Magnetic Effects of Current and Magnetism MCQs with Answers – Part 6 (NEET Physics)

Explanation: \(\textbf{Relate the solenoid's interior field to its turn number:}\) For a sufficiently long solenoid, \[ B_{\mathrm{in}}=\mu_0nI \] where \[ n=\frac NL \] Therefore, \[ B_{\mathrm{in}}=\frac{\mu_0NI}{L} \] Rearranging, \[ N=\frac{B_{\mathrm{in}}L}{\mu_0I} \] \(\textbf{Recover the original turn number:}\) Substituting the measured quantities, \[ N=\frac{(4\pi\times10^{-4})(0.50)} {(4\pi\times10^{-7})(1.0)} \] \[ N=\frac{2\pi\times10^{-4}}{4\pi\times10^{-7}} \] \[ N=500 \] \(\textbf{Determine the cross-sectional area from the dipole moment:}\) The solenoid magnetic moment is \[ m=NIA \] Thus, \[ A=\frac{m}{NI} \] Substituting, \[ A=\frac{0.50}{(500)(1.0)} \] \[ A=1.0\times10^{-3}\,\text{m}^2 \] \(\textbf{Apply the rebuilt geometry and current:}\) The changed quantities are \[ N'=N,\qquad A'=4A \] \[ L'=2L,\qquad I'=\frac I2 \] The new magnetic moment is \[ m'=N'I'A' \] \[ m'=N\left(\frac I2\right)(4A) \] \[ m'=2NIA \] Therefore, \[ \frac{m'}m=2 \] \(\textbf{Compare interior magnetic fields:}\) The rebuilt solenoid has \[ B'_{\mathrm{in}}=\frac{\mu_0N'I'}{L'} \] Substituting the changed quantities, \[ B'_{\mathrm{in}}=\frac{\mu_0N(I/2)}{2L} \] \[ B'_{\mathrm{in}}=\frac14\frac{\mu_0NI}{L} \] Thus, \[ \frac{B'_{\mathrm{in}}}{B_{\mathrm{in}}}=\frac14 \] \(\textbf{Final answer:}\) \[ \boxed{N=500,\qquad A=1.0\times10^{-3}\,\text{m}^2,\qquad \frac{m'}m=2,\qquad \frac{B'_{\mathrm{in}}}{B_{\mathrm{in}}}=\frac14} \] The two measurements determine different combinations of the solenoid parameters. The interior field depends on turns per unit length and current. The total magnetic moment depends on the total number of turns, current and enclosed area. A rebuilt solenoid can therefore have a larger magnetic moment while producing a weaker internal magnetic field. Confusing these two quantities would lead to an incorrect prediction.

Explanation: \(\textbf{Use the torque and energy measurements together:}\) For a magnetic dipole in a uniform field, \[ \tau=mB\sin\theta \] Its orientation-dependent potential energy is \[ U=-mB\cos\theta \] The measured values therefore give \[ mB\sin\theta=0.24\,\text{J} \] and \[ mB\cos\theta=0.18\,\text{J} \] Here, \(\text{N m}\) and joule are dimensionally equivalent, although torque and energy represent different physical quantities. \(\textbf{Determine the product }mB\textbf{:}\) Squaring and adding, \[ (mB)^2(\sin^2\theta+\cos^2\theta) =(0.24)^2+(0.18)^2 \] Using \[ \sin^2\theta+\cos^2\theta=1 \] we obtain \[ (mB)^2=0.0576+0.0324 \] \[ (mB)^2=0.0900 \] \[ mB=0.30\,\text{J} \] \(\textbf{Recover the magnetic moment:}\) \[ m=\frac{mB}{B} \] \[ m=\frac{0.30}{0.50} \] \[ m=0.60\,\text{A m}^2 \] \(\textbf{Determine the original orientation:}\) The sine and cosine are \[ \sin\theta=\frac{0.24}{0.30}=0.80 \] \[ \cos\theta=\frac{0.18}{0.30}=0.60 \] Because the angle is acute, \[ \theta=\tan^{-1}\left(\frac{0.80}{0.60}\right) \] \[ \theta\approx53.1^\circ \] \(\textbf{Apply physical reversal of the magnet:}\) Reversal changes \[ \vec m\longrightarrow-\vec m \] The uniform magnetic field remains unchanged. Therefore, \[ U'=-(-\vec m)\cdot\vec B \] \[ U'=-U \] Hence, \[ U'=+0.18\,\text{J} \] \(\textbf{Final answer:}\) \[ \boxed{m=0.60\,\text{A m}^2,\qquad \theta\approx53.1^\circ,\qquad U'=+0.18\,\text{J}} \] The torque measurement alone determines only \(mB\sin\theta\). The energy measurement supplies the complementary cosine relation. Combining them determines both the moment magnitude and the orientation. Reversing the magnet reverses its moment and changes the sign of the potential energy at the same physical position. Its torque vector also reverses, although its torque magnitude remains unchanged.

