Magnetic Effects of Current and Magnetism MCQs with Answers – Part 2 (NEET Physics)

Explanation: \(\textbf{Original cancellation condition:}\) Initially, solenoid \(P\) produces \[ \vec B_P=+B_0\hat k \] Solenoid \(Q\) produces \[ \vec B_Q=-B_0\hat k \] Therefore, \[ \vec B_{\mathrm{initial}}=\vec0 \] The original winding density of \(P\) is \[ n_P=\frac NL \] Its field magnitude is \[ B_0=\mu_0\frac NL I \] \(\textbf{Effect of compressing only solenoid }P\textbf{:}\) The new length is \[ L'_P=\frac{2L}{3} \] The number of turns remains \(N\). Therefore, its new winding density is \[ n'_P=\frac{N}{2L/3} \] \[ n'_P=\frac32\frac NL \] The current magnitude remains unchanged. Hence, \[ B'_P=\mu_0 n'_PI \] \[ B'_P=\mu_0\left(\frac32\frac NL\right)I \] \[ B'_P=\frac32B_0 \] Its direction remains positive \(z\). \(\textbf{Field of the unchanged solenoid:}\) Solenoid \(Q\) retains its original winding density, current and orientation. Therefore, \[ \vec B'_Q=-B_0\hat k \] \(\textbf{Calculate the resultant field:}\) \[ \vec B_{\mathrm{final}}=\vec B'_P+\vec B'_Q \] \[ \vec B_{\mathrm{final}}= \left(\frac32B_0-B_0\right)\hat k \] \[ \vec B_{\mathrm{final}}=\frac{B_0}{2}\hat k \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_{\mathrm{final}}=\frac{B_0}{2}\hat k} \] The initial zero-field condition is destroyed because only one solenoid's winding density changes. Compressing the first solenoid increases its field magnitude without reversing its direction. The second field remains unchanged and still opposes the first. The excess positive-\(z\) contribution produces the final nonzero field. This illustrates how magnetic-field cancellation depends on maintaining both equal magnitudes and opposite directions.

Explanation: The ideal infinitely long solenoid has a uniform magnetic field inside and zero field outside. The interior field is \[ \vec B=B_0\hat k \] The circulation along a closed path is \[ \oint\vec B\cdot d\vec l \] For a rectangular path whose connecting sides are perpendicular to the solenoid axis, those connecting sides contribute zero. The result is determined by the contributions from the two axial sides. For path \(P\), both axial sides lie in the same uniform interior field. One side is traversed along positive \(z\), giving \(+B_0\ell\). The opposite side is necessarily traversed along negative \(z\), giving \(-B_0\ell\). Their sum is zero. For path \(Q\), the positive-\(z\) side lies inside the solenoid and contributes \(+B_0\ell\). Its opposite axial side lies outside, where the ideal field vanishes. Therefore, its circulation is \(B_0\ell\). For path \(R\), both axial sides lie outside, so both contributions vanish. Its circulation is zero. For path \(S\), both axial sides again lie inside the uniform field. Although they are at different radial distances, their field magnitudes are equal in the ideal infinite-solenoid approximation. Their oppositely directed traversals therefore produce equal and opposite line-integral contributions. Thus, path \(S\) also has zero circulation. The result is consistent with Ampere's law because only path \(Q\) links a nonzero algebraic current through the windings. The key principle is that a nonzero magnetic field along a closed path does not necessarily produce nonzero circulation. The oriented contributions around the complete path must be added, rather than considering only one side.

Explanation: \(\textbf{Identify the field of the outer solenoid:}\) Region \(B\) lies inside solenoid \(P\) but outside solenoid \(Q\). In the ideal infinite-solenoid approximation, the inner solenoid contributes zero field outside its winding. Therefore, the field measured in region \(B\) is entirely due to \(P\). Thus, \[ B_P=+0.60\,\text{mT} \] \(\textbf{Infer the original field of the inner solenoid:}\) Region \(A\) lies inside both solenoids. The measured field is the algebraic sum of the two contributions: \[ B_A=B_P+B_Q \] Substituting the measured values, \[ -0.40=+0.60+B_Q \] where all field values are in millitesla. Therefore, \[ B_Q=-0.40-0.60 \] \[ B_Q=-1.00\,\text{mT} \] The inner solenoid originally produces a stronger field than the outer solenoid, but in the opposite direction. \(\textbf{Apply the current reversal:}\) Reversing the current in \(Q\) reverses the direction of its magnetic field without changing its magnitude. Therefore, \[ B'_Q=+1.00\,\text{mT} \] The outer solenoid is unchanged: \[ B'_P=+0.60\,\text{mT} \] \(\textbf{Calculate the new common-interior field:}\) \[ B'_A=B'_P+B'_Q \] \[ B'_A=0.60+1.00 \] \[ B'_A=+1.60\,\text{mT} \] \(\textbf{Final answer:}\) \[ \boxed{B'_A=+1.60\,\text{mT}} \] The measurement in the annular region isolates the outer solenoid's field. The measurement inside both solenoids then determines the original field of the inner solenoid. Only after separating these contributions can the effect of reversing one current be calculated. The exterior measurement confirms the ideal field model but does not itself determine either winding current. The new common-interior field is larger because the two contributions reinforce after the current reversal.

