Magnetic Effects of Current and Magnetism MCQs with Answers – Part 1 (NEET Physics)

Explanation: The Biot–Savart law gives the magnetic field produced by a small current element at an observation point. Its vector form is \[ d\vec B=\frac{\mu_0}{4\pi}\frac{I\,d\vec l\times\hat r}{r^2} \] Here, \(d\vec l\) points along the conventional current, while \(\hat r\) points from the current element toward the observation point. The current element is directed along \(+\hat i\), and the displacement toward \(P\) is along \(+\hat j\). The required cross product is \[ \hat i\times\hat j=\hat k \] Therefore, the elementary magnetic field points along the positive \(z\)-axis. The magnitude of the elementary field is \[ dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2} \] The included angle is \(90^\circ\), for which \(\sin90^\circ=1\). Hence, for the specified current, element length and distance, the elementary magnetic field has its maximum possible magnitude. The important distinction is that the magnetic field is perpendicular to both the current element and the displacement vector, rather than being directed along either of them.

Explanation: \(\textbf{Governing relation:}\) The magnitude of the elementary magnetic field is \[ dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2} \] The magnetic field depends on four relevant factors: current, element length, included angle and distance. \(\textbf{Original arrangement:}\) \[ dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin30^\circ}{r^2} \] Since \(\sin30^\circ=\frac12\), \[ dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{2r^2} \] \(\textbf{Modified arrangement:}\) The new quantities are \[ I'=2I,\qquad dl'=\frac{dl}{2},\qquad r'=2r,\qquad\theta'=90^\circ \] Substituting into the Biot–Savart relation, \[ dB'=\frac{\mu_0}{4\pi}\frac{(2I)(dl/2)\sin90^\circ}{(2r)^2} \] \[ dB'=\frac{\mu_0}{4\pi}\frac{I\,dl}{4r^2} \] \(\textbf{Field comparison:}\) \[ \frac{dB'}{dB} =\frac{I\,dl/(4r^2)}{I\,dl/(2r^2)} =\frac12 \] The doubling of current is exactly compensated by halving the element length. Changing the angle from \(30^\circ\) to \(90^\circ\) doubles the angular factor. However, doubling the distance reduces the field by a factor of four. Combining these effects gives a net reduction by a factor of two. \(\textbf{Final answer:}\) \[ \boxed{\frac{dB'}{dB}=\frac12} \]

Explanation: The magnitude of the elementary magnetic field depends on the sine of the angle between the current element and the displacement vector. For an observation point lying on the straight-line extension of the element, these two vectors are either parallel or antiparallel. Consequently, the included angle is \(0^\circ\) or \(180^\circ\). In both cases, \[ \sin\theta=0 \] Therefore, the cross product in the Biot–Savart law vanishes, making Statement I correct. This zero-field condition is an angular effect and does not require the current itself to become zero. Statement II concerns current reversal at a fixed observation point. The elementary magnetic field is directly proportional to the current when the geometry remains unchanged. Reversing the current changes the direction of the current-element vector, so the cross product reverses direction. Its magnitude remains unchanged because the magnitude of the current is unchanged. The field therefore reverses without becoming twice as large. Statement II is incorrect. A useful distinction is that changing the sign of a vector contribution changes its orientation, whereas changing the magnitude of the current changes the magnitude of that contribution.

Explanation: \(\textbf{Initial field direction:}\) The initial current element points along \(+\hat j\), while the displacement toward \(P\) is along \(+\hat k\). According to the Biot–Savart law, \[ d\vec B\propto d\vec l\times\hat r \] Therefore, \[ \hat j\times\hat k=\hat i \] The initial elementary magnetic field points along the positive \(x\)-axis. \(\textbf{Direction after the changes:}\) Reversing the current makes the current-element vector point along \(-\hat j\). Moving the observation point to the opposite side makes the displacement unit vector \(-\hat k\). The new cross product is \[ (-\hat j)\times(-\hat k)=\hat i \] Both vectors have reversed, so the two negative signs cancel in the cross product. The magnetic field therefore retains its original direction. \(\textbf{Magnitude comparison:}\) In both arrangements, the element is perpendicular to the displacement vector. The element length and distance are also unchanged. Thus, \[ \frac{dB'}{dB}=\frac{I'}{I}=\frac{2I}{I}=2 \] \(\textbf{Final answer:}\) \[ \boxed{d\vec B'=2d\vec B} \] This result illustrates why magnetic-field direction must be determined from both vectors in the cross product, rather than from the current direction alone.