Explanation: \(\textbf{Interpret the zero-field observation at P:}\) The uniform external field points along positive \(z\): \[ \vec B_e=+B_0\hat k \] The equatorial magnetic field of a short dipole points opposite to its magnetic moment. Because \(\vec m=m\hat k\), the dipole field at P points along negative \(z\). The measured resultant is zero: \[ \vec B_e+\vec B_d(P)=\vec0 \] Therefore, \[ \vec B_d(P)=-B_0\hat k \] \(\textbf{Apply the inverse-cube distance relation:}\) The equatorial dipole-field magnitude is \[ B_d(r)=\frac{\mu_0}{4\pi}\frac{m}{r^3} \] At Q, the distance is \[ r_Q=\frac{r_0}{2} \] Therefore, \[ \frac{B_d(Q)}{B_d(P)} =\left(\frac{r_0}{r_0/2}\right)^3 \] \[ \frac{B_d(Q)}{B_d(P)}=2^3 \] \[ \frac{B_d(Q)}{B_d(P)}=8 \] The moment direction is unchanged, so the dipole field remains negative: \[ \vec B_d(Q)=-8B_0\hat k \] \(\textbf{Determine the field at Q before reversal:}\) By vector superposition, \[ \vec B_Q=\vec B_e+\vec B_d(Q) \] \[ \vec B_Q=(B_0-8B_0)\hat k \] \[ \vec B_Q=-7B_0\hat k \] \(\textbf{Reverse the bar magnet:}\) Reversing the magnet reverses its magnetic moment. The equatorial field at Q consequently changes direction but not magnitude: \[ \vec B'_d(Q)=+8B_0\hat k \] The separately maintained uniform field remains positive: \[ \vec B'_Q=(B_0+8B_0)\hat k \] \[ \vec B'_Q=+9B_0\hat k \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_Q=-7B_0\hat k,\qquad \vec B'_Q=+9B_0\hat k} \] A zero-field point results from cancellation between distinct magnetic-field sources. It does not imply that either source field is absent. Moving closer to the dipole increases its field by an inverse-cube factor while the uniform background remains unchanged. Magnet reversal then converts opposition into reinforcement. The total field at Q therefore changes both magnitude and direction.

Explanation: \(\textbf{Determine the magnetic moments after longitudinal division:}\) For uniformly magnetised pieces, magnetic moment is proportional to volume when magnetisation is unchanged. Both pieces retain the original length. Their volumes are therefore proportional to their cross-sectional areas. For the smaller piece, \[ m_s=\frac14m \] For the larger piece, \[ m_l=\frac34m \] Initially, both moments point along positive \(x\). \(\textbf{Apply reversal of the smaller piece:}\) After physical reversal, \[ \vec m_s=-\frac m4\hat i \] The larger piece remains \[ \vec m_l=+\frac{3m}{4}\hat i \] Their combined magnetic moment is \[ \vec m_{\mathrm{net}}=\vec m_s+\vec m_l \] \[ \vec m_{\mathrm{net}}= \left(-\frac m4+\frac{3m}{4}\right)\hat i \] \[ \vec m_{\mathrm{net}}=\frac m2\hat i \] \(\textbf{Calculate the combined torque:}\) The torque on a magnetic dipole is \[ \vec\tau=\vec m\times\vec B \] Because both pieces experience the same uniform magnetic field, their torques add: \[ \vec\tau_{\mathrm{total}}=\vec m_{\mathrm{net}}\times\vec B \] Substituting, \[ \vec\tau_{\mathrm{total}}= \frac m2\hat i\times B\hat j \] Using \(\hat i\times\hat j=\hat k\), \[ \vec\tau_{\mathrm{total}}=+\frac12mB\hat k \] \(\textbf{Calculate the combined potential energy:}\) The orientation-dependent magnetic potential energy is \[ U=-\vec m\cdot\vec B \] The resultant moment points along \(x\), while the field points along \(y\). Their scalar product vanishes: \[ U=-\left(\frac m2\hat i\right)\cdot(B\hat j) \] \[ U=0 \] \(\textbf{Final answer:}\) \[ \boxed{m_s=\frac m4,\qquad m_l=\frac{3m}{4},\qquad \vec\tau_{\mathrm{total}}=+\frac12mB\hat k,\qquad U=0} \] Unequal longitudinal pieces have different magnetic moments because their volumes differ. Reversing the smaller piece reduces the resultant moment without eliminating it. The combined assembly consequently retains a nonzero magnetic torque in the perpendicular field. Its orientation-dependent potential energy is zero at that instant because its resultant moment is perpendicular to the field. Zero magnetic potential energy does not require zero magnetic torque.

Explanation: \(\textbf{Establish the original axial reference field:}\) For the original magnetic dipole, \[ B_0=\frac{\mu_0}{4\pi}\frac{2m}{r^3} \] The field at an axial point points along the dipole moment. Thus, the original field at P points along positive \(z\). \(\textbf{Calculate the modified axial field at P:}\) The replacement magnet has moment magnitude \[ m'=3m \] Its moment points along negative \(z\). The new axial observation distance is \[ r'_P=2r \] Therefore, the modified axial-field magnitude is \[ B'_P=\frac{\mu_0}{4\pi}\frac{2(3m)}{(2r)^3} \] \[ B'_P=\frac{\mu_0}{4\pi}\frac{6m}{8r^3} \] Comparing with \(B_0\), \[ \frac{B'_P}{B_0}=\frac{6/8}{2} \] \[ \frac{B'_P}{B_0}=\frac38 \] The new moment points along negative \(z\), so the axial field also points along negative \(z\): \[ \vec B'_P=-\frac38B_0\hat k \] \(\textbf{Calculate the modified equatorial field at Q:}\) The new equatorial observation distance is \[ r'_Q=r \] The equatorial-field magnitude is \[ B'_Q=\frac{\mu_0}{4\pi}\frac{3m}{r^3} \] Dividing by the original axial reference, \[ \frac{B'_Q}{B_0} =\frac{3m/r^3}{2m/r^3} \] \[ \frac{B'_Q}{B_0}=\frac32 \] \(\textbf{Determine the equatorial direction:}\) At an equatorial point, the short-dipole field points opposite to the dipole moment. The replacement magnet points along negative \(z\). Therefore, its equatorial field points along positive \(z\): \[ \vec B'_Q=+\frac32B_0\hat k \] \(\textbf{Final answer:}\) \[ \boxed{\vec B'_P=-\frac38B_0\hat k,\qquad \vec B'_Q=+\frac32B_0\hat k} \] The moment change and observation-distance changes affect both readings. The axial field contains a factor of two, whereas the equatorial field does not. The inverse-cube dependence makes doubling distance reduce field magnitude by a factor of eight. The new moment orientation determines the final signs independently of these magnitude factors.