Explanation: \(\textbf{Determine the circular coil's central field:}\) The central field of a circular coil having \(N\) turns is \[ B_{\mathrm{coil}}=\frac{\mu_0 NI}{2R} \] The given parameters are \[ N=5,\qquad I=2.0\,\text{A},\qquad R=0.10\,\text{m} \] Substituting, \[ B_{\mathrm{coil}}=\frac{\mu_0(5)(2.0)}{2(0.10)} \] \[ B_{\mathrm{coil}}=50\mu_0\,\text{A m}^{-1} \] In tesla, \[ B_{\mathrm{coil}}=50(4\pi\times10^{-7}) \] \[ B_{\mathrm{coil}}=2\pi\times10^{-5}\,\text{T} \] The coil current appears anticlockwise from positive \(z\). Hence, its field points along positive \(z\). \(\textbf{Apply the magnetic-field cancellation requirement:}\) The solenoid field must have the same magnitude as the circular coil's field. It must also point in the opposite direction. Therefore, \[ B_{\mathrm{solenoid}}=50\mu_0\,\text{A m}^{-1} \] with direction negative \(z\). \(\textbf{Determine the solenoid current:}\) For the ideal long solenoid, \[ B_{\mathrm{solenoid}}=\mu_0 nI_s \] The winding density is \[ n=1000\,\text{m}^{-1} \] Therefore, \[ \mu_0(1000)I_s=50\mu_0\,\text{A m}^{-1} \] Cancelling \(\mu_0\), \[ 1000I_s=50\,\text{A} \] \[ I_s=0.05\,\text{A} \] \(\textbf{Direction check:}\) The solenoid current must circulate clockwise when viewed from positive \(z\). This produces an interior field along negative \(z\), opposite to the circular coil's field. \(\textbf{Final answer:}\) \[ \boxed{I_s=0.05\,\text{A},\qquad\vec B_s\parallel-\hat k} \] The circular coil and solenoid have different central-field formulas. The circular coil's field depends on its turn number and radius, whereas the ideal solenoid's field depends on its winding density. The required solenoid current is determined by equating the two field magnitudes. The current orientation must be chosen independently to ensure cancellation rather than reinforcement.

Explanation: For an ideal long solenoid, \[ B=\mu_0 nI \] where \[ n=\frac NL \] The current magnitudes are equal for all arrangements. Therefore, equal magnetic-field magnitudes require equal winding densities. To obtain cancellation, their field directions must also be opposite. \(\textbf{Winding density of arrangement P:}\) \[ n_P=\frac{400}{0.20} \] \[ n_P=2000\,\text{m}^{-1} \] Its interior field points along positive \(z\). \(\textbf{Winding density of arrangement Q:}\) \[ n_Q=\frac{600}{0.40} \] \[ n_Q=1500\,\text{m}^{-1} \] Its interior field points along negative \(z\). \(\textbf{Winding density of arrangement R:}\) \[ n_R=\frac{800}{0.40} \] \[ n_R=2000\,\text{m}^{-1} \] Its field points along negative \(z\). \(\textbf{Winding density of arrangement S:}\) \[ n_S=\frac{600}{0.30} \] \[ n_S=2000\,\text{m}^{-1} \] Its field points along positive \(z\). \(\textbf{Compare the cancellation conditions:}\) Arrangements \(P\), \(R\), and \(S\) have equal field magnitudes. However, \(P\) and \(S\) produce fields in the same direction. Arrangement \(R\) produces its field in the opposite direction to \(P\). Therefore, \(P\) and \(R\) give equal and opposite magnetic-field contributions. Arrangement \(Q\) has a smaller winding density and cannot cancel the field of \(P\) or \(S\) at the same current magnitude. Among the listed pairs, only \(P\) and \(R\) satisfy both requirements. This comparison illustrates that different total turn numbers and lengths can produce identical ideal magnetic-field magnitudes. The relevant quantity is the number of turns per unit length, not the total number of turns considered separately. The field orientation must then be checked to distinguish cancellation from reinforcement.