Explanation: \(\textbf{Field due to the first element:}\) For \(P_1\), the current element points along \(+\hat i\). The displacement from \(P_1\) to \(O\) points along \(+\hat j\). Thus, \[ \hat i\times\hat j=\hat k \] The first element produces a field along \(+z\). Since the included angle is \(90^\circ\), its magnitude is \[ dB_1=\frac{\mu_0}{4\pi}\frac{I\,dl}{a^2} \] \(\textbf{Required direction of the second field:}\) At \(P_2\), the displacement toward the origin is along \(-\hat j\). To cancel the first field, the second contribution must point along \(-\hat k\). A current along \(+\hat i\) gives \[ \hat i\times(-\hat j)=-\hat k \] Therefore, the current in \(P_2\) must be reversed from its initial direction. \(\textbf{Required magnitude:}\) The second element is at distance \(2a\) from the origin. If its new current magnitude is \(I_2\), \[ dB_2=\frac{\mu_0}{4\pi}\frac{I_2\,dl}{(2a)^2} \] For cancellation, \[ dB_2=dB_1 \] Substituting, \[ \frac{\mu_0}{4\pi}\frac{I_2\,dl}{4a^2} = \frac{\mu_0}{4\pi}\frac{I\,dl}{a^2} \] Cancelling the common factors, \[ \frac{I_2}{4}=I \] \[ I_2=4I \] \(\textbf{Final answer:}\) The second element must carry current \(4I\) along \(+x\), producing a field equal in magnitude and opposite in direction to the field of \(P_1\). Increasing the current alone would not achieve cancellation if the direction remained incorrect. Similarly, reversing the current without compensating for the fourfold distance-squared factor would leave a nonzero resultant field.

Explanation: \(\textbf{Contribution from element }P\textbf{:}\) The current in \(P\) is along \(+\hat i\), while the displacement from \(P\) to \(O\) is along \(+\hat j\). Hence, \[ \hat i\times\hat j=\hat k \] The field contribution from \(P\) is \[ d\vec B_P=B_0\hat k \] Its magnitude is \(B_0\) because the current element and displacement are perpendicular. \(\textbf{Initial contribution from element }Q\textbf{:}\) The current in \(Q\) is initially along \(+\hat j\). The displacement from \(Q\) to \(O\) is along \(+\hat k\). Therefore, \[ \hat j\times\hat k=\hat i \] Its initial contribution is \[ d\vec B_Q=B_0\hat i \] The two initial field contributions are perpendicular, not oppositely directed. \(\textbf{Change in the second current:}\) Reversing the current in \(Q\) reverses its field contribution. Doubling the current doubles the magnitude of that contribution. Thus, \[ d\vec B'_Q=-2B_0\hat i \] \(\textbf{Vector superposition:}\) The field from \(P\) remains unchanged. The final resultant is \[ \vec B_{\mathrm{net}}=d\vec B_P+d\vec B'_Q \] \[ \vec B_{\mathrm{net}}=B_0\hat k-2B_0\hat i \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_{\mathrm{net}}=B_0(\hat k-2\hat i)} \] Its magnitude would be \(\sqrt5B_0\), but magnitude alone would not specify the requested vector. Adding the two field magnitudes algebraically would be incorrect because the contributions are perpendicular. The negative \(x\)-component records the current reversal, while the positive \(z\)-component comes from the unchanged first element.

