Magnetic Effects of Current and Magnetism MCQs with Answers – Part 3 (NEET Physics)

Explanation: \(\textbf{Determine the conventional current direction:}\) Electrons drift along negative \(x\). Because electrons carry negative charge, conventional current is directed opposite to their drift. Therefore, conventional current flows along positive \(x\). \(\textbf{Calculate the current from charge transport:}\) The magnitude of steady current in the conductor is \[ I=neAv_d \] where \(n\) is free-electron number density, \(e\) is charge magnitude, \(A\) is cross-sectional area and \(v_d\) is drift speed. Substitute the values: \[ I=(5.0\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})(1.0\times10^{-4}) \] Group the numerical factors: \[ I=(5.0)(1.6)\times10^{28-19-6-4} \] \[ I=8.0\times10^{-1} \] \[ I=0.80\,\text{A} \] \(\textbf{Apply the conductor-force relation:}\) The current-direction length vector is \[ \vec L=0.25\hat i\,\text{m} \] The external magnetic field is \[ \vec B=0.50\hat k\,\text{T} \] The magnetic force is \[ \vec F=I(\vec L\times\vec B) \] Substituting, \[ \vec F=0.80(0.25\hat i\times0.50\hat k) \] Using \[ \hat i\times\hat k=-\hat j \] we obtain \[ \vec F=-0.80(0.25)(0.50)\hat j \] \[ \vec F=-0.10\hat j\,\text{N} \] \(\textbf{Microscopic consistency check:}\) The drift velocity of each electron points along negative \(x\). Its velocity–field cross product points along positive \(y\). Multiplication by negative electron charge reverses the force to negative \(y\). This agrees with the force obtained using conventional current. \(\textbf{Final answer:}\) \[ \boxed{\vec F=-0.10\hat j\,\text{N}} \] The current magnitude is determined by electron number density, cross-sectional area and drift speed. The current direction is opposite to electron drift. The macroscopic conductor-force formula uses the conventional current direction. The numerical and directional calculations together connect microscopic charge transport with the magnetic force on a conducting segment.

Explanation: \(\textbf{Determine the field produced by solenoid }P\textbf{:}\) The interior field of an ideal long air-core solenoid is \[ B=\mu_0nI \] For solenoid \(P\), \[ n_P=600\,\text{m}^{-1},\qquad I_P=1.0\,\text{A} \] Therefore, \[ B_P=\mu_0(600)(1.0) \] \[ B_P=600\mu_0\,\text{A m}^{-1} \] Its specified direction is positive \(y\): \[ \vec B_P=600\mu_0\hat j\,\text{A m}^{-1} \] \(\textbf{Determine the field produced by solenoid }Q\textbf{:}\) Similarly, \[ B_Q=\mu_0(400)(1.0) \] \[ B_Q=400\mu_0\,\text{A m}^{-1} \] Its direction is positive \(z\): \[ \vec B_Q=400\mu_0\hat k\,\text{A m}^{-1} \] \(\textbf{Superpose the two uniform fields:}\) At the common interior, \[ \vec B_{\mathrm{net}}=\vec B_P+\vec B_Q \] \[ \vec B_{\mathrm{net}}=\mu_0(600\hat j+400\hat k)\,\text{A m}^{-1} \] The two magnetic fields are perpendicular rather than oppositely directed. They must therefore be added as vectors. \(\textbf{Calculate the force on the rod:}\) The rod carries current along positive \(x\). Its current-direction length vector is \[ \vec L=0.20\hat i\,\text{m} \] The rod current is \[ I_r=2.0\,\text{A} \] The force is \[ \vec F=I_r(\vec L\times\vec B_{\mathrm{net}}) \] Substituting, \[ \vec F=2.0[0.20\hat i\times\mu_0(600\hat j+400\hat k)] \] \[ \vec F=0.40\mu_0[600(\hat i\times\hat j)+400(\hat i\times\hat k)] \] Using \[ \hat i\times\hat j=\hat k,\qquad \hat i\times\hat k=-\hat j \] we obtain \[ \vec F=0.40\mu_0(600\hat k-400\hat j) \] \[ \vec F=\mu_0(-160\hat j+240\hat k)\,\text{N} \] \(\textbf{Determine the force magnitude:}\) \[ F=\mu_0\sqrt{(-160)^2+(240)^2} \] \[ F=\mu_0\sqrt{25600+57600} \] \[ F=\mu_0\sqrt{83200} \] \[ F=80\sqrt{13}\mu_0\,\text{N} \] \(\textbf{Final answer:}\) \[ \boxed{\vec F=\mu_0(-160\hat j+240\hat k)\,\text{N},\qquad F=80\sqrt{13}\mu_0\,\text{N}} \] The two solenoids produce field components along different axes. The rod current is perpendicular to each magnetic-field component, but the two resulting force components also point along different axes. The negative-\(y\) force arises from the positive-\(z\) magnetic-field component. The positive-\(z\) force arises from the positive-\(y\) magnetic-field component. The force magnitude must therefore be obtained by vector addition rather than by adding the individual force magnitudes directly.

