\) The bare galvanometer has resistance \[ G=80\,\Omega \] Therefore, \[ V_g=I_gG \] \[ V_g=(100\times10^{-6})(80) \] \[ V_g=8.0\,\text{mV} \] \(\textbf{Determine how the replacement spring changes sensitivity:}\) For a radial-field galvanometer, \[ S_I=\frac{NAB}{C} \] The new torsional constant is \[ C'=2C \] All magnetic and coil parameters remain unchanged. Therefore, \[ S'_I=\frac{S_I}{2} \] \[ S'_I=2000\,\text{rad A}^{-1} \] \(\textbf{Determine the new full-scale quantities:}\) The permissible maximum angle is still \(0.40\,\text{rad}\). Thus, \[ I'_g=\frac{0.40}{2000} \] \[ I'_g=200\,\mu\text{A} \] The electrical resistance is unchanged. Hence, \[ V'_g=I'_gG \] \[ V'_g=(200\times10^{-6})(80) \] \[ V'_g=16\,\text{mV} \] \(\textbf{Final answer:}\) \[ \boxed{I_g=100\,\mu\text{A},\quad V_g=8.0\,\text{mV};\qquad I'_g=200\,\mu\text{A},\quad V'_g=16\,\text{mV}} \] A stiffer spring produces less deflection for the same current. Therefore, current sensitivity decreases, but the current required to reach the same maximum angular deflection increases. The full-scale voltage increases because coil resistance remains unchanged. Greater full-scale measuring range and greater sensitivity are not equivalent properties.

Explanation: \(\textbf{Identify the complete electrical resistance:}\) The source has an internal resistance \[ r=50\,\Omega \] The galvanometer resistance is \[ G=100\,\Omega \] When connected directly to the source, the source's internal resistance and galvanometer resistance are in series. Therefore, \[ R_{\mathrm{total}}=r+G \] \[ R_{\mathrm{total}}=50+100 \] \[ R_{\mathrm{total}}=150\,\Omega \] \(\textbf{Calculate the actual current through the coil:}\) The source emf is \[ \mathcal E=12\times10^{-3}\,\text{V} \] Using Ohm's law for the complete circuit, \[ I=\frac{\mathcal E}{r+G} \] Substituting, \[ I=\frac{12\times10^{-3}}{150} \] \[ I=8.0\times10^{-5}\,\text{A} \] \[ I=80\,\mu\text{A} \] \(\textbf{Determine the galvanometer deflection:}\) The current sensitivity is \[ S_I=\frac{\phi}{I} \] Therefore, \[ \phi=S_I I \] \[ \phi=(2500)(80\times10^{-6}) \] \[ \phi=0.20\,\text{rad} \] \(\textbf{Determine the coil's terminal voltage:}\) The potential difference across the galvanometer is \[ V_g=IG \] \[ V_g=(80\times10^{-6})(100) \] \[ V_g=8.0\times10^{-3}\,\text{V} \] \[ V_g=8.0\,\text{mV} \] \(\textbf{Check the remaining source voltage:}\) The potential drop across the source's internal resistance is \[ V_r=Ir \] \[ V_r=(80\times10^{-6})(50) \] \[ V_r=4.0\,\text{mV} \] Thus, \[ V_g+V_r=8.0+4.0=12\,\text{mV} \] \(\textbf{Final answer:}\) \[ \boxed{\phi=0.20\,\text{rad},\qquad V_g=8.0\,\text{mV}} \] The intrinsic current sensitivity of the galvanometer has not changed. However, the source's internal resistance reduces the current entering the moving coil. The emf of a real source must not automatically be treated as the terminal voltage across a connected instrument. Ignoring internal resistance would overestimate the galvanometer deflection.