Explanation: Statement I is correct because the Biot–Savart law contains the vector product \(d\vec l\times\hat r\). A cross product is perpendicular to the plane formed by the two vectors entering it. Therefore, every nonzero elementary magnetic field is perpendicular to its own current-element vector and displacement vector. Statement II is also correct. For an unchanged conductor, reversing the current reverses the direction of every current-element vector. Consequently, every elementary magnetic field reverses direction while retaining its magnitude. Vector addition then gives a total field opposite to the original total field. Statement III is incorrect because cancellation between two selected contributions says nothing conclusive about contributions from the remaining elements. Other parts of the conductor may still produce a nonzero resultant magnetic field at the observation point. Complete cancellation requires the vector sum of all contributions to vanish. Statement IV is correct because magnetic fields obey superposition. Mathematically, the field of the complete conductor is obtained from \[ \vec B=\int d\vec B \] This integral represents vector addition over the current-carrying conductor. It cannot generally be replaced by adding elementary field magnitudes because different elements may produce fields in different directions. The important reasoning step is to distinguish a local elementary contribution from the resultant field obtained after including the entire conductor.

Explanation: \(\textbf{Given:}\) \[ N=5,\qquad I=2\,\text{A},\qquad R=0.10\,\text{m} \] The permeability of free space is \[ \mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1} \] \(\textbf{Magnetic field at the centre:}\) For a circular coil of \(N\) closely wound turns, \[ B=\frac{\mu_0 NI}{2R} \] Every current element of a circular turn produces an elementary magnetic field directed along the same axial direction at the centre. Consequently, the contributions reinforce rather than cancel. The contributions from identical turns also add, producing the factor \(N\). \(\textbf{Substitution:}\) \[ B=\frac{(4\pi\times10^{-7})(5)(2)}{2(0.10)} \] \[ B=\frac{40\pi\times10^{-7}}{0.20} \] \[ B=200\pi\times10^{-7}\,\text{T} \] \[ B=2\pi\times10^{-5}\,\text{T} \] \(\textbf{Direction check:}\) The magnetic field at the centre is perpendicular to the plane of the coil. According to the right-hand rule, a clockwise current as viewed from \(+z\) produces a field directed away from that observer. Therefore, the direction is along the negative \(z\)-axis. \(\textbf{Final answer:}\) \[ \boxed{\vec B=-2\pi\times10^{-5}\hat k\,\text{T}} \] Using the single-turn formula without the factor \(N\) would underestimate the field. The sign of the field must be determined independently from the specified viewing direction.

Explanation: \(\textbf{Central-field dependence:}\) The field at the centre of a circular coil is \[ B=\frac{\mu_0 NI}{2R} \] Its magnitude is directly proportional to the number of turns and the current, but inversely proportional to the radius. For the second coil, \[ N'=2N,\qquad I'=3I,\qquad R'=\frac{3R}{2} \] \(\textbf{Comparison of magnitudes:}\) \[ \frac{B'}{B} = \frac{N'I'/R'}{NI/R} \] Substituting the changed parameters, \[ \frac{B'}{B} = \frac{(2N)(3I)}{3R/2}\times\frac{R}{NI} \] \[ \frac{B'}{B} = \frac{6}{3/2}=4 \] Therefore, the second field has four times the original magnitude. \(\textbf{Direction:}\) Reversing the circulation sense reverses the normal direction of the magnetic field. The increases in turn number and current change the magnitude but do not independently reverse the field. Hence, the final field points opposite to the initial field. \(\textbf{Final answer:}\) If the original field vector is \(\vec B\), then \[ \boxed{\vec B'=-4\vec B} \] A common mistake is to multiply the changes in turn number and current while forgetting that the larger radius reduces the central field. Another mistake is to infer the direction from the field magnitude, even though the circulation sense must be considered separately.

