Magnetic Effects of Current and Magnetism MCQs with Answers – Part 7 (NEET Physics)

Explanation: \(\textbf{Determine the axial directions of the two source fields:}\) The axial field of a short magnetic dipole points along its magnetic moment on both sides of the source. Therefore, P produces axial field along positive \(x\). Q produces axial field along negative \(x\). Their contributions oppose everywhere on the common axis except at the excluded source centres. Cancellation occurs where their magnitudes are equal. \(\textbf{Establish the field-magnitude relation:}\) For a short magnetic dipole, \[ B_a=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] At an arbitrary axial position \(x\), the source distances are \[ r_P=|x| \] and \[ r_Q=|x-3a| \] Equal field magnitudes require \[ \frac{m}{|x|^3}=\frac{8m}{|x-3a|^3} \] Cancelling \(m\) and taking positive cube roots, \[ |x-3a|=2|x| \] \(\textbf{Check the interval between the magnets:}\) For \[ 0\lt x\lt 3a \] the distances are \[ |x|=x,\qquad |x-3a|=3a-x \] Thus, \[ 3a-x=2x \] \[ 3a=3x \] \[ x=a \] This position belongs to the required interval. \(\textbf{Check the region to the left of P:}\) For \[ x\lt 0 \] the distances are \[ |x|=-x,\qquad |x-3a|=3a-x \] Therefore, \[ 3a-x=2(-x) \] \[ 3a=-x \] \[ x=-3a \] This point also belongs to its assumed region. \(\textbf{Check the region to the right of Q:}\) For \[ x\gt 3a \] the condition becomes \[ x-3a=2x \] \[ x=-3a \] This contradicts the assumed region, so there is no cancellation point to the right of Q. \(\textbf{Final answer:}\) \[ \boxed{x=a\qquad\text{and}\qquad x=-3a} \] The magnetic moments are oppositely directed, so their axial fields oppose on both sides as well as between the magnets. Because their moment magnitudes are unequal, cancellation can occur at different distances from the two sources. The absolute values in the inverse-cube field relation generate different equations in different spatial intervals. Checking the physical interval of each solution is essential. The two cancellation positions follow within the specified point-dipole approximation.

Explanation: The magnetic torque magnitude on a dipole in a uniform field is \[ \tau=mB\sin\theta \] At the initial angle, \[ \tau=(0.30)(0.40)\sin30^\circ \] \[ \tau=(0.12)\left(\frac12\right) \] \[ \tau=0.060\,\text{N m} \] Statement I is correct. The orientation-dependent potential energy is \[ U=-mB\cos\theta \] At \(30^\circ\), the energy is negative. At \(180^\circ\), the energy is positive and reaches its maximum possible value for the given moment and field. Magnetic torque acts towards lower potential energy. The slow externally imposed rotation from \(30^\circ\) towards \(180^\circ\) therefore requires positive work by the external agent, while magnetic torque does negative work. The fact that torque vanishes at the endpoint does not determine the sign of work over the entire motion. Statement II is incorrect. At exactly \(180^\circ\), the magnetic moment and field are antiparallel. Their cross product vanishes, so this is an equilibrium orientation. However, a small disturbance causes the magnetic torque to turn the dipole farther away from the antiparallel orientation and towards lower energy. Statement III is correct. At the parallel orientation, the torque also vanishes, but the energy is \[ U_{\parallel}=-mB \] This is a negative minimum, not zero. Statement IV is incorrect. Zero torque can occur at either the maximum or minimum of magnetic potential energy. Stability distinguishes these two equilibrium orientations. The correct interpretation must consider torque direction and potential-energy variation, not only the torque magnitude at an endpoint.

Explanation: \(\textbf{Use the axial zero-field observation:}\) At positive axial position P, the dipole field points along its moment, positive \(z\). The separate external field points along negative \(z\). Their resultant vanishes: \[ \vec B_d(P)+\vec B_e=\vec0 \] Therefore, \[ \vec B_d(P)=+B_0\hat k \] \(\textbf{Relate axial and equatorial field magnitudes:}\) The short-dipole axial field magnitude is \[ B_a(r)=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] The equatorial field magnitude is \[ B_q(s)=\frac{\mu_0}{4\pi}\frac{m}{s^3} \] Here, the equatorial point is at \[ s=\frac r2 \] Hence, \[ \frac{B_q(r/2)}{B_a(r)} =\frac{m/(r/2)^3}{2m/r^3} \] \[ \frac{B_q(r/2)}{B_a(r)}=4 \] The dipole field at Q therefore has magnitude \(4B_0\), not \(8B_0\). \(\textbf{Determine the original resultant at Q:}\) The equatorial field points opposite to the positive-\(z\) moment: \[ \vec B_d(Q)=-4B_0\hat k \] The external field is \[ \vec B_e=-B_0\hat k \] Thus, \[ \vec B_Q=(-4B_0-B_0)\hat k \] \[ \vec B_Q=-5B_0\hat k \] \(\textbf{Reverse the magnetic dipole:}\) After reversal, the dipole moment points along negative \(z\). Its equatorial field at the fixed point Q reverses: \[ \vec B'_d(Q)=+4B_0\hat k \] The external field does not change. Therefore, \[ \vec B'_Q=(4B_0-B_0)\hat k \] \[ \vec B'_Q=+3B_0\hat k \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_Q=-5B_0\hat k,\qquad \vec B'_Q=+3B_0\hat k} \] The axial zero-field observation fixes the dipole's field magnitude at distance \(r\). Moving to the equatorial plane introduces a factor of one-half in the dipole-field formula. Halving the observation distance simultaneously introduces an inverse-cube factor of eight. Together, these effects make the equatorial dipole field four times the original axial magnitude. The independently maintained external field must then be included with its unchanged sign. Reversing the magnet reverses only the dipole contribution.