Explanation: \(\textbf{Determine the original electromagnetic force:}\) The Lorentz-force relation is \[ \vec F=q(\vec E+\vec v\times\vec B) \] For the original velocity and magnetic field, \[ \vec v\times\vec B =v(\hat i+\hat j)\times B_0\hat k \] Using \[ \hat i\times\hat k=-\hat j,\qquad \hat j\times\hat k=\hat i \] we obtain \[ \vec v\times\vec B=vB_0(\hat i-\hat j) \] Therefore, \[ \vec F_1=q[E_0\hat i+vB_0(\hat i-\hat j)] \] \[ \vec F_1=q[(E_0+vB_0)\hat i-vB_0\hat j] \] \(\textbf{Apply both reversals:}\) In the second trial, \[ \vec v'=-\vec v,\qquad\vec B'=-\vec B \] The new magnetic cross product is \[ \vec v'\times\vec B'=(-\vec v)\times(-\vec B) \] \[ \vec v'\times\vec B'=\vec v\times\vec B \] The electric field is unchanged. Consequently, \[ \vec F_2=q(\vec E+\vec v'\times\vec B') \] \[ \vec F_2=\vec F_1 \] \(\textbf{Compare instantaneous kinetic-energy rates:}\) The instantaneous rate of change of kinetic energy is \[ P=\vec F\cdot\vec v \] The magnetic-force contribution to this scalar product is always zero. Thus, \[ P=q\vec E\cdot\vec v \] Initially, \[ P_1=qE_0v \] After velocity reversal, \[ P_2=q\vec E\cdot(-\vec v) \] \[ P_2=-qE_0v \] Therefore, \[ P_2=-P_1 \] The instantaneous force remains unchanged, but its power reverses sign because the velocity reverses. This distinction is possible because force and power depend differently on velocity. The magnetic contribution depends on the cross product of velocity and field, while electric power depends on the dot product of electric field and velocity. The unchanged force vector does not imply an unchanged rate of energy transfer.

Explanation: \(\textbf{Separate axial and perpendicular motion:}\) The magnetic field is directed along positive \(z\). The magnetic force cannot change the component of velocity parallel to the field. However, the electric field is also directed along \(z\), so it produces axial acceleration. The axial acceleration is \[ a_z=\frac{qE_0}{m} \] \[ a_z=+4.0\,\text{m s}^{-2} \] The initial axial velocity is \[ v_{z0}=-8.0\,\text{m s}^{-1} \] \(\textbf{Find when the velocity becomes perpendicular to the field:}\) The velocity is perpendicular to the magnetic field when its axial component becomes zero. Using \[ v_z=v_{z0}+a_zt \] the condition is \[ 0=-8.0+4.0t \] Therefore, \[ t=2.0\,\text{s} \] \(\textbf{Determine the axial displacement:}\) For constant axial acceleration, \[ z=v_{z0}t+\frac12a_zt^2 \] Substituting, \[ z=(-8.0)(2.0)+\frac12(4.0)(2.0)^2 \] \[ z=-16+8 \] \[ z=-8.0\,\text{m} \] The particle is still on the negative-\(z\) side of its starting point when its axial velocity first becomes zero. \(\textbf{Calculate the transverse circular radius:}\) The electric field has no perpendicular component. Therefore, the perpendicular speed remains \[ v_\perp=6.0\,\text{m s}^{-1} \] The angular frequency of the circular projection is \[ \omega=\frac{qB_0}{m}=5.0\,\text{rad s}^{-1} \] The circular radius is \[ r=\frac{v_\perp}{\omega} \] \[ r=\frac{6.0}{5.0} \] \[ r=1.2\,\text{m} \] \(\textbf{Final answer:}\) \[ \boxed{z=-8.0\,\text{m},\qquad r=1.2\,\text{m}} \] The electric field initially slows the particle's motion along negative \(z\). At the turning instant, the parallel velocity is zero, but the perpendicular circular motion continues. Afterwards, the particle accelerates along positive \(z\). The circular radius remains unchanged because the perpendicular speed and magnetic-field magnitude remain constant. This motion differs from an ordinary constant-pitch helix because the axial velocity changes continuously.