Explanation: \(\textbf{Constraint imposed by fixed wire length:}\) The total length of wire used to form \(N\) circular turns is \[ L=N(2\pi R) \] The original radius is therefore \[ R=\frac{L}{2\pi N} \] In the rewound coil, the number of turns becomes \(2N\), but the total wire length remains \(L\). Thus, \[ L=(2N)(2\pi R') \] Comparing the two wire-length relations, \[ N(2\pi R)=2N(2\pi R') \] \[ R'=\frac{R}{2} \] \(\textbf{Effect on the central field:}\) Initially, \[ B=\frac{\mu_0 NI}{2R} \] After rewinding, \[ B'=\frac{\mu_0(2N)I}{2(R/2)} \] Simplifying, \[ B'=\frac{2\mu_0 NI}{R} \] Therefore, \[ \frac{B'}{B} = \frac{2\mu_0 NI/R}{\mu_0 NI/(2R)} =4 \] The field becomes four times as large because the number of turns doubles while the radius simultaneously halves. \(\textbf{Magnetic orientation:}\) The new winding sense is opposite to the original winding sense. By the right-hand rule, this reverses the direction of the central magnetic field. \(\textbf{Final answer:}\) \[ \boxed{\vec B'=-4\vec B} \] The fixed-wire-length condition is essential: doubling the number of turns cannot leave the radius unchanged if the same wire is used. Treating the radius as fixed would predict only a twofold increase and would violate the stated construction constraint.

Explanation: For a circular coil of \(N\) identical turns, the axial magnetic field magnitude at distance \(x\) from the centre is \[ B(x)=\frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}} \] At the centre, \(x=0\), so \[ B_0=\frac{\mu_0 NI}{2R} \] Dividing the axial expression by the central expression gives \[ \frac{B(x)}{B_0} = \frac{R^3}{(R^2+x^2)^{3/2}} \] Statement I is correct because the axial expression depends on \(x^2\). Consequently, equal positive and negative axial displacements give equal field magnitudes. For Statement II, put \(x=R\): \[ \frac{B(R)}{B_0} = \frac{R^3}{(2R^2)^{3/2}} = \frac{1}{2\sqrt2} \] Therefore, Statement II is correct. For Statement III, put \(x=2R\): \[ \frac{B(2R)}{B_0} = \frac{R^3}{(5R^2)^{3/2}} = \frac{1}{5\sqrt5} \] Hence, Statement III is also correct. Statement IV is incorrect because the axial magnetic field of a circular current loop has the same axial direction on both sides of the loop. Although the observation point crosses the plane of the coil, the current circulation sense remains unchanged. The transverse contributions from different current elements cancel by symmetry, leaving the axial contributions reinforcing. The sign of the field is controlled by the current circulation, not by whether \(x\) is positive or negative.

Explanation: \(\textbf{Field produced by the first coil:}\) The magnetic field on the axis of a circular coil is \[ B(x)=\frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}} \] The observation point for the first coil is at \(x=R\). Therefore, \[ B_1=\frac{\mu_0 NIR^2}{2(R^2+R^2)^{3/2}} \] Simplifying the denominator, \[ B_1=\frac{\mu_0 NIR^2}{2(2R^2)^{3/2}} \] \[ B_1=\frac{\mu_0 NI}{4\sqrt2 R} \] \(\textbf{Field required at the centre of the second coil:}\) The second coil has radius \(R_2=2R\), while its turn number remains \(N\). At its centre, \[ B_2=\frac{\mu_0 NI_2}{2R_2} \] Substituting \(R_2=2R\), \[ B_2=\frac{\mu_0 NI_2}{4R} \] \(\textbf{Equal-field condition:}\) The question requires \(B_2=B_1\). Hence, \[ \frac{\mu_0 NI_2}{4R} = \frac{\mu_0 NI}{4\sqrt2 R} \] Cancelling the common factors gives \[ I_2=\frac{I}{\sqrt2} \] \(\textbf{Final answer:}\) \[ \boxed{I_2=\frac{I}{\sqrt2}} \] The first field must be calculated using the axial-field formula because its observation point is away from the centre. The second field must instead be calculated using the centre-field relation. Applying the centre formula to both coils would miss the axial-distance factor and produce an incorrect current. The answer is smaller than \(I\) because the second coil is being compared at its centre with the weaker off-centre field of the first coil.