Explanation: \(\textbf{Recover the magnetic moment from the measured torque:}\) The torque in a uniform field is \[ \vec\tau=\vec m\times\vec B_e \] Initially, \[ \vec m=m\hat k \] Therefore, \[ \vec\tau=m\hat k\times(0.20\hat i+0.20\sqrt3\hat k) \] The field component parallel to \(\hat k\) gives zero torque. Using \(\hat k\times\hat i=\hat j\), \[ \vec\tau=0.20m\hat j \] The measured torque is \[ 0.20m=0.60 \] \[ m=3.0\,\text{A m}^2 \] \(\textbf{Determine the initial probe geometry:}\) P lies along positive \(x\), perpendicular to the original magnetic moment along positive \(z\). It is therefore an equatorial point of the short dipole. The equatorial field has magnitude \[ B_{\mathrm{eq}}=\frac{\mu_0}{4\pi}\frac{m}{r^3} \] At \(r=0.10\,\text{m}\), \[ B_{\mathrm{eq}}=10^{-7}\frac{3.0}{(0.10)^3} \] \[ B_{\mathrm{eq}}=3.0\times10^{-4}\,\text{T} \] The equatorial field points opposite to the moment: \[ \vec B_P=-3.0\times10^{-4}\hat k\,\text{T} \] \(\textbf{Analyse the rotation of the magnet:}\) The magnet is rotated so that its moment becomes \[ \vec m'=3.0\hat i\,\text{A m}^2 \] The fixed probe P now lies on the positive axial line of this new dipole orientation. The short-dipole axial field is \[ B_{\mathrm{ax}}=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] Substituting, \[ B'_{\mathrm{ax}}=10^{-7}\frac{2(3.0)}{(0.10)^3} \] \[ B'_{\mathrm{ax}}=6.0\times10^{-4}\,\text{T} \] The axial field points along the new moment: \[ \vec B'_P=+6.0\times10^{-4}\hat i\,\text{T} \] \(\textbf{Final answer:}\) \[ \boxed{m=3.0\,\text{A m}^2,\qquad \vec B_P=-3.0\times10^{-4}\hat k\,\text{T},\qquad \vec B'_P=+6.0\times10^{-4}\hat i\,\text{T}} \] The uniform-field torque first determines the otherwise unknown magnetic moment. Removing that field allows the probe to measure the bar magnet's own field without an external contribution. Rotating the magnet changes the probe from an equatorial to an axial observation position. The field consequently changes both direction and magnitude. At equal distance, the axial dipole field is twice the equatorial field magnitude.

Explanation: The torque magnitude of a magnetic dipole in a uniform field is \[ \tau=mB\sin\theta \] For a small displacement \(\delta\) from parallel alignment, P makes an angle approximately \(\delta\) with the field. Its torque magnitude is approximately \(mB\delta\). For the same displacement from antiparallel alignment, Q makes an angle approximately \(180^\circ-\delta\). Because \[ \sin(180^\circ-\delta)=\sin\delta \] Q's torque magnitude is approximately the same as P's. Statement I is correct. However, the directions of the torques relative to the imposed displacements differ. The torque on P tends to decrease its small angle and return it towards parallel alignment. Its small displacement therefore produces a restoring torque. The torque on Q instead tends to increase the departure from its antiparallel position. Q is consequently in an unstable equilibrium configuration. Statement II is correct. Small stable harmonic oscillations require a restoring torque approximately proportional to the negative angular displacement. That condition holds for P but not for Q near its original orientation. Equal torque magnitudes do not establish equal stability. Statement III is incorrect. Magnetic potential energy is \[ U=-mB\cos\theta \] At exactly parallel alignment, this energy is at its minimum, \(-mB\). At exactly antiparallel alignment, it is at its maximum, \(+mB\). The torque vanishes at both exact orientations because the cross product of moment and field is zero. Statement IV is correct. The comparison shows why equilibrium and stability are different physical properties. Potential-energy variation and torque direction determine the response to a small disturbance.