Explanation: The magnetic field inside a finite solenoid follows the solenoid axis near its central region. The specified interior field points along positive \(x\). Therefore, the field lines travel internally from the left end towards the right end. The right end acts as the magnetic north end of the solenoid. The left end acts as its magnetic south end. Magnetic field lines emerge from the north end and return through the exterior region to the south end. Consequently, the field immediately outside the right end, close to the central axis, continues along positive \(x\). The field does not suddenly reverse direction at the physical end of the winding. Instead, the field lines spread outward and curve through the surrounding space. Away from the end regions, the returning external field lines travel generally from right to left. They eventually approach the left south end and enter the solenoid. Inside the solenoid, the lines continue from left to right, completing closed curves. This continuity distinguishes magnetic field lines from electric field lines that may begin or terminate on electric charges. A finite solenoid therefore possesses an exterior magnetic field even though the ideal infinite-solenoid approximation predicts zero exterior field. The approximate uniformity of the central interior field does not extend to the fringing region near the ends. The direction at the right end must be inferred from the established interior field orientation, not merely from the presence of the winding.

Explanation: \(\textbf{Long-solenoid comparison:}\) The magnetic field of an ideal long solenoid is \[ B_s=\mu_0 nI \] Here, \(n\) is the number of turns per unit length. The modified solenoid has the same winding density and the same current magnitude. Therefore, \[ B'_s=\mu_0 nI \] Hence, \[ B'_s=B_s \] The radius does not appear explicitly in the ideal long-solenoid field expression. The assumed construction is important: increasing the radius must not accidentally change the winding density or current. \(\textbf{Circular-loop comparison:}\) Now consider the separate circular current loop. The magnetic field at the centre of a single circular turn is \[ B_c=\frac{\mu_0 I}{2R} \] After doubling its radius, \[ R'=2R \] The current remains \(I\). Therefore, \[ B'_c=\frac{\mu_0 I}{2(2R)} \] \[ B'_c=\frac12\frac{\mu_0 I}{2R} \] Thus, \[ B'_c=\frac{B_c}{2} \] The circular-loop field decreases because it depends inversely on its radius at fixed current. The ideal solenoid field is independent of radius at fixed winding density and current. These conclusions concern different current distributions and different geometrical constraints. The difference is not a contradiction between Biot–Savart law and Ampere's law. Both laws describe the same magnetic-field physics, but the complete current arrangements have different geometrical dependencies. The result also illustrates why a formula derived for one field-producing system cannot be transferred to another without checking its assumptions.

Explanation: \(\textbf{Separate the solenoid and background contributions:}\) Let the original positive-current solenoid field be \(B_s\). Let the signed uniform background field be \(B_{\mathrm{bg}}\). The first measurement gives \[ B_s+B_{\mathrm{bg}}=11\,\mu\text{T} \] Reversing the current reverses the solenoid field but does not affect the background. Therefore, the second measurement gives \[ -B_s+B_{\mathrm{bg}}=-5\,\mu\text{T} \] \(\textbf{Determine the background field:}\) Adding the equations, \[ 2B_{\mathrm{bg}}=6\,\mu\text{T} \] \[ B_{\mathrm{bg}}=+3\,\mu\text{T} \] \(\textbf{Determine the original solenoid field:}\) Substitute into the first equation: \[ B_s+3=11 \] \[ B_s=8\,\mu\text{T} \] Thus, the solenoid alone produces a field of \(8\,\mu\text{T}\) along positive \(z\) when its current is \(+I\). \(\textbf{Apply the final current change:}\) The ideal solenoid field is \[ B_s=\mu_0 nI \] With the winding density unchanged, the field is directly proportional to current. At current \(+I/2\), \[ B'_s=\frac{B_s}{2} \] \[ B'_s=4\,\mu\text{T} \] The current direction remains positive, so the field points along positive \(z\). \(\textbf{Add the unchanged background field:}\) \[ B_{\mathrm{final}}=B'_s+B_{\mathrm{bg}} \] \[ B_{\mathrm{final}}=4+3 \] \(\textbf{Final answer:}\) \[ \boxed{B_{\mathrm{final}}=+7\,\mu\text{T}} \] The two opposite-current measurements separate the current-dependent solenoid field from the constant background contribution. The final current is half the original positive current, so the solenoid contribution is halved without changing direction. The background field remains unchanged. Applying the current factor to the complete initial probe reading would incorrectly change the background contribution as well. The result follows from combining current reversal, experimental inference and magnetic-field superposition.