Explanation: \(\textbf{Extract the transverse moment components from torque:}\) The magnetic-torque vector is \[ \vec\tau=\vec m\times\vec B \] Substituting the uniform field, \[ \vec\tau=(m_x\hat i+m_y\hat j+m_z\hat k)\times(0.50\hat k) \] Using the Cartesian cross products, \[ \vec\tau=0.50(m_y\hat i-m_x\hat j) \] The measured torque is \[ \vec\tau=0.10\hat i-0.20\hat j \] Equating the \(x\)-components, \[ 0.50m_y=0.10 \] \[ m_y=0.20\,\text{A m}^2 \] Equating the \(y\)-components, \[ -0.50m_x=-0.20 \] \[ m_x=0.40\,\text{A m}^2 \] \(\textbf{Recover the field-parallel component from energy:}\) The magnetic potential energy is \[ U=-\vec m\cdot\vec B \] With the field along positive \(z\), \[ U=-0.50m_z \] The measured value is \(-0.15\,\text{J}\): \[ -0.50m_z=-0.15 \] \[ m_z=0.30\,\text{A m}^2 \] Thus, \[ \vec m=(0.40\hat i+0.20\hat j+0.30\hat k)\,\text{A m}^2 \] \(\textbf{Apply the field reversal:}\) The new field is \[ \vec B'=-0.50\hat k\,\text{T} \] The dipole remains fixed. Therefore, \[ \vec\tau'=\vec m\times(-\vec B) \] \[ \vec\tau'=-\vec\tau \] Hence, \[ \vec\tau'=(-0.10\hat i+0.20\hat j)\,\text{N m} \] \(\textbf{Check the energy change:}\) The energy after field reversal becomes \[ U'=-\vec m\cdot(-\vec B) \] \[ U'=-U=+0.15\,\text{J} \] \(\textbf{Final answer:}\) \[ \boxed{\vec m=(0.40\hat i+0.20\hat j+0.30\hat k)\,\text{A m}^2,\qquad \vec\tau'=(-0.10\hat i+0.20\hat j)\,\text{N m}} \] The magnetic torque reveals only the moment components perpendicular to the field. The moment component parallel to the field produces no torque. The energy measurement supplies this missing component through a scalar product. Together, torque and energy reconstruct the complete moment vector. Field reversal then reverses both the torque vector and the sign of the orientation-dependent potential energy.

Explanation: \(\textbf{Identify the observation geometry relative to P:}\) Dipole P lies on the negative \(x\)-axis and has magnetic moment along positive \(x\). The origin is on its positive axial side. The distant axial field is \[ \vec B_{P,O}=\frac{\mu_0}{4\pi}\frac{2m}{a^3}\hat i \] Using the defined quantity b, \[ \vec B_{P,O}=+2b\hat i \] \(\textbf{Identify the observation geometry relative to Q:}\) Dipole Q lies on the positive \(y\)-axis, while its moment points along positive \(x\). The origin lies in its equatorial plane. The equatorial field points opposite to the magnetic moment: \[ \vec B_{Q,O}=-\frac{\mu_0}{4\pi}\frac{m}{a^3}\hat i \] \[ \vec B_{Q,O}=-b\hat i \] \(\textbf{Calculate the initial resultant field:}\) \[ \vec B_O=\vec B_{P,O}+\vec B_{Q,O} \] \[ \vec B_O=(2b-b)\hat i \] \[ \vec B_O=b\hat i \] \(\textbf{Determine the initial test-needle torque:}\) The test needle has moment \[ \vec m_t=m_t\hat j \] Therefore, \[ \vec\tau=\vec m_t\times\vec B_O \] \[ \vec\tau=m_t\hat j\times b\hat i \] Using \(\hat j\times\hat i=-\hat k\), \[ \vec\tau=-m_tb\hat k \] \(\textbf{Reverse dipole Q:}\) Only Q's equatorial field changes direction: \[ \vec B'_{Q,O}=+b\hat i \] The contribution from P remains \(+2b\hat i\). Hence, \[ \vec B'_O=(2b+b)\hat i \] \[ \vec B'_O=3b\hat i \] The new test-needle torque is \[ \vec\tau'=m_t\hat j\times3b\hat i \] \[ \vec\tau'=-3m_tb\hat k \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_O=b\hat i,\qquad \vec\tau=-m_tb\hat k,\qquad \vec B'_O=3b\hat i,\qquad \vec\tau'=-3m_tb\hat k} \] The origin is axial for one dipole but equatorial for the other. Their initial contributions therefore oppose despite the dipoles having identical moment directions. Reversing the equatorial source converts opposition into reinforcement. The field direction at O remains positive \(x\), but its magnitude triples. The unchanged test needle consequently experiences a torque of the same direction and three times the magnitude.

Explanation: \(\textbf{Determine the energy graph slope:}\) The magnetic potential energy is \[ U=-mB\cos\theta \] Define the horizontal graph variable \[ x=\cos\theta \] Then, \[ U=(-mB)x \] The graph slope is therefore \[ s=-mB \] Substituting, \[ s=-(0.50)(0.40) \] \[ s=-0.20\,\text{J} \] \(\textbf{Determine the initial orientation energy:}\) The rotation begins at \[ \theta_i=120^\circ \] Therefore, \[ U_i=-0.20\cos120^\circ \] \[ U_i=-0.20\left(-\frac12\right) \] \[ U_i=+0.100\,\text{J} \] \(\textbf{Determine the final orientation energy:}\) The final angle is \[ \theta_f=30^\circ \] Thus, \[ U_f=-0.20\cos30^\circ \] \[ U_f=-0.20\left(\frac{\sqrt3}{2}\right) \] \[ U_f=-0.10\sqrt3\,\text{J} \] \(\textbf{Calculate the work by magnetic torque:}\) The magnetic work is the negative change in magnetic potential energy: \[ W_B=-(U_f-U_i) \] Substituting, \[ W_B=-(-0.10\sqrt3-0.100) \] \[ W_B=0.10(1+\sqrt3)\,\text{J} \] \[ W_B\approx0.273\,\text{J} \] \(\textbf{Compare endpoint torque magnitudes:}\) The magnetic-torque magnitude is \[ \tau=mB\sin\theta \] Initially, \[ \tau_i=mB\sin120^\circ=\frac{\sqrt3}{2}mB \] Finally, \[ \tau_f=mB\sin30^\circ=\frac12mB \] Thus, \[ \frac{\tau_f}{\tau_i} =\frac{1/2}{\sqrt3/2} \] \[ \frac{\tau_f}{\tau_i}=\frac1{\sqrt3} \] \(\textbf{Final answer:}\) \[ \boxed{s=-0.20\,\text{J},\qquad W_B=0.10(1+\sqrt3)\,\text{J}\approx0.273\,\text{J},\qquad \frac{\tau_f}{\tau_i}=\frac1{\sqrt3}} \] The energy graph has negative slope because alignment with the field lowers magnetic potential energy. During the specified rotation, the dipole moves towards a more favourable orientation. Magnetic torque consequently does positive work overall. The final torque magnitude is smaller than the initial torque magnitude, even though the total work is positive. The work depends on the energy change, not merely on the two endpoint torque magnitudes.