Explanation: \(\textbf{Field produced by the inner coil:}\) The inner coil has \(2\) turns and radius \(R\). Its central field magnitude is \[ B_1=\frac{\mu_0(2)I}{2R} \] \[ B_1=\frac{\mu_0 I}{R} \] Since its current appears anticlockwise from the positive \(z\)-axis, its field points along \(+z\). \(\textbf{Initial field of the outer coil:}\) Initially, the outer coil has \(3\) turns and radius \(3R\). Thus, \[ B_{2,\mathrm{initial}}=\frac{\mu_0(3)I}{2(3R)} \] \[ B_{2,\mathrm{initial}}=\frac{\mu_0 I}{2R} \] Its clockwise current produces a field along \(-z\). Therefore, the two fields initially oppose, but their magnitudes are unequal. \(\textbf{Effect of reducing the outer radius:}\) The outer radius becomes \[ R'_2=\frac{3R}{2} \] The current and turn number remain unchanged. Its new central field is \[ B'_2=\frac{\mu_0(3)I}{2(3R/2)} \] \[ B'_2=\frac{\mu_0 I}{R} \] The magnitude of the outer coil's field has doubled because its radius has been halved. Its direction remains \(-z\), since its circulation sense has not changed. \(\textbf{Resultant field:}\) The two final contributions are \[ \vec B_1=\frac{\mu_0 I}{R}\hat k \] \[ \vec B'_2=-\frac{\mu_0 I}{R}\hat k \] Therefore, \[ \vec B_{\mathrm{net}}=\vec B_1+\vec B'_2=\vec 0 \] The resultant field is zero because the contributions are now equal in magnitude and opposite in direction. Before reducing the radius, the inner coil produced the stronger field. The geometry change makes the opposing contributions exactly equal without requiring any current reversal.

Explanation: \(\textbf{Initial axial contributions:}\) Both coils have the same radius, turn number and current magnitude. Their centres are symmetrically placed at equal distances \(R\) from the midpoint \(O\). The axial magnetic field of a circular coil is \[ B(x)=\frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}} \] At the midpoint, the distance from either coil centre is \(R\). Therefore, each coil initially contributes the same field magnitude \[ b=\frac{\mu_0 NIR^2}{2(2R^2)^{3/2}} \] \[ b=\frac{\mu_0 NI}{4\sqrt2R} \] Both currents are anticlockwise when viewed from \(+z\), so both axial fields at \(O\) point along \(+z\). The initial resultant is \[ \vec B_{\mathrm{initial}}=b\hat k+b\hat k \] \[ \vec B_{\mathrm{initial}}=2b\hat k \] \(\textbf{Field after the current change:}\) The coil centred at \(z=-R\) remains unchanged and continues to contribute \(b\hat k\). The current in the other coil is reversed, so its field direction becomes \(-z\). Its current magnitude is also tripled, making its field magnitude \(3b\). Thus, \[ \vec B_{\mathrm{final}}=b\hat k-3b\hat k \] \[ \vec B_{\mathrm{final}}=-2b\hat k \] Comparing with the original field, \[ \frac{\vec B_{\mathrm{final}}}{\vec B_{\mathrm{initial}}} = \frac{-2b\hat k}{2b\hat k} =-1 \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_{\mathrm{final}}=-\vec B_{\mathrm{initial}}} \] The magnitude of the resultant remains unchanged, but its direction reverses. The equal initial contributions arise from the symmetry of the two axial positions. The final result follows from vector superposition after correctly accounting for both the reversal and the threefold increase of one current. Simply subtracting the current magnitudes without considering their corresponding field contributions would obscure the physical reason for this result.