Explanation: The expression \[ B=\mu_0 nI \] is the ideal magnetic-field relation for a sufficiently long, closely wound solenoid in air or vacuum. It is most accurate near the central interior when the solenoid length is large compared with its transverse dimensions. The absence of length and diameter from the ideal formula does not mean these geometrical quantities are irrelevant to the validity of the approximation. For solenoid \(P\), the length is \(1.0\,\text{m}\) and the diameter is \(0.020\,\text{m}\). Its length-to-diameter ratio is \[ \frac{L_P}{D_P}=\frac{1.0}{0.020}=50 \] For solenoid \(Q\), \[ \frac{L_Q}{D_Q}=\frac{0.20}{0.10}=2 \] For solenoid \(R\), \[ \frac{L_R}{D_R}=\frac{0.80}{0.040}=20 \] Solenoids \(P\) and \(R\) are much longer relative to their diameters than solenoid \(Q\). Their central interior regions therefore more closely resemble the ideal long-solenoid geometry. Solenoid \(Q\) has a length only twice its diameter, so effects associated with its finite ends cannot be neglected as confidently. A finite solenoid has field spreading near its ends and a nonzero exterior field. Consequently, the ideal formula should not be treated as an exact result for all three constructions. Equal winding density and current predict equal fields only within the validity of the same ideal approximation. The information supplied does not justify declaring the three actual measured fields exactly equal. Nor does it justify applying a new short-solenoid formula without further physical information. The correct evaluation is that the long-solenoid approximation is better justified for \(P\) and \(R\) than for \(Q\). The key reasoning requirement is to verify the assumptions behind an ideal relation before using it to make exact physical claims.

Explanation: Ampere's circuital law relates the circulation around a closed path to the algebraic current linked by that path. For the anticlockwise traversal, positive-\(z\) currents are counted as positive. The first section contributes \(+I\), while the returning section contributes \(-I\). Therefore, the total linked current is zero. Statement I is correct. A zero circulation, however, does not imply that the magnetic field vanishes at every point within the path. At \(O\), the first wire carries current along positive \(z\), and the observation point lies to its right. The right-hand rule gives a magnetic field along positive \(y\). For the return wire, the current points along negative \(z\), and \(O\) lies to its left. Its magnetic field also points along positive \(y\). Thus, the two contributions reinforce rather than cancel. Statement II is incorrect, and Statement IV is correct. Now suppose the return section is moved while remaining inside \(C\). Its distance from \(O\) changes, so its magnetic-field contribution at \(O\) generally changes according to the inverse-distance relation. The algebraic current enclosed by \(C\), however, remains \(I-I=0\). Therefore, the circulation remains zero. Statement III is correct. The physical connection between the two sections explains why their currents have equal magnitudes and opposite axial directions. It does not require their magnetic fields to cancel at a selected observation point. The essential distinction is between linked-current cancellation around a complete path and magnetic-field superposition at a particular location.

Explanation: \(\textbf{Translate the first three measurements:}\) For steady currents, \[ \oint\vec B\cdot d\vec l=\mu_0 I_{\mathrm{enc}} \] The first three paths have identical anticlockwise traversal conventions. Thus, positive current is along positive \(z\). The first measurement gives \[ I_P+I_Q=5\,\text{A} \] The second gives \[ I_Q+I_R=-2\,\text{A} \] The third gives \[ I_Q=-3\,\text{A} \] \(\textbf{Recover the current in wire }P\textbf{:}\) Substitute \(I_Q=-3\,\text{A}\) into the first equation: \[ I_P-3=5 \] \[ I_P=+8\,\text{A} \] \(\textbf{Recover the current in wire }R\textbf{:}\) The second measurement gives \[ -3+I_R=-2 \] \[ I_R=+1\,\text{A} \] \(\textbf{Determine the net linked current:}\) The signed current along positive \(z\) through all three wires is \[ I_{\mathrm{total},+z}=I_P+I_Q+I_R \] \[ I_{\mathrm{total},+z}=8-3+1 \] \[ I_{\mathrm{total},+z}=+6\,\text{A} \] \(\textbf{Apply the reversed path orientation:}\) Path \(C_4\) is traversed clockwise when viewed from positive \(z\). Its positive surface normal therefore points along negative \(z\). Consequently, the signed current for this path is \[ I_{\mathrm{enc},C_4}=-6\,\text{A} \] Ampere's law gives \[ C_4=\mu_0(-6\,\text{A}) \] \(\textbf{Final answer:}\) \[ \boxed{C_4=-6\mu_0\,\text{A}} \] The three initial measurements permit the individual signed currents to be reconstructed. The fourth path encloses their algebraic sum, but its opposite traversal direction changes the sign of the circulation. The current through wire \(Q\) must not be counted twice when combining the overlapping first two paths. An equivalent calculation is to use the combination \(C_1+C_2-C_3\) for an anticlockwise path enclosing all three wires and then reverse its sign. This provides an independent check of the result.