Explanation: \(\textbf{Calculate the gravitational torque about the pivot:}\) The small mass is attached at \[ \vec r=0.10\hat i\,\text{m} \] Gravity acts along negative \(y\): \[ \vec F_g=-mg\hat j \] The force magnitude is \[ mg=(0.020)(10) \] \[ mg=0.20\,\text{N} \] Therefore, \[ \vec\tau_g=\vec r\times\vec F_g \] \[ \vec\tau_g=(0.10\hat i)\times(-0.20\hat j) \] \[ \vec\tau_g=-0.020\hat k\,\text{N m} \] The negative-\(z\) direction represents clockwise torque when viewed from positive \(z\). \(\textbf{Determine the balancing magnetic torque:}\) Initially, the magnet has moment \[ \vec m=+0.40\hat i\,\text{A m}^2 \] The uniform magnetic field is \[ \vec B=B\hat j \] Its magnetic torque is \[ \vec\tau_m=\vec m\times\vec B \] \[ \vec\tau_m=+0.40B\hat k \] For equilibrium, \[ \vec\tau_m+\vec\tau_g=\vec0 \] Thus, \[ 0.40B=0.020 \] \[ B=0.050\,\text{T} \] \(\textbf{Reverse the magnetic moment:}\) The new moment is \[ \vec m'=-0.40\hat i\,\text{A m}^2 \] The external field and arm orientation remain unchanged. Therefore, \[ \vec\tau'_m=\vec m'\times\vec B \] \[ \vec\tau'_m=-0.40(0.050)\hat k \] \[ \vec\tau'_m=-0.020\hat k\,\text{N m} \] The gravitational torque also remains negative: \[ \vec\tau_g=-0.020\hat k\,\text{N m} \] Their resultant is \[ \vec\tau_{\mathrm{net}}=-0.040\hat k\,\text{N m} \] \(\textbf{Final answer:}\) \[ \boxed{B=0.050\,\text{T},\qquad \vec\tau_{\mathrm{net,after}}=-0.040\hat k\,\text{N m}} \] Initially, magnetic torque opposes the torque produced by the off-centre weight. The balance condition determines the applied magnetic-field magnitude. Reversing the magnet changes the magnetic torque direction but does not change the gravitational torque. The two torques then reinforce instead of cancelling. The resultant clockwise torque is twice the magnitude of either individual torque.

Explanation: A permanent bar magnet has a magnetic moment associated with its existing magnetisation. Its moment direction is fixed relative to the magnet's body unless its magnetisation or physical orientation changes. Suppose that moment initially points parallel to the applied uniform field. The parallel orientation is stable because small angular disturbances generate restoring magnetic torque. If the external field reverses while the bar magnet has not yet rotated, its moment becomes antiparallel to the new field. The exact antiparallel configuration has zero magnetic torque but is unstable under a small angular displacement. An ordinary isotropic weak paramagnet responds differently. Its bulk magnetisation is induced by the applied field and is approximately proportional to it: \[ \vec M=\chi\vec H \] For paramagnetism, the susceptibility is small and positive. Reversing the applied field therefore reverses the induced magnetisation in the laboratory frame. The weak paramagnet does not retain a fixed permanent dipole moment pointing in the old direction under the conditions described. Its induced moment remains aligned with the new applied field. Treating its induced moment as though it were a mechanically fixed permanent moment leads to an incorrect stability conclusion. The two systems may initially have similarly directed moments, but their behaviour during field reversal is governed by different physical mechanisms. The distinction is between a pre-existing permanent magnetic moment and a reversible field-induced moment.

Explanation: The apparent clockwise or anticlockwise sense of a current depends on the side from which its loop is viewed. An observer on the opposite side looks along the opposite viewing direction. Therefore, the same physical circulation that appears anticlockwise from positive \(z\) appears clockwise from negative \(z\). Statement I is correct. The magnetic moment of a planar current loop is determined by the right-hand grip rule. Curling the fingers of the right hand in the conventional-current direction makes the thumb point along the loop's magnetic moment. For the initially anticlockwise current viewed from positive \(z\), the thumb points along positive \(z\). An equivalent bar magnet must consequently have its north end towards positive \(z\). Statement II is correct. The direction of the loop's moment depends on its current direction as well as on the orientation of its plane. Reversing the current while keeping the loop fixed reverses the moment vector. Statement III is incorrect. At sufficiently distant axial observation points, the loop behaves as a magnetic dipole. Its axial magnetic field is directed along its magnetic moment on either side of the loop. When the current reverses, the dipole moment and the axial magnetic-field vectors reverse. Statement IV is correct. Changing the observer's side does not physically reverse a current. Changing the current itself does reverse the magnetic moment. These are different operations even though both can change a verbal description involving clockwise and anticlockwise directions.