Explanation: \(\textbf{Find the coil currents needed for the same deflection:}\) Current sensitivity is \[ S_I=\frac{\phi}{I} \] Therefore, \[ I=\frac{\phi}{S_I} \] For instrument P, \[ I_P=\frac{0.30}{6000} \] \[ I_P=50\,\mu\text{A} \] For instrument Q, \[ I_Q=\frac{0.30}{12000} \] \[ I_Q=25\,\mu\text{A} \] The more current-sensitive instrument requires less current for the same angular deflection. \(\textbf{Compare the required terminal voltages:}\) For P, \[ V_P=I_PG_P \] \[ V_P=(50\times10^{-6})(60) \] \[ V_P=3.0\,\text{mV} \] For Q, \[ V_Q=I_QG_Q \] \[ V_Q=(25\times10^{-6})(120) \] \[ V_Q=3.0\,\text{mV} \] The instruments therefore have the same voltage sensitivity: \[ S_{V,P}=\frac{6000}{60}=100\,\text{rad V}^{-1} \] \[ S_{V,Q}=\frac{12000}{120}=100\,\text{rad V}^{-1} \] \(\textbf{Compare their electrical powers:}\) The dissipated power is \[ P=VI=I^2G \] For instrument P, \[ P_P=(3.0\times10^{-3})(50\times10^{-6}) \] \[ P_P=1.50\times10^{-7}\,\text{W} \] For instrument Q, \[ P_Q=(3.0\times10^{-3})(25\times10^{-6}) \] \[ P_Q=7.50\times10^{-8}\,\text{W} \] Therefore, \[ \frac{P_P}{P_Q}=2 \] \(\textbf{Final answer:}\) \[ \boxed{V_P=V_Q=3.0\,\text{mV},\qquad \frac{P_P}{P_Q}=2} \] Equal voltage sensitivity means that equal applied voltages produce equal deflections. It does not mean that the instruments carry equal currents or dissipate equal powers. Instrument Q requires half the current because its current sensitivity is twice as large. Its doubled resistance compensates electrically, producing the same terminal voltage. At that common voltage, Q consequently dissipates half as much power.

Explanation: \(\textbf{Relate current sensitivity to instrument parameters:}\) For a moving-coil galvanometer in a radial magnetic field, \[ S_I=\frac{NAB}{C} \] The redesigned instrument has \[ N'=2N,\qquad B'=3B \] Its area is unchanged. Therefore, \[ S'_I=\frac{(2N)A(3B)}{C'} \] The ratio of current sensitivities is \[ \frac{S'_I}{S_I} =\frac{6NAB/C'}{NAB/C} \] \[ \frac{S'_I}{S_I}=6\frac{C}{C'} \] \(\textbf{Impose the required voltage-sensitivity change:}\) The voltage sensitivity of a bare galvanometer is \[ S_V=\frac{S_I}{G} \] The redesigned resistance is \[ G'=2G \] The required condition is \[ S'_V=2S_V \] Substituting the voltage-sensitivity expressions, \[ \frac{S'_I}{2G}=2\frac{S_I}{G} \] Multiplying by \(2G\), \[ S'_I=4S_I \] Thus, \[ \frac{S'_I}{S_I}=4 \] \(\textbf{Determine the required torsional constant:}\) Combine this condition with the magnetic sensitivity relation: \[ 6\frac{C}{C'}=4 \] Rearranging, \[ \frac{C'}{C}=\frac64 \] \[ \frac{C'}{C}=\frac32 \] \(\textbf{Verify the voltage-sensitivity target:}\) With the chosen spring, \[ \frac{S'_I}{S_I}=4 \] The coil resistance doubles. Therefore, \[ \frac{S'_V}{S_V} =\frac{S'_I/S_I}{G'/G} \] \[ \frac{S'_V}{S_V}=\frac42=2 \] The required voltage-sensitivity condition is satisfied. \(\textbf{Final answer:}\) \[ \boxed{\frac{C'}{C}=\frac32,\qquad \frac{S'_I}{S_I}=4} \] Doubling the turn number and tripling the field would increase current sensitivity sixfold if the spring were unchanged. However, the simultaneous increase in electrical resistance reduces the gain in voltage sensitivity. A stronger restoring spring is therefore required to obtain the specified voltage-sensitivity increase of two. The design constraints must be applied together rather than treating magnetic and electrical improvements independently.