Explanation: The central magnetic field of a circular coil is \[ B=\frac{\mu_0 NI}{2R} \] To compare the four coils, take the reference field magnitude as \[ B_{\mathrm{ref}}=\frac{\mu_0 I}{2R} \] The field of each coil is determined by multiplying this reference value by the appropriate factor \(NI/R\), expressed relative to the reference quantities. For coil \(P\), \[ B_P=\frac{\mu_0(2)I}{2R}=2B_{\mathrm{ref}} \] Its direction is \(+z\) because its current appears anticlockwise. For coil \(Q\), \[ B_Q=\frac{\mu_0(3)I}{2(2R)}=\frac32 B_{\mathrm{ref}} \] Its direction is also \(+z\). For coil \(R\), \[ B_R=\frac{\mu_0(4)I}{2(2R)}=2B_{\mathrm{ref}} \] Its direction is \(-z\) because the current appears clockwise. For coil \(S\), \[ B_S=\frac{\mu_0(1)(3I)}{2R}=3B_{\mathrm{ref}} \] Its direction is \(-z\). Therefore, the central-field vectors of coils \(P\) and \(R\) are \[ \vec B_P=2B_{\mathrm{ref}}\hat k,\qquad \vec B_R=-2B_{\mathrm{ref}}\hat k \] These fields have equal magnitudes and opposite directions. The remaining listed pairs either have unequal field magnitudes or the same field direction. A correct comparison requires checking both the magnitude factor \(NI/R\) and the circulation sense. Equal values of current alone do not guarantee equal central fields, because radius and number of turns also matter.

Explanation: \(\textbf{Given measurements:}\) The number of turns and central magnetic field are \[ N=60,\qquad B_0=2\pi\times10^{-5}\,\text{T} \] The axial observation distance is \[ x=12\,\text{cm}=0.12\,\text{m} \] The measured field ratio is \[ \frac{B(x)}{B_0}=\frac{1}{2\sqrt2} \] The radius of the coil is not directly given, so it must first be inferred from the axial measurement. \(\textbf{Stage 1 — Determine the coil radius:}\) The axial magnetic field of a circular coil is \[ B(x)=\frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}} \] At the centre, \[ B_0=\frac{\mu_0 NI}{2R} \] Taking the ratio eliminates the unknown current and number of turns: \[ \frac{B(x)}{B_0} = \frac{R^3}{(R^2+x^2)^{3/2}} \] Using the measured ratio, \[ \frac{R^3}{(R^2+x^2)^{3/2}} = \frac{1}{2\sqrt2} \] Since \(2\sqrt2=2^{3/2}\), raising both sides to the power \(2/3\) gives \[ \frac{R^2}{R^2+x^2}=\frac12 \] Rearranging, \[ 2R^2=R^2+x^2 \] \[ R^2=x^2 \] As radius is positive, \[ R=x=0.12\,\text{m} \] \(\textbf{Stage 2 — Calculate the current:}\) Now use the measured central magnetic field: \[ B_0=\frac{\mu_0 NI}{2R} \] Solving for the current, \[ I=\frac{2RB_0}{\mu_0 N} \] Substituting the known values, \[ I=\frac{2(0.12)(2\pi\times10^{-5})}{(4\pi\times10^{-7})(60)} \] Simplifying the numerator, \[ 2(0.12)(2\pi\times10^{-5}) = 4.8\pi\times10^{-6} \] Simplifying the denominator, \[ (4\pi\times10^{-7})(60) = 2.4\pi\times10^{-5} \] Therefore, \[ I=\frac{4.8\pi\times10^{-6}}{2.4\pi\times10^{-5}} \] \[ I=2\times10^{-1}\,\text{A} \] \(\textbf{Physical check:}\) At an axial distance equal to the coil radius, the magnetic field is indeed \(B_0/(2\sqrt2)\). The measured field is weaker away from the centre, consistent with the axial-field relation. The calculated current also has the correct SI unit because the central-field formula relates magnetic field, permeability, turn number and coil radius to current. \(\textbf{Final answer:}\) \[ \boxed{I=0.20\,\text{A}} \] The essential reasoning is to use the field ratio to determine the unknown geometry before using the absolute field measurement to determine the current. Attempting to calculate current immediately from the central magnetic field would leave the unknown radius unresolved.