Explanation: \(\textbf{Field from the first wire:}\) The displacement from wire \(P\) to \(S\) is \[ \vec r_P=a\hat i+a\hat j \] The squared distance is \[ r_P^2=2a^2 \] For an infinitely long wire carrying positive-\(z\) current, the magnetic-field vector can be written as \[ \vec B=\frac{\mu_0 I}{2\pi r^2}(\hat k\times\vec r) \] Therefore, \[ \vec B_P=\frac{\mu_0 I}{2\pi(2a^2)} \left[\hat k\times(a\hat i+a\hat j)\right] \] Using the cross-product relations, \[ \hat k\times\hat i=\hat j,\qquad \hat k\times\hat j=-\hat i \] we obtain \[ \vec B_P=\frac{\mu_0 I}{4\pi a}(-\hat i+\hat j) \] Thus, \[ \vec B_P=K(-\hat i+\hat j) \] \(\textbf{Field from the second wire:}\) The displacement from wire \(Q\) to \(S\) is \[ \vec r_Q=-a\hat i+a\hat j \] Its squared magnitude is also \(2a^2\). The current in \(Q\) is \(3I\). Therefore, \[ \vec B_Q=\frac{\mu_0(3I)}{2\pi(2a^2)} \left[\hat k\times(-a\hat i+a\hat j)\right] \] \[ \vec B_Q=3K(-\hat i-\hat j) \] \(\textbf{Vector addition:}\) \[ \vec B_{\mathrm{net}}=\vec B_P+\vec B_Q \] \[ \vec B_{\mathrm{net}}=K(-\hat i+\hat j)+3K(-\hat i-\hat j) \] \[ \vec B_{\mathrm{net}}=K(-4\hat i-2\hat j) \] \(\textbf{Magnitude and direction check:}\) The resultant lies in the third quadrant of the \(xy\)-plane. Its magnitude is \[ B_{\mathrm{net}}=K\sqrt{4^2+2^2} \] \[ B_{\mathrm{net}}=2\sqrt5K \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_{\mathrm{net}}=K(-4\hat i-2\hat j)} \] The two magnetic-field contributions are neither parallel nor antiparallel. Their \(x\)-components reinforce, while their \(y\)-components partially cancel. This problem cannot be solved by adding or subtracting only the field magnitudes. The positions of the wires determine the field directions, and the unequal current magnitudes determine the relative component strengths.

Explanation: \(\textbf{Associate each measured component with its source:}\) Both currents flow along positive \(z\). The origin lies to the left of wire \(P\), which is located at positive \(x\). By the right-hand rule, the field produced by \(P\) at the origin points along negative \(y\). Its vector is \[ \vec B_P=-\frac{\mu_0 I_P}{2\pi a}\hat j \] The origin lies below wire \(Q\), which is located at positive \(y\). For current along positive \(z\), the field at the origin points along positive \(x\). Thus, \[ \vec B_Q=+\frac{\mu_0 I_Q}{2\pi a}\hat i \] \(\textbf{Use the measured resultant vector:}\) By superposition, \[ \vec B_O=\frac{\mu_0}{2\pi a} (I_Q\hat i-I_P\hat j) \] The measured field is \[ \vec B_O=\frac{\mu_0}{2\pi a} (6\hat i-4\hat j)\,\text{A} \] Comparing independent Cartesian components gives \[ I_Q=6\,\text{A} \] and \[ I_P=4\,\text{A} \] \(\textbf{Determine the circulation:}\) The Amperian path encloses both wires. Because both currents are along positive \(z\), their signed sum is \[ I_{\mathrm{enc}}=I_P+I_Q \] \[ I_{\mathrm{enc}}=4+6 \] \[ I_{\mathrm{enc}}=10\,\text{A} \] Ampere's law gives \[ \oint\vec B\cdot d\vec l=\mu_0I_{\mathrm{enc}} \] \(\textbf{Final answer:}\) \[ \boxed{I_P=4\,\text{A},\quad I_Q=6\,\text{A},\quad\oint\vec B\cdot d\vec l=10\mu_0\,\text{A}} \] The local field measurement determines the individual currents because their magnetic-field contributions lie along different Cartesian axes. The closed-path circulation then follows from the algebraic sum of those currents. Using only the magnitude of the measured resultant would not separately determine the two source currents. The directional information contained in the measured vector is essential.