Explanation: \(\textbf{Infer resistance from the voltage and power measurements:}\) The initial terminal voltage is \[ V_1=2.0\times10^{-3}\,\text{V} \] The measured dissipated power is \[ P_1=0.040\,\mu\text{W} \] \[ P_1=4.0\times10^{-8}\,\text{W} \] For a resistive moving coil, \[ P=\frac{V^2}{G} \] Therefore, \[ G=\frac{V_1^2}{P_1} \] Substituting, \[ G=\frac{(2.0\times10^{-3})^2}{4.0\times10^{-8}} \] \[ G=\frac{4.0\times10^{-6}}{4.0\times10^{-8}} \] \[ G=100\,\Omega \] \(\textbf{Recover the original current and sensitivity:}\) The original current is \[ I_1=\frac{V_1}{G} \] \[ I_1=\frac{2.0\times10^{-3}}{100} \] \[ I_1=20\,\mu\text{A} \] Current sensitivity is \[ S_I=\frac{\phi_1}{I_1} \] \[ S_I=\frac{0.20}{20\times10^{-6}} \] \[ S_I=10000\,\text{rad A}^{-1} \] \(\textbf{Determine the current with the added series resistance:}\) The second source has voltage \[ V_2=1.0\times10^{-3}\,\text{V} \] The galvanometer and series resistor each have resistance \(100\,\Omega\). Therefore, \[ R_{\mathrm{total}}=100+100 \] \[ R_{\mathrm{total}}=200\,\Omega \] The current is \[ I_2=\frac{1.0\times10^{-3}}{200} \] \[ I_2=5.0\,\mu\text{A} \] \(\textbf{Predict the new deflection:}\) \[ \phi_2=S_I I_2 \] \[ \phi_2=(10000)(5.0\times10^{-6}) \] \[ \phi_2=0.050\,\text{rad} \] \(\textbf{Final answer:}\) \[ \boxed{G=100\,\Omega,\quad S_I=10000\,\text{rad A}^{-1},\quad \phi_2=0.050\,\text{rad}} \] The initial power measurement supplies independent information about the coil resistance. The measured deflection and recovered current then determine intrinsic current sensitivity. The external series resistor reduces the current supplied by the second voltage source but does not change that intrinsic sensitivity. The final deflection must be calculated from the actual current, not from the entire applied voltage divided by the bare coil resistance alone.

Explanation: \(\textbf{Compare the total lengths of winding wire:}\) For a circular coil with \(N\) turns and radius \(R\), the approximate winding-wire length is \[ \ell_w=2\pi NR \] For P, \[ \ell_P=2\pi(200)(0.050) \] \[ \ell_P=20\pi\,\text{m} \] For Q, \[ \ell_Q=2\pi(100)(0.10) \] \[ \ell_Q=20\pi\,\text{m} \] Thus, both moving coils use equal lengths of the same conducting wire. Their electrical resistances are equal because resistivity and wire cross-sectional area are also the same. \(\textbf{Compare their magnetic moments per unit current:}\) The area of one circular turn is \[ A=\pi R^2 \] For radial-field galvanometers, \[ S_I=\frac{NAB}{C} \] Because \(B\) and \(C\) are identical, current sensitivity is proportional to \(NR^2\). Hence, \[ \frac{S_{I,Q}}{S_{I,P}} =\frac{N_QR_Q^2}{N_PR_P^2} \] Substituting, \[ \frac{S_{I,Q}}{S_{I,P}} =\frac{100(0.10)^2}{200(0.050)^2} \] \[ \frac{S_{I,Q}}{S_{I,P}} =\frac{1.0}{0.50} \] \[ \frac{S_{I,Q}}{S_{I,P}}=2 \] \(\textbf{Determine the voltage-sensitivity ratio:}\) The voltage sensitivity is \[ S_V=\frac{S_I}{G} \] The winding resistances are equal: \[ G_Q=G_P \] Therefore, \[ \frac{S_{V,Q}}{S_{V,P}} =\frac{S_{I,Q}}{S_{I,P}} \] \[ \frac{S_{V,Q}}{S_{V,P}}=2 \] \(\textbf{Final answer:}\) \[ \boxed{\frac{S_{I,Q}}{S_{I,P}}=2,\qquad \frac{S_{V,Q}}{S_{V,P}}=2} \] Equal total winding-wire length fixes the coil resistance under the stated construction assumptions. However, the larger radius of Q increases the enclosed area per turn. Its halved turn count does not completely compensate for that area increase. Consequently, Q has twice the current sensitivity. Because the resistances remain equal, its voltage sensitivity also doubles. The comparison illustrates why turn number alone cannot determine galvanometer sensitivity.