Explanation: The Biot–Savart law describes the magnetic field contributed by an individual current element. For an element at perpendicular distance \(r\), \[ dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{r^2} \] In a circular loop of radius \(R\), every current element is at distance \(R\) from the centre. Moreover, the current element is perpendicular to the radius joining it to the centre, so the angular factor is unity. Every elementary magnetic-field contribution at the centre points along the same normal to the loop. The total field is therefore obtained by integrating the elementary magnitudes over the complete circumference: \[ B=\frac{\mu_0 I}{4\pi R^2}\int dl \] The circumference of the circular loop is \[ \int dl=2\pi R \] Consequently, \[ B=\frac{\mu_0 I}{2R} \] The elementary field contains an inverse-square distance factor, but the length of the complete current path increases in direct proportion to \(R\). These two dependencies combine to give the inverse-radius dependence of the total central field. Therefore, doubling the radius at fixed current reduces the central magnetic field to half. Both the Assertion and Reason are correct, and the Reason explains precisely why the total field has a different radius dependence from an elementary contribution. The essential distinction is between the contribution of a fixed infinitesimal element and the integrated field of an entire circular conductor.

Explanation: \(\textbf{Current element and observation geometry:}\) The current element is directed along the positive \(z\)-axis. Therefore, \[ d\vec l=dl\,\hat k \] The displacement vector from the element to the observation point is \[ \vec r=3a\hat i+4a\hat j \] Its magnitude is \[ r=\sqrt{(3a)^2+(4a)^2} \] \[ r=5a \] Hence, the displacement unit vector is \[ \hat r=\frac35\hat i+\frac45\hat j \] \(\textbf{Apply the vector Biot–Savart law:}\) \[ d\vec B=\frac{\mu_0 I}{4\pi r^2}(d\vec l\times\hat r) \] Substituting the vectors, \[ d\vec B= \frac{\mu_0 I\,dl}{4\pi(25a^2)} \left[\hat k\times\left(\frac35\hat i+\frac45\hat j\right)\right] \] Using the cross-product relations, \[ \hat k\times\hat i=\hat j \] \[ \hat k\times\hat j=-\hat i \] The cross product becomes \[ \hat k\times\hat r = \frac35\hat j-\frac45\hat i \] Therefore, \[ d\vec B= \frac{\mu_0 I\,dl}{4\pi(25a^2)} \left(-\frac45\hat i+\frac35\hat j\right) \] \[ d\vec B= \frac{\mu_0 I\,dl}{4\pi(125a^2)} (-4\hat i+3\hat j) \] \(\textbf{Magnitude and direction check:}\) The displacement vector lies in the \(xy\)-plane, while the current element lies along \(z\). Thus, the included angle is \(90^\circ\), and the elementary field magnitude must be \[ dB=\frac{\mu_0 I\,dl}{4\pi(25a^2)} \] The magnitude of the vector coefficient \(-4\hat i+3\hat j\) is \(5\). Consequently, the derived vector has precisely the required magnitude. Its direction is perpendicular to both the positive \(z\)-axis and the displacement vector toward \(P\). \(\textbf{Final answer:}\) \[ \boxed{d\vec B=\frac{\mu_0 I\,dl}{500\pi a^2}(-4\hat i+3\hat j)} \] The factor \(125a^2\) in the denominator of the equivalent expression arises from both the inverse-square distance dependence and the components of the displacement unit vector.