Explanation: \(\textbf{Calculate the electric force on the charged rod:}\) The rod's net charge is \[ Q=5.0\times10^{-3}\,\text{C} \] The external electric field points upward. Therefore, \[ \vec F_E=Q\vec E \] \[ \vec F_E=(5.0\times10^{-3})(40)\hat k \] \[ \vec F_E=+0.20\hat k\,\text{N} \] \(\textbf{Determine the initial magnetic force:}\) For a straight current-carrying conductor, \[ \vec F_B=I(\vec L\times\vec B) \] Initially, \[ \vec L=0.50\hat i\,\text{m} \] and \[ \vec B=+0.30\hat j\,\text{T} \] Thus, \[ \vec F_{B1}=2.0[0.50\hat i\times(+0.30\hat j)] \] \[ \vec F_{B1}=+0.30\hat k\,\text{N} \] \(\textbf{Check the initial force balance:}\) The rod's weight is \[ mg=(0.050)(10) \] \[ mg=0.50\,\text{N} \] It acts downward: \[ \vec W=-0.50\hat k\,\text{N} \] The sum of the three vertical forces is \[ \vec F_{\mathrm{initial}}=(+0.20+0.30-0.50)\hat k \] \[ \vec F_{\mathrm{initial}}=\vec0 \] This confirms that the stated initial equilibrium is physically consistent. \(\textbf{Determine the force after current reversal:}\) The electric force and weight are unchanged because the charge, electric field and mass remain the same. Reversing the current reverses the magnetic-force direction without changing its magnitude. \[ \vec F_{B2}=-0.30\hat k\,\text{N} \] The new net force is \[ \vec F_{\mathrm{net,2}}=(+0.20-0.30-0.50)\hat k \] \[ \vec F_{\mathrm{net,2}}=-0.60\hat k\,\text{N} \] \(\textbf{Calculate the acceleration:}\) Applying Newton's second law immediately after the reversal, \[ \vec a=\frac{\vec F_{\mathrm{net,2}}}{m} \] \[ \vec a=\frac{-0.60\hat k}{0.050} \] \[ \vec a=-12\hat k\,\text{m s}^{-2} \] \(\textbf{Final answer:}\) \[ \boxed{\vec a=-12\hat k\,\text{m s}^{-2}} \] The negative sign indicates a downward acceleration. The current reversal changes only the magnetic contribution; the electric force remains upward while the weight remains downward. The initial magnetic force is upward and helps balance the rod, whereas after reversal it acts downward. The resulting acceleration is twice the magnetic-force change divided by the rod mass, consistent with the initial zero-force condition.