Explanation: \(\textbf{Simplify the point-dipole prediction:}\) The magnetic moment of the coil is \[ m=NI\pi R^2 \] Its distant axial dipole field is \[ B_{\mathrm{dipole}}=\frac{\mu_0}{4\pi}\frac{2m}{x^3} \] Substituting the magnetic moment, \[ B_{\mathrm{dipole}}=\frac{\mu_0NIR^2}{2x^3} \] \(\textbf{Form a comparison independent of current and turn number:}\) Divide the approximate field by the exact circular-coil field: \[ \frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}} =\frac{\mu_0NIR^2/(2x^3)} {\mu_0NIR^2/[2(R^2+x^2)^{3/2}]} \] The common factors cancel: \[ \frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}} =\frac{(R^2+x^2)^{3/2}}{x^3} \] \(\textbf{Apply the specified observation distance:}\) At \[ x=2R \] the ratio becomes \[ \frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}} =\frac{(R^2+4R^2)^{3/2}}{(2R)^3} \] \[ \frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}} =\frac{(5R^2)^{3/2}}{8R^3} \] \[ \frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}} =\frac{5\sqrt5}{8} \] \[ \frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}}\approx1.398 \] \(\textbf{Determine the percentage difference:}\) The approximation predicts a larger field. The percentage excess relative to the exact value is \[ \delta=\left(\frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}}-1\right)\times100 \] \[ \delta=\left(\frac{5\sqrt5}{8}-1\right)\times100 \] \[ \delta\approx39.8\% \] \(\textbf{Final answer:}\) \[ \boxed{\frac{B_{\mathrm{dipole}}}{B_{\mathrm{exact}}} =\frac{5\sqrt5}{8}\approx1.398,\qquad \text{percentage overestimate}\approx39.8\%} \] The point-dipole field is the far-field approximation to the exact circular-loop field. It assumes the observation distance is sufficiently large compared with the coil radius. At only twice the radius, the neglected finite-size contribution is not small. The approximate expression consequently overestimates the exact axial field in this situation. As the observation distance becomes much greater than the radius, the ratio approaches unity. The two formulas are therefore compatible in their common far-field limit, not interchangeable at every distance.

Explanation: \(\textbf{Identify the original magnetic-moment relation:}\) For a uniformly magnetised bar, \[ m=MV \] where \(M\) is the magnetisation magnitude and \(V\) the volume. The original volume is \[ V=AL \] Thus, \[ m=MAL \] In the elementary pole representation, \[ m=pL \] so the effective pole strength is proportional to cross-sectional area when magnetisation is unchanged. \(\textbf{Analyse the transverse cut in P:}\) Each transverse half has length \[ L_P=\frac L2 \] Its cross-sectional area remains \(A\). Consequently, the effective pole strength remains \[ p_P=p \] The magnetic moment is \[ m_P=p_P L_P \] \[ m_P=p\frac L2 \] \[ m_P=\frac m2 \] \(\textbf{Analyse the longitudinal cut in Q:}\) Each longitudinal half retains length \(L\), but its area becomes \(A/2\). Its effective pole strength therefore becomes \[ p_Q=\frac p2 \] The magnetic moment is \[ m_Q=p_QL \] \[ m_Q=\frac p2L \] \[ m_Q=\frac m2 \] \(\textbf{Compare distant axial fields:}\) At an observation distance much greater than the dimensions of either piece, the axial field is \[ B_a=\frac{\mu_0}{4\pi}\frac{2m_{\mathrm{piece}}}{r^3} \] The two pieces have equal magnetic moments: \[ m_P=m_Q=\frac m2 \] Therefore, at the same sufficiently large distance, \[ \frac{B_{a,P}}{B_{a,Q}}=\frac{m_P}{m_Q} \] \[ \frac{B_{a,P}}{B_{a,Q}}=1 \] \(\textbf{Final answer:}\) \[ \boxed{m_P=m_Q=\frac m2,\qquad p_P=p,\qquad p_Q=\frac p2,\qquad B_{a,P}=B_{a,Q}\ \text{at equal distant axial positions}} \] Both cutting operations halve the volume and hence halve the magnetic moment under the specified unchanged magnetisation. They affect length and cross-sectional area differently, so the effective pole strengths differ. Neither cut produces an isolated north or south pole. Each piece remains a complete smaller magnet. The distant magnetic field depends primarily on the magnetic dipole moment rather than on the separate pole-strength representation. The qualification about distant observation is needed because differently shaped pieces can have different near-field distributions.

Explanation: Magnetic field lines provide a visual representation of the direction and relative strength of a magnetic field. Their tangent at a point gives the direction of the resultant magnetic field at that point, provided the field is nonzero. Outside a bar magnet, magnetic field lines are conventionally drawn from the north end towards the south end. Inside the magnet, they continue from south towards north, completing closed curves. Drawing P is therefore consistent with the continuity of magnetic field lines. Two distinct magnetic field lines cannot cross at a point having a definite nonzero magnetic field. At an intersection with different tangents, the field would be assigned two different directions at one point. That contradicts the uniqueness of the local resultant field vector. Drawing Q is consequently impossible as described. A family of approximately straight, parallel, equally spaced field lines with the same arrow direction is a standard representation of an approximately uniform magnetic field. Drawing R is physically meaningful. Field-line spacing represents relative field strength within an appropriately constructed diagram. It is not a rule that magnetic-field magnitude must remain constant as one follows an individual curved line. Field magnitude can vary along a magnetic field line. The student has confused a line's directional role with a claim about constant field magnitude. The three descriptions must be evaluated using direction, continuity, uniqueness and spatial variation together.