Explanation: The magnetic polarity of a current-carrying solenoid follows from the conventional current direction when viewed from an end. If the current appears anticlockwise, that observed end behaves as a north pole. If the current appears clockwise, that observed end behaves as a south pole. The observer at the left sees clockwise current. Therefore, the left end behaves as a south pole. Statement I is correct. The opposite right end behaves as a north pole. The magnetic dipole moment of the equivalent solenoid points from its effective south end towards its north end. Thus, the solenoid's magnetic moment points towards the right. Statement II is correct. At a sufficiently distant axial point, the solenoid's field can be treated using the magnetic-dipole approximation. The axial field points along the source magnetic moment on both axial sides. Reversing the solenoid current reverses the solenoid's magnetic moment. Its distant axial magnetic field consequently reverses direction. Statement III is correct. The separate bar magnet has its own magnetic moment determined by its magnetisation and physical orientation. Merely reversing the current of a nearby solenoid does not automatically reverse a mechanically held bar magnet's moment. The problem does not describe remagnetisation of that bar magnet. Statement IV is incorrect. A field source and a responding magnetic dipole must be treated as distinct objects. Changing the source current reverses its magnetic field. It does not, by itself, establish a physical reversal of another permanent magnet. The observer's viewing direction is essential for assigning the original solenoid polarity correctly.

Explanation: \(\textbf{Determine the magnetic field at the equatorial point:}\) For a short magnetic dipole, \[ B_{\mathrm{eq}}=\frac{\mu_0}{4\pi}\frac{m}{r^3} \] The magnetic moment points along positive \(z\). At an equatorial point, the field points opposite to the magnetic moment. The observation distance is \[ r=0.20\,\text{m} \] Therefore, \[ B_{\mathrm{eq}}=10^{-7}\frac{0.80}{(0.20)^3} \] \[ B_{\mathrm{eq}}=10^{-7}\frac{0.80}{0.008} \] \[ B_{\mathrm{eq}}=1.0\times10^{-5}\,\text{T} \] Hence, \[ \vec B_P=-1.0\times10^{-5}\hat k\,\text{T} \] \(\textbf{Calculate the original Lorentz force:}\) The magnetic force on a moving charge is \[ \vec F=q(\vec v\times\vec B) \] Substituting, \[ \vec F_1=(2.0\times10^{-6})(3.0\times10^3\hat i) \times(-1.0\times10^{-5}\hat k) \] Using \[ \hat i\times(-\hat k)=+\hat j \] we obtain \[ \vec F_1=+6.0\times10^{-8}\hat j\,\text{N} \] \(\textbf{Reverse the bar magnet:}\) After reversal, the magnetic moment points along negative \(z\). Its equatorial field at P points along positive \(z\): \[ \vec B'_P=+1.0\times10^{-5}\hat k\,\text{T} \] The particle velocity remains positive \(x\). Therefore, \[ \vec F_2=(2.0\times10^{-6})(3.0\times10^3\hat i) \times(+1.0\times10^{-5}\hat k) \] \[ \vec F_2=-6.0\times10^{-8}\hat j\,\text{N} \] \(\textbf{Check the instantaneous magnetic power:}\) Magnetic force is perpendicular to particle velocity: \[ \vec F\cdot\vec v=0 \] Therefore, \[ P_B=\vec F\cdot\vec v=0 \] in both arrangements. The magnetic force may change the direction of velocity but cannot directly change the particle's kinetic energy. \(\textbf{Final answer:}\) \[ \boxed{\vec F_1=+6.0\times10^{-8}\hat j\,\text{N},\qquad \vec F_2=-6.0\times10^{-8}\hat j\,\text{N},\qquad P_{B,1}=P_{B,2}=0} \] The particle lies in the dipole's equatorial plane, where the field opposes the moment. The field reverses when the bar magnet is reversed. The resulting Lorentz force also reverses for an unchanged charge and velocity. However, the force remains perpendicular to the motion, so its instantaneous power is zero in both cases. The calculation gives the instantaneous force at P, not a complete trajectory in the nonuniform dipole field.