Explanation: \(\textbf{Determine the magnetic field at the initial position:}\) The field of an infinitely long straight wire at radial distance \(r\) is \[ B(r)=\frac{\mu_0I}{2\pi r} \] At points on the positive \(x\)-axis, positive-\(z\) current produces a field along positive \(y\). At \(P=(a,0,0)\), \[ \vec B_P=\frac{\mu_0I}{2\pi a}\hat j \] Define \[ B_P=\frac{\mu_0I}{2\pi a} \] \(\textbf{Apply the initial cancellation condition:}\) The particle moves along positive \(z\). Its magnetic-force direction is \[ \hat k\times\hat j=-\hat i \] Therefore, the magnetic force opposes the electric force. Zero total force requires \[ qE_0=qv_0B_P \] Cancelling \(q\), \[ E_0=v_0B_P \] \(\textbf{Determine the field at the second position:}\) Point \(Q\) is twice as far from the wire. Therefore, \[ B_Q=\frac{\mu_0I}{2\pi(2a)} \] \[ B_Q=\frac{B_P}{2} \] The field direction remains positive \(y\). \(\textbf{Calculate the force at unchanged speed:}\) The electric force remains \[ \vec F_E=+qE_0\hat i \] The magnetic force at \(Q\) is \[ \vec F_{B,Q}=-qv_0B_Q\hat i \] \[ \vec F_{B,Q}=-\frac{qv_0B_P}{2}\hat i \] Using \(v_0B_P=E_0\), \[ \vec F_{B,Q}=-\frac{qE_0}{2}\hat i \] Hence, \[ \vec F_Q=+\frac{qE_0}{2}\hat i \] \(\textbf{Find the new cancellation speed:}\) Let the required speed be \(v'\). For zero force, \[ E_0=v'B_Q \] \[ v'=\frac{E_0}{B_P/2} \] \[ v'=2\frac{E_0}{B_P} \] \[ v'=2v_0 \] \(\textbf{Final answer:}\) \[ \boxed{\vec F_Q=+\frac{qE_0}{2}\hat i,\qquad v'=2v_0} \] The electric field remains uniform, while the wire's magnetic field decreases inversely with distance. Moving the particle farther from the wire therefore weakens only the magnetic contribution. At unchanged speed, the electric force becomes dominant. Doubling the speed restores cancellation at twice the radial distance. This combines the spatial dependence of a wire-generated field with the velocity dependence of the magnetic Lorentz force.

Explanation: \(\textbf{Determine the circular coil's axial magnetic field:}\) For an \(N\)-turn circular coil, \[ B(z)=\frac{\mu_0NIR^2}{2(R^2+z^2)^{3/2}} \] The field at the centre is \[ B_0=\frac{\mu_0NI}{2R} \] Therefore, the axial field may be expressed as \[ B(z)=B_0\frac{R^3}{(R^2+z^2)^{3/2}} \] \(\textbf{Evaluate the field at }z=R\textbf{:}\) Substitute \(z=R\): \[ B(R)=B_0\frac{R^3}{(2R^2)^{3/2}} \] \[ B(R)=\frac{B_0}{2\sqrt2} \] The field points along positive \(z\), as it does at the centre. \(\textbf{Find the required positive-}x\textbf{ speed:}\) For positive charge moving along positive \(x\) in a positive-\(z\) field, the magnetic force points along negative \(y\). It can therefore oppose the positive-\(y\) electric force. The zero-force condition is \[ qE_0=qv'B(R) \] Hence, \[ v'=\frac{E_0}{B(R)} \] Substituting, \[ v'=\frac{E_0}{B_0/(2\sqrt2)} \] \[ v'=2\sqrt2\frac{E_0}{B_0} \] \[ v'=2\sqrt2v_0 \] \(\textbf{Reverse the coil current:}\) Current reversal reverses the magnetic field at every axial point. At \(z=R\), the coil field now points along negative \(z\). For a particle moving along positive \(x\), the magnetic force consequently points along positive \(y\). The electric force also points along positive \(y\). Thus, the two forces reinforce rather than oppose one another. No positive speed along positive \(x\) can produce cancellation after this reversal. The positive-\(x\) velocity direction would itself have to be reconsidered. The problem requires both the spatial variation of the circular-coil field and the vector direction condition for electromagnetic-force cancellation.