Explanation: \(\textbf{Calculate the solenoid's initial dipole moment:}\) The magnetic moment of a solenoid is \[ m_s=NIA \] Substituting, \[ m_s=(200)(1.0)(1.0\times10^{-3}) \] \[ m_s=0.20\,\text{A m}^2 \] Its initial direction is negative \(x\): \[ \vec m_s=-0.20\hat i\,\text{A m}^2 \] \(\textbf{Determine the original resultant moment:}\) The bar magnet has \[ \vec m_b=+0.20\hat i\,\text{A m}^2 \] Thus, \[ \vec m_{\mathrm{net},1}=\vec m_s+\vec m_b \] \[ \vec m_{\mathrm{net},1}=\vec0 \] In the stipulated distant-dipole approximation, the combined dipole field at P consequently vanishes. The uniform-field torque on the rigid assembly is also zero because its resultant moment is zero. \(\textbf{Reverse the solenoid current:}\) Current reversal changes the solenoid moment to \[ \vec m'_s=+0.20\hat i\,\text{A m}^2 \] The bar magnet remains unchanged. Therefore, \[ \vec m_{\mathrm{net},2}=+0.20\hat i+0.20\hat i \] \[ \vec m_{\mathrm{net},2}=+0.40\hat i\,\text{A m}^2 \] \(\textbf{Calculate the distant equatorial field:}\) At an equatorial position, the short-dipole field points opposite to the moment: \[ \vec B_{\mathrm{eq}}=-\frac{\mu_0}{4\pi}\frac{\vec m_{\mathrm{net},2}}{r^3} \] At \[ r=0.20\,\text{m} \] we obtain \[ \vec B_P=-10^{-7}\frac{0.40}{(0.20)^3}\hat i \] \[ \vec B_P=-10^{-7}\frac{0.40}{0.008}\hat i \] \[ \vec B_P=-5.0\times10^{-6}\hat i\,\text{T} \] \(\textbf{Determine the torque in the independently applied field:}\) The external uniform field is \[ \vec B_e=0.10\hat j\,\text{T} \] The torque on the combined moment is \[ \vec\tau=\vec m_{\mathrm{net},2}\times\vec B_e \] \[ \vec\tau=(0.40\hat i)\times(0.10\hat j) \] \[ \vec\tau=+0.040\hat k\,\text{N m} \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_{P,1}=\vec0,\quad\vec\tau_1=\vec0,\quad \vec B_{P,2}=-5.0\times10^{-6}\hat i\,\text{T},\quad \vec\tau_2=+0.040\hat k\,\text{N m}} \] A finite current-carrying solenoid and a bar magnet can be equivalent magnetic dipoles at sufficiently distant observation points. Their moments initially cancel even though each source remains magnetically active. Reversing only the solenoid current changes cancellation into reinforcement. The resultant moment then determines both the distant dipole field and the torque in an independent uniform field. A vanishing net dipole moment does not establish that the local magnetic field is zero at all points near the separate sources.

Explanation: \(\textbf{Determine the field at the midpoint:}\) The midpoint lies on the common axis of both short dipoles. The axial field of a short dipole points along its magnetic moment on either axial side. The left moment points along positive \(x\), giving \[ \vec B_{L,O}=+b\hat i \] The right moment points along negative \(x\), giving \[ \vec B_{R,O}=-b\hat i \] The moments have equal magnitudes, and O is equally distant from both magnets. Therefore, \[ \vec B_O=\vec B_{L,O}+\vec B_{R,O} \] \[ \vec B_O=\vec0 \] \(\textbf{Determine the two distances to P:}\) The point P is at \(x=-2a\). Its distance from the left magnet at \(-a\) is \[ r_L=a \] Its distance from the right magnet at \(+a\) is \[ r_R=3a \] \(\textbf{Calculate the initial contributions at P:}\) The left magnet produces axial field along positive \(x\): \[ \vec B_{L,P}=+b\hat i \] The right magnet produces axial field along negative \(x\). Its distance is three times greater: \[ \vec B_{R,P}=-\frac{b}{3^3}\hat i \] \[ \vec B_{R,P}=-\frac b{27}\hat i \] The initial resultant is \[ \vec B_P=\left(1-\frac1{27}\right)b\hat i \] \[ \vec B_P=+\frac{26}{27}b\hat i \] \(\textbf{Reverse the right magnet:}\) The right dipole moment now points along positive \(x\). Its axial field at P reverses direction but retains the same magnitude: \[ \vec B'_{R,P}=+\frac b{27}\hat i \] The left source field remains unchanged. Therefore, \[ \vec B'_P=\left(1+\frac1{27}\right)b\hat i \] \[ \vec B'_P=+\frac{28}{27}b\hat i \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_O=\vec0,\qquad \vec B_P=+\frac{26}{27}b\hat i,\qquad \vec B'_P=+\frac{28}{27}b\hat i} \] Two facing north ends do not produce additive fields at their exact midpoint in this symmetric arrangement. The opposite magnetic moments give equal and opposite axial contributions there. At P, the observation distances differ, so the inverse-cube dependence makes the contributions unequal. Reversing one magnet changes the weaker contribution from opposing to reinforcing. The field at P therefore increases without changing direction. The entire calculation uses the stipulated distant-dipole approximation for each magnet.