Explanation: \(\textbf{Identify the perpendicular field components:}\) The solenoid produces a uniform axial magnetic field. The straight wire produces an azimuthal magnetic field around the same axis. At any point away from the axis, these two fields are perpendicular. The solenoid field magnitude is \[ B_s=\mu_0nI_s \] Substituting, \[ B_s=\mu_0(500)(2.0) \] \[ B_s=1000\mu_0\,\text{A m}^{-1} \] \(\textbf{Field of the coaxial straight wire:}\) At radial distance \(r\), \[ B_w=\frac{\mu_0I_w}{2\pi r} \] The wire current is \[ I_w=2\pi\,\text{A} \] Therefore, \[ B_w=\frac{\mu_0(2\pi)}{2\pi r} \] \[ B_w=\frac{\mu_0}{r}\,\text{A} \] where \(r\) is expressed in metres. \(\textbf{Apply the required field direction:}\) Let \(\theta\) be the angle between the resultant field and the positive \(z\)-axis. Since the axial and azimuthal components are perpendicular, \[ \tan\theta=\frac{B_w}{B_s} \] For \(\theta=45^\circ\), \[ \tan45^\circ=1 \] Therefore, \[ B_w=B_s \] Substituting the expressions, \[ \frac{\mu_0I_w}{2\pi r}=\mu_0nI_s \] Cancel \(\mu_0\): \[ r=\frac{I_w}{2\pi nI_s} \] \[ r=\frac{2\pi}{2\pi(500)(2.0)} \] \[ r=1.0\times10^{-3}\,\text{m} \] \[ r=1.0\,\text{mm} \] This lies inside the specified solenoid radius. \(\textbf{Calculate the circular-path circulation:}\) The counterclockwise path encloses the central straight-wire current. The axial magnetic field of the solenoid is perpendicular to the circular path element everywhere. Therefore, the solenoid contributes zero to this particular circulation integral. Ampere's law gives \[ \oint\vec B\cdot d\vec l=\mu_0I_w \] \[ \oint\vec B\cdot d\vec l=2\pi\mu_0\,\text{A} \] \(\textbf{Final answer:}\) \[ \boxed{r=1.0\,\text{mm},\qquad\oint\vec B\cdot d\vec l=2\pi\mu_0\,\text{A}} \] The resultant field direction is controlled by the ratio of azimuthal and axial field components. The circulation around the coaxial circular path depends only on the tangential wire field. The solenoid affects the resultant field angle without contributing to the circulation around that particular circle. This illustrates the distinction between a magnetic-field vector and its component along a chosen integration path.

Explanation: \(\textbf{Write the two axial magnetic fields:}\) The magnetic field produced by the circular coil on its axis is \[ B_c(z)=B_0\frac{R^3}{(R^2+z^2)^{3/2}} \] Its direction is positive \(z\) on both sides of the coil. The solenoid produces a uniform opposing field: \[ \vec B_s=-\frac{B_0}{8}\hat k \] Therefore, the signed resultant is \[ B_{\mathrm{net},z}(z) =B_0\frac{R^3}{(R^2+z^2)^{3/2}}-\frac{B_0}{8} \] \(\textbf{Determine the zero-field locations:}\) For cancellation, \[ B_{\mathrm{net},z}(z)=0 \] Thus, \[ B_0\frac{R^3}{(R^2+z^2)^{3/2}}=\frac{B_0}{8} \] Cancel the nonzero central field: \[ \frac{R^3}{(R^2+z^2)^{3/2}}=\frac18 \] Rearranging, \[ (R^2+z^2)^{3/2}=8R^3 \] Raise both sides to the power \(2/3\): \[ R^2+z^2=4R^2 \] Therefore, \[ z^2=3R^2 \] \[ z=\pm\sqrt3R \] Both signs are valid because the axial coil field has equal magnitudes at symmetric positions. \(\textbf{Determine the resultant field at the centre:}\) At \(z=0\), the coil produces \(+B_0\hat k\). The solenoid produces \(-B_0\hat k/8\). Therefore, \[ \vec B_{\mathrm{net}}(0) =\left(B_0-\frac{B_0}{8}\right)\hat k \] \[ \vec B_{\mathrm{net}}(0)=+\frac78B_0\hat k \] \(\textbf{Final answer:}\) \[ \boxed{z=\pm\sqrt3R,\qquad\vec B_{\mathrm{net}}(0)=+\frac78B_0\hat k} \] The coil field decreases with increasing absolute axial distance, whereas the ideal solenoid field remains constant in the specified region. Cancellation therefore occurs at two symmetric axial positions rather than at the centre. Closer to the coil, the positive coil field dominates. Farther away within the ideal region, the negative solenoid field can dominate. This provides an example of a magnetic-field direction changing with position even though neither source current reverses.