Explanation: \(\textbf{Determine the original interaction direction:}\) Two long parallel currents flowing in the same direction attract each other. Therefore, the wires initially attract. The magnitude of their force per unit length is \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] The current magnitudes are \(I\) and \(2I\), and their separation is \(2a\). Thus, \[ \left(\frac FL\right)_{\mathrm{initial}} =\frac{\mu_0(I)(2I)}{2\pi(2a)} \] \[ \left(\frac FL\right)_{\mathrm{initial}} =\frac{\mu_0I^2}{2\pi a} \] \(\textbf{Determine the original midpoint magnetic field:}\) At \(O\), each wire is at distance \(a\). Let \[ b=\frac{\mu_0I}{2\pi a} \] The left positive-\(z\) wire produces \(+b\hat j\). The right positive-\(z\) wire produces \(-2b\hat j\). Therefore, \[ \vec B_{O,\mathrm{initial}}=b\hat j-2b\hat j \] \[ \vec B_{O,\mathrm{initial}}=-b\hat j \] Its magnitude is \(b\). \(\textbf{Reverse the right-wire current:}\) The two currents now flow in opposite directions. Therefore, the wires repel. The magnitudes of both currents and their separation remain unchanged. Hence, the interaction-force magnitude per unit length remains \(F\). \(\textbf{Calculate the new midpoint field:}\) Reversing the right-wire current reverses its magnetic field at \(O\). The two contributions now point along positive \(y\): \[ \vec B_{O,\mathrm{final}}=b\hat j+2b\hat j \] \[ \vec B_{O,\mathrm{final}}=3b\hat j \] Therefore, \[ \frac{B_{O,\mathrm{final}}}{B_{O,\mathrm{initial}}}=3 \] The interaction changes from attraction to repulsion while keeping the same force magnitude. The midpoint field increases because the two local field contributions change from opposing to reinforcing. The force between the wires and the resultant field at a selected point are different physical quantities. They must not be inferred from one another merely by using the same current directions.

Explanation: \(\textbf{Apply the parallel-wire force relation:}\) For two sufficiently long parallel wires, \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] The currents are \[ I_1=3.0\,\text{A},\qquad I_2=5.0\,\text{A} \] The initial separation is \[ d_1=0.050\,\text{m} \] \(\textbf{Calculate the original force per unit length:}\) \[ \left(\frac FL\right)_1 =(2.0\times10^{-7})\frac{(3.0)(5.0)}{0.050} \] \[ \left(\frac FL\right)_1 =(2.0\times10^{-7})(300) \] \[ \left(\frac FL\right)_1=6.0\times10^{-5}\,\text{N m}^{-1} \] \(\textbf{Calculate the force on the specified length:}\) The length is \[ L=0.20\,\text{m} \] Therefore, \[ F_1=\left(\frac FL\right)_1L \] \[ F_1=(6.0\times10^{-5})(0.20) \] \[ F_1=1.2\times10^{-5}\,\text{N} \] The currents initially flow in the same direction, so this force is attractive. \(\textbf{Apply the changed separation and current direction:}\) The new separation is \[ d_2=\frac{0.050}{2}=0.025\,\text{m} \] The current magnitudes are unchanged. Because force per unit length is inversely proportional to separation, \[ F_2=F_1\frac{d_1}{d_2} \] \[ F_2=(1.2\times10^{-5})\frac{0.050}{0.025} \] \[ F_2=2.4\times10^{-5}\,\text{N} \] Reversing the second current makes the currents oppositely directed. Thus, the final force is repulsive. \(\textbf{Final answer:}\) \[ \boxed{F_1=1.2\times10^{-5}\,\text{N}\ \text{attractive},\qquad F_2=2.4\times10^{-5}\,\text{N}\ \text{repulsive}} \] The first calculation requires distinguishing force per unit length from the force on a specified wire length. Halving the separation doubles the force magnitude. Reversing one current changes the interaction direction but not the magnitude arising from the current product. These effects must be considered independently.

Explanation: \(\textbf{Determine the initial force from the leftmost wire:}\) The middle and leftmost wires carry currents in the same direction. They therefore attract. The leftmost wire lies to the left of the middle wire, so this force points along negative \(x\). Their separation is \(a\). The magnitude per unit length is \[ f_L=\frac{\mu_0(2I)(I)}{2\pi a} \] Using the reference value, \[ f_L=2f_0 \] Therefore, \[ \vec f_L=-2f_0\hat i \] \(\textbf{Determine the initial force from the rightmost wire:}\) The rightmost wire lies at \(x=3a\), while the middle wire lies at \(x=a\). Their separation is \[ d_R=3a-a=2a \] Both currents initially flow along positive \(z\), so the force is attractive. The middle wire is therefore pulled towards positive \(x\). Its magnitude per unit length is \[ f_R=\frac{\mu_0(4I)(I)}{2\pi(2a)} \] \[ f_R=2f_0 \] Hence, \[ \vec f_R=+2f_0\hat i \] \(\textbf{Find the original resultant:}\) \[ \vec f_{\mathrm{initial}}=\vec f_L+\vec f_R \] \[ \vec f_{\mathrm{initial}}=-2f_0\hat i+2f_0\hat i \] \[ \vec f_{\mathrm{initial}}=\vec0 \] The middle wire experiences two nonzero forces that cancel. \(\textbf{Reverse the rightmost current:}\) The middle and rightmost currents now have opposite directions. Their interaction becomes repulsive. Because the rightmost wire lies to the right, it repels the middle wire towards negative \(x\). The rightmost interaction magnitude remains \(2f_0\). Therefore, \[ \vec f'_R=-2f_0\hat i \] The leftmost interaction remains \[ \vec f_L=-2f_0\hat i \] \(\textbf{Calculate the new resultant:}\) \[ \vec f_{\mathrm{final}}=-2f_0\hat i-2f_0\hat i \] \[ \vec f_{\mathrm{final}}=-4f_0\hat i \] \(\textbf{Final answer:}\) \[ \boxed{\vec f_{\mathrm{initial}}=\vec0,\qquad \vec f_{\mathrm{final}}=-4f_0\hat i} \] Initially, the two outer wires exert equal and opposite forces on the middle wire. Reversing the rightmost current changes its contribution from attraction to repulsion. Both interactions then push or pull the middle wire towards negative \(x\). Zero initial resultant force does not mean that the individual magnetic interactions vanish.