Explanation: A magnetic field-line drawing must be consistent with both the direction and continuity of the magnetic field. For a bar magnet, external field lines are conventionally directed from north to south. Inside the magnet, their continuations run from south to north. Sketch P therefore describes a continuous magnetic field-line pattern. Sketch R represents a region of approximately uniform magnetic field because its lines are parallel, equally spaced and similarly directed. Sketch Q is inconsistent with the usual field-line representation. At a point where the magnetic field is nonzero, the resultant field has one definite direction. If two distinct field lines crossed with different tangents, they would assign different field directions to the same point. Sketch S is also inconsistent with magnetic field-line continuity. Ordinary magnetic field lines do not terminate permanently at an isolated north or south magnetic charge. A line emerging near a north end must continue through the surrounding field and eventually form part of a closed magnetic path. The Assertion is therefore true. The Reason states the physical requirements responsible for accepting P and R and rejecting Q and S. It correctly explains the Assertion. Field lines are representations of the magnetic field, not physical threads. Their construction must respect the underlying field rather than merely produce a visually plausible picture.

Explanation: The magnetic polarity of a solenoid follows from the conventional-current direction viewed from one of its ends. Anticlockwise current as viewed from an end makes that end behave as a north pole. Observer P therefore identifies the right end as north. The left end acts as south, so observer Q sees clockwise current. Statement I is correct. Inside a long solenoid, the magnetic field is directed along its magnetic moment, from the effective south end towards the north end. The original interior field therefore points along positive \(x\). Outside the solenoid, the field lines complete their paths from north towards south. At a point on the axis to the left of the solenoid, the external axial dipole field points along the magnetic moment, also positive \(x\). The distinction is between the direction along an axial field line at a particular point and the direction of the complete external field-line path. Statement II is correct. Reversing current reverses the solenoid's effective north and south ends. It also reverses the magnetic moment and magnetic field of the solenoid. The separate bar magnet does not necessarily reverse its moment when the solenoid current changes. It would need to rotate physically or undergo a change in its own magnetisation. Statement III is correct. A current-carrying solenoid can resemble a bar magnet in its distant magnetic-dipole field. That analogy does not imply identical magnetic-field distributions at every interior point. A solenoid contains electric currents arranged in turns, while a permanent bar magnet is magnetised material. The exact internal fields depend on the source structures and material properties. Statement IV is incorrect. The magnetic dipole analogy must be used with its geometrical and far-field qualifications. It does not convert two different magnetic sources into physically identical systems.

Explanation: \(\textbf{Express the moment-matching condition:}\) The magnetic dipole moment of a circular coil is \[ m=NI\pi R^2 \] Both coil moments point in the same direction. Equal moment magnitudes therefore require \[ N_PI_PR_P^2=N_QI_QR_Q^2 \] The common factor \(\pi\) cancels. \(\textbf{Determine the unknown turn number:}\) For P, \[ N_P=80,\qquad I_P=1.0\,\text{A},\qquad R_P=0.050\,\text{m} \] For Q, \[ I_Q=2.0\,\text{A},\qquad R_Q=0.10\,\text{m} \] Thus, \[ 80(1.0)(0.050)^2=N_Q(2.0)(0.10)^2 \] \[ 80(0.0025)=N_Q(0.020) \] \[ 0.20=0.020N_Q \] \[ N_Q=10 \] \(\textbf{Compare the central fields:}\) The central magnetic induction of an \(N\)-turn circular coil is \[ B_c=\frac{\mu_0NI}{2R} \] Therefore, \[ \frac{B_{c,Q}}{B_{c,P}} =\frac{N_QI_Q/R_Q}{N_PI_P/R_P} \] Substituting, \[ \frac{B_{c,Q}}{B_{c,P}} =\frac{(10)(2.0)/0.10}{(80)(1.0)/0.050} \] \[ \frac{B_{c,Q}}{B_{c,P}}=\frac{200}{1600} \] \[ \frac{B_{c,Q}}{B_{c,P}}=\frac18 \] \(\textbf{Compare the external-field torques:}\) For perpendicular magnetic moment and external field, \[ \tau_{\max}=mB_e \] The two coils were designed to have the same magnetic moment. They are placed in the same external field at the same relative orientation. Therefore, \[ \frac{\tau_Q}{\tau_P}=\frac{m_Q}{m_P} \] \[ \frac{\tau_Q}{\tau_P}=1 \] \(\textbf{Final answer:}\) \[ \boxed{N_Q=10,\qquad \frac{B_{c,Q}}{B_{c,P}}=\frac18,\qquad \frac{\tau_Q}{\tau_P}=1} \] Matching magnetic dipole moments does not require matching the magnetic fields at the coil centres. The moment depends on \(NIR^2\), whereas the central field depends on \(NI/R\). The larger coil can therefore have the same moment but a much smaller central field. Both coils experience the same maximum torque in an identical external field because that torque depends on their equal magnetic moments. The problem distinguishes a coil's field-producing geometry from its overall dipole response.