Explanation: \(\textbf{Determine the original resultant magnetic moment:}\) The two magnets have moments \[ \vec m_1=m\hat i \] and \[ \vec m_2=m\hat j \] Because they form one rigid assembly, their moments add vectorially: \[ \vec m_{\mathrm{net},1}=m(\hat i+\hat j) \] \(\textbf{Calculate the original total torque:}\) The torque in a uniform field is \[ \vec\tau=\vec m_{\mathrm{net}}\times\vec B \] The external field points along positive \(x\): \[ \vec B=B\hat i \] Therefore, \[ \vec\tau_1=m(\hat i+\hat j)\times B\hat i \] \[ \vec\tau_1=mB(\hat j\times\hat i) \] \[ \vec\tau_1=-mB\hat k \] \(\textbf{Determine the original potential energy:}\) The orientation-dependent energy is \[ U=-\vec m_{\mathrm{net}}\cdot\vec B \] Thus, \[ U_1=-m(\hat i+\hat j)\cdot B\hat i \] \[ U_1=-mB \] Only the moment component parallel to the magnetic field contributes to this energy. \(\textbf{Reverse the second magnet:}\) The second moment becomes \[ \vec m'_2=-m\hat j \] The first magnet remains unchanged. Therefore, \[ \vec m_{\mathrm{net},2}=m(\hat i-\hat j) \] \(\textbf{Calculate the changed torque:}\) \[ \vec\tau_2=m(\hat i-\hat j)\times B\hat i \] \[ \vec\tau_2=-mB(\hat j\times\hat i) \] \[ \vec\tau_2=+mB\hat k \] Its magnitude remains \(mB\). \(\textbf{Calculate the changed energy:}\) \[ U_2=-m(\hat i-\hat j)\cdot B\hat i \] \[ U_2=-mB \] \(\textbf{Final answer:}\) \[ \boxed{\vec\tau_1=-mB\hat k,\quad \vec\tau_2=+mB\hat k,\quad U_1=U_2=-mB,\quad |\vec\tau_1|=|\vec\tau_2|=mB} \] Reversing the perpendicular magnet changes the component of the total moment responsible for torque. The component parallel to the field remains unchanged. Consequently, the magnetic torque reverses while the orientation-dependent potential energy stays the same. Torque and potential energy depend on different vector products of magnetic moment and magnetic field. The distinction is especially clear when the assembly contains mutually perpendicular dipoles.

Explanation: Magnetic fields obey the principle of vector superposition. The resultant field at a point is obtained by adding the field due to the bar magnet and the separately produced external field. If those two vectors have equal magnitudes and opposite directions, their sum is zero. Statement I is correct. At a point where the resultant magnetic field is zero, a small ideal compass experiences no magnetic aligning torque from that field. There is therefore no unique nonzero local field direction for the compass to follow. The usual tangent interpretation of a magnetic field line does not assign a unique direction at such a zero-field point. Statement II is incorrect. Magnetic field lines represent the resultant field of the complete source configuration. Adding a uniform external field changes the field vector at points where the bar magnet already produces a field. This can change both the direction and the magnitude of the resultant field. The corresponding total-field line pattern can consequently differ from the isolated magnet's pattern. Statement III is correct. Reversing the bar magnet reverses its magnetic dipole moment and its source-field contribution. However, the independent external field is not reversed. If the original resultant is \[ \vec B_{\mathrm{total}}=\vec B_{\mathrm{magnet}}+\vec B_e \] then after reversal it becomes \[ \vec B'_{\mathrm{total}}=-\vec B_{\mathrm{magnet}}+\vec B_e \] This is generally not equal to \[ -\vec B_{\mathrm{total}} \] because the external field remains unchanged. Statement IV is correct. A zero-field point is a property of the combined configuration rather than evidence that either field source has disappeared. The field-line drawing must represent the actual resultant, not the isolated contributions considered independently. Likewise, reversing one source is not equivalent to reversing every field present.

Explanation: \(\textbf{Determine the magnetic energy scale:}\) The orientation-dependent potential energy is \[ U=-mB\cos\theta \] The product of moment and field is \[ mB=(0.40)(0.50) \] \[ mB=0.20\,\text{J} \] \(\textbf{Calculate the energy at the original orientation:}\) Initially, \[ \theta_i=120^\circ \] Therefore, \[ U_i=-0.20\cos120^\circ \] \[ U_i=-0.20\left(-\frac12\right) \] \[ U_i=+0.10\,\text{J} \] \(\textbf{Determine the work during the first rotation:}\) The dipole rotates until its moment becomes perpendicular to the field: \[ \theta_f=90^\circ \] Its orientation energy is then \[ U_f=-mB\cos90^\circ \] \[ U_f=0 \] The magnetic work during this motion is \[ W_{B,1}=-(U_f-U_i) \] \[ W_{B,1}=-(0-0.10) \] \[ W_{B,1}=+0.10\,\text{J} \] \(\textbf{Determine the torque after field reversal:}\) At the instant of reversal, the dipole is held perpendicular to the field. Reversing the field preserves this perpendicular relation. The torque magnitude is \[ \tau=mB\sin90^\circ \] \[ \tau=(0.40)(0.50) \] \[ \tau=0.20\,\text{N m} \] The field reversal changes the torque direction. \(\textbf{Predict the subsequent spontaneous motion:}\) When the dipole is released again, the magnetic torque tends to align its moment with the NEW field direction. The orientation parallel to the reversed field is the stable equilibrium. At the perpendicular orientation, the torque magnitude is maximum, not zero. The dipole therefore begins rotating towards alignment with the reversed field rather than remaining stationary. \(\textbf{Final answer:}\) \[ \boxed{W_{B,1}=+0.10\,\text{J},\qquad \tau_{\mathrm{after\ reversal}}=0.20\,\text{N m}} \] The first spontaneous rotation lowers magnetic potential energy, so magnetic torque does positive work. Stopping the dipole mechanically changes its kinetic energy but does not alter the orientation-energy formula. Field reversal changes the direction in which magnetic torque tends to rotate the held dipole. A perpendicular dipole has maximum torque magnitude, whereas parallel and antiparallel orientations have zero torque. The stable direction after reversal is parallel to the reversed external field.