Explanation: The magnetic field of an infinitely long straight wire at perpendicular distance \(d\) is \[ B=\frac{\mu_0I}{2\pi d} \] The field produced by wire \(Q\), which carries current \(3I\), at wire \(P\) is \[ B_{P\leftarrow Q}=\frac{\mu_0(3I)}{2\pi d} \] The field produced by wire \(P\) at wire \(Q\) is \[ B_{Q\leftarrow P}=\frac{\mu_0I}{2\pi d} \] Therefore, the first field is three times the second. Statement I is correct. The force per unit length on wire \(P\) arises from its own current \(I\) interacting with the field produced by \(Q\). Its magnitude is \[ f_P=I B_{P\leftarrow Q} \] \[ f_P=\frac{3\mu_0I^2}{2\pi d} \] For wire \(Q\), \[ f_Q=3I B_{Q\leftarrow P} \] \[ f_Q=\frac{3\mu_0I^2}{2\pi d} \] Thus, both interaction-force magnitudes are equal. Statement II is incorrect, while Statement III is correct. This equality is consistent with Newton's third law. The currents initially flow in the same direction, so the wires attract. Reversing both current directions preserves their relative current directions. They therefore continue to attract. Statement IV is correct. The different source-field magnitudes do not produce unequal mutual force magnitudes. Each wire's force depends on the product of its own current and the field generated by the other wire. The same product \(I_1I_2\) appears in both interaction-force calculations. This distinction is important when interpreting magnetic fields and interaction forces in systems containing unequal currents.

Explanation: \(\textbf{Determine the pairwise interactions:}\) All three currents flow in the same direction. Therefore, every pair of wires attracts. Each pair is separated by the same distance \(a\). Consequently, each pairwise force per unit length has magnitude \[ f_0=\frac{\mu_0I^2}{2\pi a} \] \(\textbf{Resolve the forces on one selected wire:}\) Choose any vertex of the equilateral triangle. The other two wires attract the selected wire along the two sides meeting at that vertex. These two force vectors have equal magnitude \(f_0\). The angle between the two sides of an equilateral triangle is \(60^\circ\). Therefore, the resultant magnitude is \[ f_{\mathrm{net}}= \sqrt{f_0^2+f_0^2+2f_0^2\cos60^\circ} \] Using \[ \cos60^\circ=\frac12 \] we obtain \[ f_{\mathrm{net}}= f_0\sqrt{2+1} \] \[ f_{\mathrm{net}}=\sqrt3f_0 \] The resultant lies along the angle bisector of the two equal force vectors. In an equilateral triangle, this angle bisector passes through the circumcentre. Thus, the selected wire is pulled towards \(O\). \(\textbf{Determine the field at the circumcentre:}\) Each wire is at the same distance from \(O\). Therefore, each produces a magnetic field of equal magnitude at \(O\). The three field vectors have directions separated by \(120^\circ\), owing to the threefold rotational symmetry. Their vector sum is zero: \[ \vec B_O=\vec0 \] This cancellation of magnetic fields at the circumcentre does not imply that the wires experience zero force. The forces on a given wire are evaluated at that wire's own position, where the fields from the other two wires do not cancel. \(\textbf{Final answer:}\) \[ \boxed{f_{\mathrm{net}}=\sqrt3f_0\ \text{towards }O,\qquad \vec B_O=\vec0} \] The problem involves two different vector additions at different locations. At the circumcentre, symmetry cancels the three magnetic-field vectors. At a vertex, the two attractive interaction forces reinforce along the inward angle bisector. Equal current directions produce attraction between each pair, but the complete three-wire configuration still requires vector superposition.