Explanation: \(\textbf{Express the opposing magnetic moments:}\) The magnetic moment of a circular coil is \[ \vec m=NI\pi R^2\hat n \] The inner current appears anticlockwise from positive \(z\), giving a positive-\(z\) moment. The outer current appears clockwise from the same side, giving a negative-\(z\) moment. Their moment magnitudes are \[ m_1=80(1.0)\pi R^2 \] and \[ m_2=20I_2\pi(2R)^2 \] \(\textbf{Impose cancellation of the resultant moment:}\) The opposite moments must have equal magnitudes: \[ 80(1.0)\pi R^2=20I_2\pi(4R^2) \] Cancelling the common factors, \[ 80=80I_2 \] \[ I_2=1.0\,\text{A} \] \(\textbf{Calculate the inner coil's central field:}\) For a circular coil, \[ B_c=\frac{\mu_0NI}{2R} \] The inner coil produces field along positive \(z\): \[ \vec B_1=+\frac{\mu_0(80)(1.0)}{2(0.050)}\hat k \] \[ \vec B_1=+800\mu_0\hat k \] \(\textbf{Calculate the outer coil's central field:}\) The outer current is clockwise, so its field at the centre points along negative \(z\). Its radius is \(0.10\,\text{m}\): \[ \vec B_2=-\frac{\mu_0(20)(1.0)}{2(0.10)}\hat k \] \[ \vec B_2=-100\mu_0\hat k \] \(\textbf{Add the two central fields:}\) \[ \vec B_{\mathrm{centre}}=\vec B_1+\vec B_2 \] \[ \vec B_{\mathrm{centre}}=(800-100)\mu_0\hat k \] \[ \vec B_{\mathrm{centre}}=700(4\pi\times10^{-7})\hat k \] \[ \vec B_{\mathrm{centre}}=+2.8\pi\times10^{-4}\hat k\,\text{T} \] \(\textbf{Final answer:}\) \[ \boxed{I_2=1.0\,\text{A},\qquad \vec B_{\mathrm{centre}}=+2.8\pi\times10^{-4}\hat k\,\text{T}} \] The two coils have equal and opposite magnetic dipole moments after adjustment. Their combined leading distant dipole field consequently vanishes. Their central fields do not cancel because central field and dipole moment depend differently on coil radius. The smaller inner coil produces a stronger field at the common centre for the specified turn-current combination. Zero total magnetic dipole moment therefore does not imply zero magnetic field at every point around the assembly.

Explanation: The magnetic moment of a bar magnet points from its effective south pole towards its north pole. Initially, that direction is positive \(x\). At an axial point on the extension beyond the north pole, the magnetic field points away from the north pole. It therefore points along positive \(x\). The north-seeking end of a compass follows the direction of the local magnetic field. Thus, the compass at P initially points right. At Q, which lies in the perpendicular bisector plane, the short-dipole field points opposite to the magnetic moment. The initial equatorial field consequently points along negative \(x\). The compass at Q initially points left. Rotating the magnet through \(180^\circ\) reverses its magnetic moment. The axial and equatorial dipole fields both reverse their laboratory directions. The compass at P then points left, while the compass at Q points right. For a short magnetic dipole, \[ B_{\mathrm{axial}}=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] and \[ B_{\mathrm{equatorial}}=\frac{\mu_0}{4\pi}\frac{m}{r^3} \] At equal distances, the axial magnitude is therefore twice the equatorial magnitude. The factor-of-two comparison is independent of the sign or orientation of the moment. Compass directions require vector reasoning, whereas the field-strength comparison requires the correct axial and equatorial formulas. Both must be applied together.