Explanation: \(\textbf{Identify the vertical forces on the upper wire:}\) The antiparallel currents produce magnetic repulsion. Since the upper wire is above the lower one, its magnetic force is upward. The insulating supports also pull upward, while gravity acts downward. For equilibrium, the forces per unit length satisfy \[ \frac{T}{\ell}+\frac{F_m}{\ell}=\lambda g \] For two long parallel wires separated by \(d\), \[ \frac{F_m}{\ell}=\frac{\mu_0I_1I_2}{2\pi d} \] Therefore, \[ \frac{T}{\ell}=\lambda g-\frac{\mu_0I_1I_2}{2\pi d} \] \(\textbf{Extract the unknown lower-wire current:}\) Subtract the support-force readings at the two upper-wire currents. The gravitational contribution cancels: \[ 0.020-0.010=\frac{\mu_0I_1}{2\pi d}(20-10) \] Substituting, \[ 0.010=\frac{(2.0\times10^{-7})I_1}{4.0\times10^{-3}}(10) \] \[ 0.010=(5.0\times10^{-4})I_1 \] \[ I_1=20\,\text{A} \] \(\textbf{Recover the upper wire's linear mass density:}\) The magnetic force per unit length at \(I_2=10\,\text{A}\) is \[ \frac{F_m}{\ell}=\frac{(2.0\times10^{-7})(20)(10)}{4.0\times10^{-3}} \] \[ \frac{F_m}{\ell}=0.010\,\text{N m}^{-1} \] The measured support force is \(0.020\,\text{N m}^{-1}\). Thus, \[ \lambda g=0.020+0.010 \] \[ \lambda g=0.030\,\text{N m}^{-1} \] \[ \lambda=\frac{0.030}{10} \] \[ \lambda=3.0\times10^{-3}\,\text{kg m}^{-1} \] \(\textbf{Determine the support-unloading current:}\) At the limiting condition, the supports exert zero upward force: \[ \frac{T}{\ell}=0 \] The upward magnetic force then balances the weight: \[ \frac{\mu_0I_1I_{2,c}}{2\pi d}=\lambda g \] Therefore, \[ I_{2,c}=\frac{(0.030)(4.0\times10^{-3})}{(2.0\times10^{-7})(20)} \] \[ I_{2,c}=30\,\text{A} \] \(\textbf{Final answer:}\) \[ \boxed{I_1=20\,\text{A},\qquad \lambda=3.0\times10^{-3}\,\text{kg m}^{-1},\qquad I_{2,c}=30\,\text{A}} \] Increasing the upper-wire current increases the upward magnetic repulsion. The insulating supports consequently need to provide less upward force. The difference between two support-force measurements eliminates the unknown weight and determines the source current. The extrapolated zero-support condition is meaningful only while the stipulated wire separation and orientation are maintained. Beyond that condition, the described upward-only supports cannot maintain the same equilibrium without an additional restraint.

Explanation: \(\textbf{Determine the field produced by P at Q:}\) For a long straight wire, \[ B=\frac{\mu_0I}{2\pi d} \] The current in P points along positive \(z\). At Q, which lies to the right of P, P's magnetic field points along positive \(y\). Its magnitude is \[ B_{P\to Q}=\frac{(2.0\times10^{-7})(12)}{3.0\times10^{-3}} \] \[ B_{P\to Q}=8.0\times10^{-4}\,\text{T} \] Hence, \[ \vec B_{P\to Q}=+8.0\times10^{-4}\hat j\,\text{T} \] \(\textbf{Find the field that eliminates Q's force:}\) The force per unit length on Q is \[ \vec f_Q=I_Q\hat k\times(\vec B_{P\to Q}+\vec B_e) \] Both fields are transverse to the wire. For the force to vanish, their resultant at Q must vanish: \[ \vec B_e=-\vec B_{P\to Q} \] Therefore, \[ \vec B_e=-8.0\times10^{-4}\hat j\,\text{T} \] \(\textbf{Determine Q's magnetic field at P:}\) P lies to the left of Q. A positive-\(z\) current produces a negative-\(y\) field at a point to its left. The field magnitude is \[ B_{Q\to P}=\frac{(2.0\times10^{-7})(4.0)}{3.0\times10^{-3}} \] \[ B_{Q\to P}=\frac{8.0}{3}\times10^{-4}\,\text{T} \] Thus, \[ \vec B_{Q\to P}=-\frac83\times10^{-4}\hat j\,\text{T} \] \(\textbf{Combine both fields acting on P:}\) The external field also points along negative \(y\). Therefore, \[ \vec B_{\mathrm{at}\ P}= -\left(\frac83+8\right)\times10^{-4}\hat j \] \[ \vec B_{\mathrm{at}\ P}=-\frac{32}{3}\times10^{-4}\hat j\,\text{T} \] The force per unit length is \[ \vec f_P=I_P\hat k\times\vec B_{\mathrm{at}\ P} \] \[ \vec f_P=12\hat k\times\left(-\frac{32}{3}\times10^{-4}\hat j\right) \] Using \(\hat k\times(-\hat j)=+\hat i\), \[ \vec f_P=+1.28\times10^{-2}\hat i\,\text{N m}^{-1} \] \(\textbf{Final answer:}\) \[ \boxed{\vec B_e=-8.0\times10^{-4}\hat j\,\text{T},\qquad \vec f_P=+1.28\times10^{-2}\hat i\,\text{N m}^{-1}} \] The two parallel currents attract each other, but the external field acts independently on both wires. Its value is chosen to cancel the magnetic field at Q due to P. That same external field does not cancel the field at P due to Q, since the two currents have different magnitudes and the wire-to-wire field directions are opposite. Consequently, zero total force on Q does not imply zero total force on P. Newton's third law still applies to the mutual wire–wire forces; the additional external-field forces change the overall balance.

Explanation: A long straight wire carrying positive-\(z\) current produces magnetic field along positive \(y\) at points to its right and negative \(y\) at points to its left. First consider arrangement P, where both currents point along positive \(z\). At the left wire, the field due to the right wire points along negative \(y\). At the right wire, the field due to the left wire points along positive \(y\). These two source fields have equal magnitudes but opposite directions. A uniform external field has the same vector at both locations. It cannot cancel both oppositely directed source fields simultaneously. Now consider arrangement Q. The left current points along positive \(z\), while the right current points along negative \(z\). At the right wire, the field from the left wire points along positive \(y\). At the left wire, the reversed right-wire current also produces a field along positive \(y\). Because the current magnitudes and separation are equal, these two fields have the same magnitude: \[ B_w=\frac{\mu_0I}{2\pi d} \] A uniform external field \[ \vec B_e=-\frac{\mu_0I}{2\pi d}\hat j \] cancels the field due to the other wire at each location. The resultant transverse magnetic field acting on each current is therefore zero. Both total magnetic forces vanish. The Assertion is correct. The Reason correctly identifies the field-direction symmetry that permits simultaneous cancellation in Q but prevents it in P. This does not mean that the mutual wire–wire forces cease to exist. In Q, the two currents still repel each other, while external-field forces cancel those mutual forces individually. The distinction is between the mutual interaction and the total magnetic force including an independent external source.