Explanation: \(\textbf{Force on the middle wire due to the leftmost wire:}\) The leftmost current is \(+I\), and the middle current is \(+2I\). They flow in the same direction and therefore attract. Since the leftmost wire lies at \(x=0\), the force on the middle wire points along negative \(x\). Their separation is \(a\). The force per unit length is \[ f_L=\frac{\mu_0(I)(2I)}{2\pi a} \] \[ f_L=2f_0 \] Hence, \[ \vec f_L=-2f_0\hat i \] \(\textbf{Initial force from the rightmost wire:}\) The middle wire lies at \(x=a\), while the rightmost wire lies at \(x=3a\). Their separation is \[ d_R=2a \] Initially, their currents are \(+2I\) and \(-3I\). Opposite currents repel. The rightmost wire therefore repels the middle wire towards negative \(x\). The force magnitude per unit length is \[ f_R=\frac{\mu_0(2I)(3I)}{2\pi(2a)} \] \[ f_R=3f_0 \] Thus, \[ \vec f_R=-3f_0\hat i \] \(\textbf{Initial resultant:}\) \[ \vec f_{\mathrm{initial}} =-2f_0\hat i-3f_0\hat i \] \[ \vec f_{\mathrm{initial}}=-5f_0\hat i \] \(\textbf{Reverse the rightmost current:}\) After reversal, the rightmost current becomes \(+3I\). The middle and rightmost currents now flow in the same direction. They therefore attract, and the force from the rightmost wire points along positive \(x\). Its magnitude remains \(3f_0\): \[ \vec f'_R=+3f_0\hat i \] The leftmost contribution remains \(-2f_0\hat i\). Therefore, \[ \vec f_{\mathrm{final}}=-2f_0\hat i+3f_0\hat i \] \[ \vec f_{\mathrm{final}}=+f_0\hat i \] \(\textbf{Final answer:}\) \[ \boxed{\vec f_{\mathrm{initial}}=-5f_0\hat i,\qquad \vec f_{\mathrm{final}}=+f_0\hat i} \] Initially, both outer wires exert forces towards negative \(x\) on the middle wire. Reversing the rightmost current changes its interaction from repulsion to attraction. The rightmost force then opposes the leftmost force and has the larger magnitude. The resultant therefore reverses direction and becomes smaller. The calculation requires using signed force directions, not simply adding the magnitudes of pairwise interactions.

Explanation: \(\textbf{Relate the plotted slope to the current product:}\) The force per unit length between two long parallel wires is \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] For a fixed wire length \(L\), \[ F=\frac{\mu_0I_1I_2L}{2\pi}\left(\frac1d\right) \] Therefore, a graph of \(F\) against \(1/d\) is a straight line through the origin. Its slope is \[ s=\frac{\mu_0I_1I_2L}{2\pi} \] \(\textbf{Determine the unknown current:}\) Rearranging, \[ I_2=\frac{s}{[\mu_0/(2\pi)]I_1L} \] The measured slope is \[ s=1.2\times10^{-6}\,\text{N m} \] Substituting, \[ I_2=\frac{1.2\times10^{-6}} {(2.0\times10^{-7})(4.0)(0.25)} \] The denominator is \[ (2.0\times10^{-7})(4.0)(0.25) =2.0\times10^{-7} \] Therefore, \[ I_2=6.0\,\text{A} \] \(\textbf{Calculate the force at the specified separation:}\) The final separation is \[ d=0.20\,\text{m} \] The force magnitude on the \(0.25\,\text{m}\) length is \[ F=\frac{\mu_0}{2\pi}\frac{I_1I_2L}{d} \] Substituting, \[ F=(2.0\times10^{-7})\frac{(4.0)(6.0)(0.25)}{0.20} \] \[ F=(2.0\times10^{-7})(30) \] \[ F=6.0\times10^{-6}\,\text{N} \] \(\textbf{Determine the interaction character:}\) The original currents flowed in the same direction, so their interaction was attractive. Reversing the second current makes their directions opposite. The final interaction is therefore repulsive. \(\textbf{Final answer:}\) \[ \boxed{I_2=6.0\,\text{A},\qquad F=6.0\times10^{-6}\,\text{N}\ \text{repulsive}} \] The graph slope contains the product of the two currents and the specified wire length. It does not directly give force per unit length. The current reversal changes the direction of the interaction but not its magnitude at a given separation. The graph measurement and changed-current condition must therefore be interpreted separately.