Explanation: \(\textbf{Separate the perpendicular magnetic fields:}\) The solenoid produces an axial field \[ \vec B_s=\mu_0nI\,\hat k \] The uniform background field is \[ \vec B_{\mathrm{bg}}=B_*\hat i \] These components are perpendicular. Therefore, the squared resultant magnitude is \[ B_{\mathrm{net}}^2=B_*^2+(\mu_0nI)^2 \] This is a quadratic function of signed current and is symmetric about \(I=0\). \(\textbf{Interpret the graph minimum:}\) At zero current, the solenoid produces no field. Therefore, \[ B_{\min}^2=B_*^2 \] The graph states that at \(I=I_0\), \[ B_{\mathrm{net}}^2=5B_{\min}^2 \] Substituting, \[ B_*^2+(\mu_0nI_0)^2=5B_*^2 \] Rearranging, \[ (\mu_0nI_0)^2=4B_*^2 \] Since \(I_0\gt 0\), \[ \mu_0nI_0=2B_* \] Thus, the axial solenoid field is twice the perpendicular background field. \(\textbf{Determine the resultant angle:}\) The angle \(\theta\) is measured from the positive \(z\)-axis. Therefore, \[ \tan\theta=\frac{B_{\perp}}{B_{\parallel}} \] Here, \[ B_{\perp}=B_*,\qquad B_{\parallel}=2B_* \] Consequently, \[ \tan\theta=\frac{B_*}{2B_*} \] \(\textbf{Final answer:}\) \[ \boxed{\tan\theta=\frac12} \] The graph does not represent the signed axial field itself; it represents the squared magnitude of the vector resultant. Its nonzero minimum arises from the perpendicular background field. At the specified current, the ordinate determines the ratio of perpendicular and axial components. The angle is then found using vector geometry rather than by subtracting magnetic-field magnitudes. The positive-current condition establishes the positive axial direction used in defining \(\theta\).

Explanation: \(\textbf{Initial perpendicular field components:}\) The first solenoid produces \[ \vec B_x=B_0\hat i \] The second solenoid produces \[ \vec B_y=B_0\hat j \] The initial resultant is \[ \vec B_{\mathrm{initial}}=B_0\hat i+B_0\hat j \] Because the components are perpendicular, \[ B_{\mathrm{initial}}=\sqrt{B_0^2+B_0^2} \] \[ B_{\mathrm{initial}}=\sqrt2B_0 \] \(\textbf{Apply the first solenoid's current reversal:}\) Reversing the current reverses the direction of the interior magnetic field. Its magnitude remains unchanged. Therefore, \[ \vec B'_x=-B_0\hat i \] \(\textbf{Apply the second solenoid's current change:}\) For an ideal long solenoid, \[ B=\mu_0nI \] The winding density is unchanged. Doubling the current doubles the field magnitude. Because the current direction is not reversed, \[ \vec B'_y=2B_0\hat j \] \(\textbf{Calculate the final resultant:}\) \[ \vec B_{\mathrm{final}}=-B_0\hat i+2B_0\hat j \] Its magnitude is \[ B_{\mathrm{final}} =\sqrt{(-B_0)^2+(2B_0)^2} \] \[ B_{\mathrm{final}}=\sqrt5B_0 \] \(\textbf{Compare the two magnitudes:}\) \[ \frac{B_{\mathrm{final}}}{B_{\mathrm{initial}}} = \frac{\sqrt5B_0}{\sqrt2B_0} \] \(\textbf{Final answer:}\) \[ \boxed{\frac{B_{\mathrm{final}}}{B_{\mathrm{initial}}}=\frac{\sqrt5}{\sqrt2}} \] The currents alter the magnetic-field vectors independently. The first reversal changes a component's sign but not its magnitude. The second change doubles the perpendicular component without changing its sign. The correct resultant follows from vector addition, not the algebraic addition of field magnitudes. The final field points into the second quadrant of the \(xy\)-plane, which provides an independent directional check.