Explanation: \(\textbf{Identify the quantity to maximise:}\) The attractive force per unit length between two long parallel currents is \[ f=\frac{F}{\ell}=\frac{\mu_0I_1I_2}{2\pi d} \] The wire separation is fixed. Therefore, maximising the force requires maximising the positive product \[ I_1I_2 \] The currents cannot be independently increased without limit because their combined resistive power is constrained. \(\textbf{Use the power constraint:}\) The total power is \[ P=R_1I_1^2+R_2I_2^2 \] For two nonnegative terms, \[ R_1I_1^2+R_2I_2^2\geq2\sqrt{R_1R_2}\,I_1I_2 \] Equality occurs when the two terms are equal: \[ R_1I_1^2=R_2I_2^2 \] \(\textbf{Determine the optimal current ratio:}\) Substituting the given resistances, \[ 2I_1^2=8I_2^2 \] \[ I_1^2=4I_2^2 \] The currents are taken as positive magnitudes. Therefore, \[ \frac{I_1}{I_2}=2 \] \(\textbf{Determine the maximum current product:}\) At the optimum, each resistor dissipates half the total power: \[ R_1I_1^2=R_2I_2^2=20\,\text{W} \] Thus, \[ I_1^2=\frac{20}{2}=10\,\text{A}^2 \] \[ I_2^2=\frac{20}{8}=2.5\,\text{A}^2 \] The current product is \[ I_1I_2=\sqrt{(10)(2.5)} \] \[ I_1I_2=5.0\,\text{A}^2 \] \(\textbf{Calculate the maximum magnetic force:}\) The separation is \[ d=2.0\times10^{-3}\,\text{m} \] Therefore, \[ f_{\max}=\frac{(2.0\times10^{-7})(5.0)}{2.0\times10^{-3}} \] \[ f_{\max}=5.0\times10^{-4}\,\text{N m}^{-1} \] \(\textbf{Final answer:}\) \[ \boxed{\frac{I_1}{I_2}=2,\qquad \left(\frac{F}{\ell}\right)_{\max}=5.0\times10^{-4}\,\text{N m}^{-1}} \] The magnetic force depends on the product of the currents, whereas resistive heating depends on their squares. When the wire resistances differ, equal currents do not maximise their product under a fixed power constraint. The optimum divides the electrical power equally between the wires, not the current. The lower-resistance wire consequently carries the greater current. The same-direction condition ensures that the maximised interaction is attractive rather than repulsive.

Explanation: \(\textbf{Identify why the midpoint initially gives zero force:}\) The two fixed outer currents have equal magnitudes and directions. At the exact midpoint, the middle wire is equally distant from both. For either direction of its own current, the two mutual magnetic forces have equal magnitudes and opposite directions. Therefore, the middle wire has zero net force at \(x=0\). However, equilibrium alone does not establish stability. \(\textbf{Determine the changed distances:}\) After displacement towards the right outer wire, \[ x=+0.0010\,\text{m} \] Its distance from the right wire is \[ d_R=a-x \] \[ d_R=0.020-0.001=0.019\,\text{m} \] Its distance from the left wire is \[ d_L=a+x \] \[ d_L=0.020+0.001=0.021\,\text{m} \] \(\textbf{Calculate the competing force magnitudes:}\) For two long parallel currents, \[ f=\frac{\mu_0I_1I_2}{2\pi d} \] The product of outer and middle current magnitudes is \[ I_oI_m=(20)(10)=200\,\text{A}^2 \] For the right wire, \[ f_R=\frac{(2.0\times10^{-7})(200)}{0.019} \] \[ f_R\approx2.1053\times10^{-3}\,\text{N m}^{-1} \] For the left wire, \[ f_L=\frac{(2.0\times10^{-7})(200)}{0.021} \] \[ f_L\approx1.9048\times10^{-3}\,\text{N m}^{-1} \] Their difference is \[ f_R-f_L\approx2.005\times10^{-4}\,\text{N m}^{-1} \] \(\textbf{Determine the force for a same-direction middle current:}\) The middle wire attracts both outer wires and is attracted by each of them. The nearer right wire exerts the stronger attraction towards positive \(x\). Thus, \[ \vec f_{\mathrm{same}}\approx+2.01\times10^{-4}\hat i\,\text{N m}^{-1} \] The force is in the direction of the small displacement. Therefore, the midpoint is unstable along the permitted axis. \(\textbf{Reverse the middle current:}\) The middle wire now repels both outer wires. The nearer right wire repels it more strongly towards negative \(x\). The force changes sign: \[ \vec f_{\mathrm{opposite}}\approx-2.01\times10^{-4}\hat i\,\text{N m}^{-1} \] This force opposes the displacement. The midpoint is therefore stable for the stated one-dimensional motion. \(\textbf{Final answer:}\) \[ \boxed{\vec f_{\mathrm{same}}\approx+2.01\times10^{-4}\hat i\,\text{N m}^{-1},\qquad \vec f_{\mathrm{opposite}}\approx-2.01\times10^{-4}\hat i\,\text{N m}^{-1}} \] The same midpoint can be an equilibrium position for either current sense. Its stability depends on whether a displacement causes a restoring or destabilising imbalance. Attraction draws the middle wire further towards whichever outer wire is nearer. Repulsion pushes it away from the nearer outer wire and back towards the midpoint. The stability conclusion is limited to motion along the line joining the three wires, as explicitly specified.