Explanation: \(\textbf{Initial observation point:}\) The current element is directed along \(+\hat i\). The displacement from the element to \(P\) is along \(+\hat j\). The included angle is \(90^\circ\). Therefore, \[ B_P=\frac{\mu_0 I\,dl}{4\pi b^2} \] The direction follows \[ \hat i\times\hat j=\hat k \] Hence, the initial field is along positive \(z\). \(\textbf{Geometry at the new observation point:}\) The displacement vector toward \(Q\) is \[ \vec r_Q=\sqrt3b\,\hat i+b\,\hat j \] The new distance is \[ r_Q=\sqrt{3b^2+b^2}=2b \] Let \(\theta\) be the angle between the positive \(x\)-axis and the displacement toward \(Q\). From the geometry, \[ \sin\theta=\frac{b}{2b}=\frac12 \] \(\textbf{New elementary magnetic field:}\) The Biot–Savart law gives \[ B_Q=\frac{\mu_0 I\,dl\sin\theta}{4\pi r_Q^2} \] Substituting, \[ B_Q=\frac{\mu_0 I\,dl(1/2)}{4\pi(2b)^2} \] \[ B_Q=\frac{\mu_0 I\,dl}{4\pi b^2}\times\frac18 \] Thus, \[ B_Q=\frac{B_P}{8} \] \(\textbf{Direction check:}\) The displacement toward \(Q\) has a positive \(y\)-component. Only this transverse component contributes to the cross product with the current element along \(x\). Therefore, the elementary field remains directed along \(+z\). The field decreases for two connected reasons: the observation distance doubles, and the included angle changes from \(90^\circ\) to \(30^\circ\). The inverse-square distance factor contributes \(1/4\), while the angular factor contributes another \(1/2\). \(\textbf{Final answer:}\) \[ \boxed{\vec B_Q=\frac{B_P}{8}\hat k} \] The angle and distance cannot be varied independently in this arrangement because both are determined by the coordinates of the observation point.

Explanation: \(\textbf{Separate the source field from the background:}\) The magnetic-field probe measures the vector sum of the background field and the contribution from the current element. Let the signed background field be \(B_{\mathrm{bg}}\). Let the initial elementary field due to current \(+I\) be \(B_e\), directed along \(+z\). The first measurement gives \[ B_{\mathrm{bg}}+B_e=7.0\,\mu\text{T} \] When the element current is reversed, its elementary field reverses direction without changing magnitude. Therefore, the second measurement gives \[ B_{\mathrm{bg}}-B_e=1.0\,\mu\text{T} \] \(\textbf{Determine the background contribution:}\) Adding the equations, \[ 2B_{\mathrm{bg}}=8.0\,\mu\text{T} \] \[ B_{\mathrm{bg}}=4.0\,\mu\text{T} \] Subtracting the second equation from the first, \[ 2B_e=6.0\,\mu\text{T} \] \[ B_e=3.0\,\mu\text{T} \] Thus, the original current element contributes \(+3.0\,\mu\text{T}\) at distance \(r\). \(\textbf{Effect of changing the probe distance:}\) For a fixed current element and fixed included angle, \[ dB\propto\frac1{r^2} \] At distance \(2r\), the elementary field becomes \[ B'_e=B_e\left(\frac{r}{2r}\right)^2 \] \[ B'_e=3.0\times\frac14 \] \[ B'_e=0.75\,\mu\text{T} \] Its direction is still positive \(z\), because the current was restored to its initial direction. \(\textbf{Calculate the new probe reading:}\) The background contribution remains \(+4.0\,\mu\text{T}\). Therefore, \[ B_{\mathrm{measured}}=B_{\mathrm{bg}}+B'_e \] \[ B_{\mathrm{measured}}=4.0+0.75 \] \(\textbf{Final answer:}\) \[ \boxed{B_{\mathrm{measured}}=+4.75\,\mu\text{T}} \] The two initial measurements are essential because they allow the constant background field to be separated from the current-dependent field. Applying the inverse-square law directly to the total \(7.0\,\mu\text{T}\) reading would incorrectly assume that the background field also changes with probe distance. Current reversal provides the information needed to distinguish the two contributions.