Explanation: \(\textbf{Express the axial measurement as a straight-line graph:}\) The axial dipole field is \[ B_d=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] It points along positive \(z\) because the dipole moment points along positive \(z\). The resultant signed field is \[ B_z=B_e+\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] Let the horizontal graph variable be \[ X=\frac1{r^3} \] Then, \[ B_z=B_e+\left(\frac{\mu_0}{4\pi}2m\right)X \] The graph is linear, with intercept \(B_e\) and slope proportional to m. \(\textbf{Calculate the horizontal coordinates:}\) At \(r_1=0.10\,\text{m}\), \[ X_1=\frac1{(0.10)^3}=1000\,\text{m}^{-3} \] At \(r_2=0.20\,\text{m}\), \[ X_2=\frac1{(0.20)^3}=125\,\text{m}^{-3} \] \(\textbf{Recover the graph slope:}\) Using microtesla for the vertical-axis readings, \[ S=\frac{30-(-5.0)}{1000-125} \] \[ S=\frac{35}{875} \] \[ S=0.040\,\mu\text{T m}^3 \] Converting to SI units, \[ S=4.0\times10^{-8}\,\text{T m}^3 \] \(\textbf{Determine the magnetic moment:}\) The theoretical slope is \[ S=\frac{\mu_0}{4\pi}2m \] Therefore, \[ m=\frac{S}{2(\mu_0/4\pi)} \] \[ m=\frac{4.0\times10^{-8}}{2.0\times10^{-7}} \] \[ m=0.20\,\text{A m}^2 \] \(\textbf{Recover the uniform-field intercept:}\) Using the first measurement, \[ 30=B_e+0.040(1000) \] \[ 30=B_e+40 \] \[ B_e=-10\,\mu\text{T} \] \(\textbf{Determine the axial cancellation distance:}\) Zero resultant field requires \[ \frac{\mu_0}{4\pi}\frac{2m}{r_0^3}=10\times10^{-6}\,\text{T} \] Substituting, \[ \frac{(10^{-7})(2)(0.20)}{r_0^3}=10^{-5} \] \[ r_0^3=4.0\times10^{-3}\,\text{m}^3 \] \[ r_0\approx0.159\,\text{m} \] \(\textbf{Final answer:}\) \[ \boxed{m=0.20\,\text{A m}^2,\qquad B_e=-10\,\mu\text{T},\qquad r_0\approx0.159\,\text{m}} \] The slope of the graph isolates the inverse-cube dipole contribution. The vertical intercept identifies the independent uniform field. The opposite signs of the two measured fields show that their resultant changes direction between the observation points. A zero-field position occurs where the axial dipole field exactly balances the negative background field. The method uses signed field data rather than treating the two measurements as positive magnitudes.

Explanation: \(\textbf{Determine the field at the axial observation point:}\) For a short magnetic dipole, the axial field is \[ B_a=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] At P, \[ r=0.10\,\text{m} \] Therefore, \[ B_P=10^{-7}\frac{2(0.40)}{(0.10)^3} \] \[ B_P=8.0\times10^{-5}\,\text{T} \] The axial field points along the magnetic moment: \[ \vec B_P=+8.0\times10^{-5}\hat k\,\text{T} \] \(\textbf{Determine the field at the equatorial point:}\) At equal distance, the equatorial dipole-field magnitude is half the axial magnitude: \[ B_q=\frac{\mu_0}{4\pi}\frac{m}{r^3} \] Substituting, \[ B_Q=10^{-7}\frac{0.40}{(0.10)^3} \] \[ B_Q=4.0\times10^{-5}\,\text{T} \] The equatorial field points opposite to the magnetic moment: \[ \vec B_Q=-4.0\times10^{-5}\hat k\,\text{T} \] \(\textbf{Calculate the instantaneous force on P:}\) The magnetic Lorentz force is \[ \vec F=q(\vec v\times\vec B) \] At P, \[ \vec F_P=(2.0\times10^{-6})(3.0\times10^3\hat j)\times(8.0\times10^{-5}\hat k) \] Using \(\hat j\times\hat k=\hat i\), \[ \vec F_P=+4.8\times10^{-7}\hat i\,\text{N} \] \(\textbf{Calculate the instantaneous force on Q:}\) At Q, the field direction is negative \(z\): \[ \vec F_Q=(2.0\times10^{-6})(3.0\times10^3\hat j)\times(-4.0\times10^{-5}\hat k) \] \[ \vec F_Q=-2.4\times10^{-7}\hat i\,\text{N} \] \(\textbf{Apply magnetic-moment reversal:}\) Reversing the bar magnet reverses its magnetic field at each fixed observation point. The particle charges and velocities remain unchanged. Consequently, \[ \vec F'_P=-\vec F_P \] \[ \vec F'_Q=-\vec F_Q \] \(\textbf{Final answer:}\) \[ \boxed{\vec F_P=+4.8\times10^{-7}\hat i\,\text{N},\qquad \vec F_Q=-2.4\times10^{-7}\hat i\,\text{N},\qquad \vec F'_P=-\vec F_P,\quad\vec F'_Q=-\vec F_Q} \] The two particles have identical charges and velocities but occupy different geometrical positions relative to the dipole. The axial and equatorial magnetic fields have opposite directions and different magnitudes. These differences produce oppositely directed instantaneous magnetic forces. Magnet reversal reverses both fields and therefore both Lorentz forces. Neither magnetic force performs work because each remains perpendicular to the particle's instantaneous velocity.