Explanation: \(\textbf{Determine the field geometry:}\) A positive-\(z\) current in the long wire produces magnetic field along positive \(y\) throughout the loop's region at \(x\gt 0\). Its magnitude decreases with distance: \[ B(x)=\frac{\mu_0I_w}{2\pi x} \] The magnetic force on a straight current-carrying segment is \[ \vec F=I_\ell(\vec L\times\vec B) \] \(\textbf{Calculate the force on the nearer side:}\) The nearer side carries current along positive \(z\), parallel to the long-wire current. The two parallel currents attract. The field there is \[ B_n=\frac{(2.0\times10^{-7})(20)}{0.010} \] \[ B_n=4.0\times10^{-4}\,\text{T} \] The force magnitude is \[ F_n=I_\ell LB_n \] \[ F_n=(5.0)(0.20)(4.0\times10^{-4}) \] \[ F_n=4.0\times10^{-4}\,\text{N} \] The force points towards the long wire: \[ \vec F_n=-4.0\times10^{-4}\hat i\,\text{N} \] \(\textbf{Calculate the force on the farther side:}\) The farther side carries current along negative \(z\), antiparallel to the long-wire current. The interaction is repulsive. At \(x=0.020\,\text{m}\), \[ B_f=\frac{(2.0\times10^{-7})(20)}{0.020} \] \[ B_f=2.0\times10^{-4}\,\text{T} \] Therefore, \[ F_f=(5.0)(0.20)(2.0\times10^{-4}) \] \[ F_f=2.0\times10^{-4}\,\text{N} \] Its direction is away from the long wire: \[ \vec F_f=+2.0\times10^{-4}\hat i\,\text{N} \] \(\textbf{Account for the horizontal connecting sides:}\) The horizontal sides carry currents in opposite \(x\)-directions. At corresponding distances from the straight wire, they experience equal and opposite forces along \(z\). Their integrated contributions cancel. Thus, only the unequal forces on the two vertical sides remain. \(\textbf{Determine the original resultant:}\) \[ \vec F_{\mathrm{net}}=\vec F_n+\vec F_f \] \[ \vec F_{\mathrm{net}}=(-4.0+2.0)\times10^{-4}\hat i \] \[ \vec F_{\mathrm{net}}=-2.0\times10^{-4}\hat i\,\text{N} \] \(\textbf{Reverse the source-wire current:}\) The source magnetic field reverses everywhere while retaining its original magnitude distribution. The loop current and geometry remain unchanged. Every magnetic-force contribution therefore reverses: \[ \vec F'_{\mathrm{net}}=-\vec F_{\mathrm{net}} \] \[ \vec F'_{\mathrm{net}}=+2.0\times10^{-4}\hat i\,\text{N} \] \(\textbf{Final answer:}\) \[ \boxed{\vec F_{\mathrm{net}}=-2.0\times10^{-4}\hat i\,\text{N},\qquad \vec F'_{\mathrm{net}}=+2.0\times10^{-4}\hat i\,\text{N}} \] The parallel nearer side is attracted more strongly than the antiparallel farther side is repelled. The unequal field strengths are responsible for the nonzero resultant force on the closed loop. The horizontal-side forces cancel by symmetry. Reversing only the source current reverses the magnetic field and hence the complete force on the loop. Unlike a closed loop in a uniform field, a loop in this nonuniform field can experience a resultant translational magnetic force.

Explanation: \(\textbf{Determine the initial mutual magnetic force:}\) For two long parallel wires, \[ f=\frac{F}{\ell}=\frac{\mu_0I_1I_2}{2\pi d} \] Using the current magnitudes, \[ I_1=I_2=20\,\text{A} \] and initial separation, \[ d=4.0\times10^{-3}\,\text{m} \] we obtain \[ f=\frac{(2.0\times10^{-7})(20)(20)}{4.0\times10^{-3}} \] \[ f=2.0\times10^{-2}\,\text{N m}^{-1} \] Because the initial currents are antiparallel, the force on each wire is directed away from the other. \(\textbf{Determine the acceleration of each wire:}\) The linear mass density is \[ \lambda=2.0\times10^{-3}\,\text{kg m}^{-1} \] For a long wire segment of length \(\ell\), both the magnetic force and mass are proportional to \(\ell\). Thus, \[ a=\frac{f}{\lambda} \] Substituting, \[ a=\frac{2.0\times10^{-2}}{2.0\times10^{-3}} \] \[ a=10\,\text{m s}^{-2} \] Each wire accelerates outward at this magnitude. \(\textbf{Determine the initial relative acceleration:}\) The two accelerations point in opposite outward directions. The separation consequently accelerates at the sum of their magnitudes: \[ \ddot d=a_1+a_2 \] \[ \ddot d=10+10 \] \[ \ddot d=20\,\text{m s}^{-2} \] \(\textbf{Determine the centre-of-mass acceleration:}\) The mutual forces are equal and opposite. They are internal forces for the two-wire system. With negligible friction and no external force along the guides, \[ \vec F_{\mathrm{external}}=\vec0 \] Therefore, \[ \vec a_{\mathrm{CM}}=\vec0 \] \(\textbf{Reverse one current:}\) In the second trial, the currents become parallel. The magnetic interaction changes from repulsion to attraction. The initial separation and current magnitudes remain unchanged, so the force magnitude on each wire remains the same. Each wire now accelerates towards the other at \(10\,\text{m s}^{-2}\). The initial separation acceleration is therefore \[ \ddot d=-20\,\text{m s}^{-2} \] where the negative sign denotes decreasing separation. \(\textbf{Final answer:}\) \[ \boxed{\ddot d_{\mathrm{opposite}}=+20\,\text{m s}^{-2},\qquad a_{\mathrm{CM}}=0,\qquad \ddot d_{\mathrm{parallel}}=-20\,\text{m s}^{-2}} \] The same mutual-force magnitude acts on both wires, but in opposite directions. The centre of mass therefore remains unaccelerated by their mutual magnetic interaction. Their relative acceleration is the sum of the individual acceleration magnitudes. Reversing one current changes the interaction from repulsive to attractive without changing its initial magnitude. The acceleration values apply only at the initial separation; as the wires subsequently move, the inverse-distance magnetic force changes.