Explanation: The magnetic force on a moving charged particle is \[ \vec F_B=q(\vec v\times\vec B) \] Its direction depends on the velocity, magnetic field and sign of the charge. Initially, the velocity is along positive \(x\), while the field is along positive \(z\). The cross product for a positive charge would point along negative \(y\). Because the actual particle has negative charge, the magnetic force points along positive \(y\), as stated. When the velocity is reversed, the cross product changes sign. The charge and magnetic field are unchanged. Therefore, the actual magnetic force also reverses and points along negative \(y\). The force does not vanish because the reversed velocity is still perpendicular to the magnetic field. The magnetic-force magnitude depends on speed rather than on whether the velocity points along positive or negative \(x\). Consequently, the instantaneous force magnitude remains unchanged. The magnetic force is always perpendicular to the instantaneous velocity. Thus, its instantaneous power is \[ P=\vec F_B\cdot\vec v=0 \] The magnetic field cannot increase or decrease the particle's kinetic energy by doing work. It can change the direction of the velocity without changing its magnitude. The speed remains constant during motion under the magnetic force alone, even though the particle accelerates because its velocity direction changes. The correct reasoning therefore requires both a charge-sign check and recognition of the perpendicular-force property.

Explanation: \(\textbf{Write the Lorentz magnetic-force relation:}\) \[ \vec F_B=q(\vec v\times\vec B) \] The velocity and magnetic field are \[ \vec v=(3\hat i+4\hat j)\times10^5\,\text{m s}^{-1} \] \[ \vec B=0.50\hat k\,\text{T} \] \(\textbf{Calculate the velocity–field cross product:}\) \[ \vec v\times\vec B =(3\hat i+4\hat j)\times10^5\times(0.50\hat k) \] \[ \vec v\times\vec B =0.50\times10^5[3(\hat i\times\hat k)+4(\hat j\times\hat k)] \] Using \[ \hat i\times\hat k=-\hat j,\qquad \hat j\times\hat k=\hat i \] we obtain \[ \vec v\times\vec B =0.50\times10^5(4\hat i-3\hat j) \] \[ \vec v\times\vec B =(2.0\hat i-1.5\hat j)\times10^5 \] \(\textbf{Include the negative charge:}\) The charge is \[ q=-2.0\times10^{-6}\,\text{C} \] Therefore, \[ \vec F_B=(-2.0\times10^{-6})(2.0\hat i-1.5\hat j)\times10^5 \] \[ \vec F_B=(-0.40\hat i+0.30\hat j)\,\text{N} \] \(\textbf{Check the magnitude:}\) The speed is \[ v=\sqrt{3^2+4^2}\times10^5 \] \[ v=5.0\times10^5\,\text{m s}^{-1} \] The velocity is perpendicular to the magnetic field. Hence, \[ F_B=|q|vB \] \[ F_B=(2.0\times10^{-6})(5.0\times10^5)(0.50) \] \[ F_B=0.50\,\text{N} \] The vector magnitude is also \[ \sqrt{(-0.40)^2+(0.30)^2}=0.50\,\text{N} \] \(\textbf{Final answer:}\) \[ \boxed{\vec F_B=(-0.40\hat i+0.30\hat j)\,\text{N}} \] The negative charge reverses the direction obtained from the velocity–field cross product. The field and velocity are perpendicular, so the force magnitude is maximum for the given speed and field strength. The independent magnitude calculation verifies the vector result. A correct answer must satisfy both the cross-product direction and the charge-sign convention.

Explanation: The magnetic force is \[ \vec F_B=q(\vec v\times\vec B) \] Initially, the particle carries negative charge. Its velocity is along positive \(x\), and its force is along positive \(z\). For a negative charge to experience force along positive \(z\), the vector \(\vec v\times\vec B\) must point along negative \(z\). With velocity along positive \(x\), this requires the relevant magnetic-field component to lie along negative \(y\). The field is stated to lie entirely in the \(yz\)-plane. A nonzero \(z\)-component of magnetic field would produce an additional force component along \(y\). Because the measured force is purely along \(z\), that field component must be zero. Thus, the original field is directed along negative \(y\). Now reverse the magnetic field. The vector \(\vec v\times\vec B\) reverses direction. At the same time, replace the original negative charge by a positive charge. This also reverses the force direction relative to the original charge-sign convention. The two direction reversals cancel. Consequently, the new magnetic force points along the same positive \(z\)-direction as before. The new charge magnitude is twice the original magnitude. The magnetic-field magnitude and particle speed remain unchanged. Therefore, the force magnitude doubles: \[ F'=2F \] The new force is along positive \(z\). This example requires distinguishing the sign of the charge from the orientation of the magnetic field. Reversing either one independently reverses the magnetic-force direction. Reversing both restores the original force direction while the increased charge magnitude changes the force magnitude.