Explanation: \(\textbf{Determine the interaction with wire }Q\textbf{:}\) Wire \(P\) carries current \(I\) along positive \(z\). Wire \(Q\) carries current \(2I\) in the same direction. Parallel currents flowing in the same direction attract. Wire \(Q\) lies at positive \(x\), a distance \(a\) from \(P\). Therefore, the force on \(P\) due to \(Q\) points along positive \(x\). Its magnitude per unit length is \[ f_{PQ}=\frac{\mu_0(I)(2I)}{2\pi a} \] \[ f_{PQ}=2f_0 \] Thus, \[ \vec f_{PQ}=+2f_0\hat i \] \(\textbf{Determine the interaction with wire }R\textbf{:}\) Wire \(R\) carries current \(3I\) along negative \(z\). Its current direction is opposite to that of wire \(P\). Therefore, wires \(P\) and \(R\) repel. Wire \(R\) lies at positive \(y\). Repulsion pushes wire \(P\) away from \(R\), along negative \(y\). The separation is again \(a\). Thus, \[ f_{PR}=\frac{\mu_0(I)(3I)}{2\pi a} \] \[ f_{PR}=3f_0 \] Therefore, \[ \vec f_{PR}=-3f_0\hat j \] \(\textbf{Combine the two force vectors:}\) The resultant magnetic force per unit length on wire \(P\) is \[ \vec f_P=\vec f_{PQ}+\vec f_{PR} \] \[ \vec f_P=2f_0\hat i-3f_0\hat j \] \[ \vec f_P=f_0(2\hat i-3\hat j) \] Its magnitude is \[ f_P=f_0\sqrt{2^2+(-3)^2} \] \[ f_P=\sqrt{13}f_0 \] \(\textbf{Final answer:}\) \[ \boxed{\vec f_P=f_0(2\hat i-3\hat j)} \] The wire at positive \(x\) attracts the selected wire, while the wire at positive \(y\) repels it. The two contributions therefore act along perpendicular directions. Their magnitudes depend on the respective products of current magnitudes. The resultant must be found by vector addition rather than ordinary scalar addition. The force is directed into the fourth quadrant of the \(xy\)-plane, providing a useful directional check.

Explanation: For two sufficiently long parallel wires, \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] In the specified ideal reference arrangement, \[ I_1=I_2=1\,\text{A} \] and \[ d=1\,\text{m} \] Using the customary vacuum relation, \[ \frac{\mu_0}{2\pi}\approx2\times10^{-7}\,\text{N A}^{-2} \] the force per unit length is approximately \[ \frac FL=2\times10^{-7}\,\text{N m}^{-1} \] Statement I is correct. For a corresponding length \(L=0.50\,\text{m}\), the ideal interaction-force magnitude is \[ F=\left(\frac FL\right)L \] \[ F=(2\times10^{-7})(0.50) \] \[ F=1\times10^{-7}\,\text{N} \] Statement II is also correct. Statement III is incorrect because it treats the historical force-based description as the present formal SI definition. The contemporary SI definition of the ampere is linked to the fixed numerical value of the elementary charge. The parallel-wire force relation remains a useful physical result and a traditional instructional benchmark. The two currents initially flow in the same direction, so their interaction is attractive. If one current reverses, their directions become opposite. Oppositely directed parallel currents repel. Because their magnitudes and separation are unchanged, the interaction-force magnitude remains unchanged. Therefore, Statement IV is correct. The distinction between force and force per unit length is essential when applying the reference relation to a specified finite observation length. The ideal-wire assumptions and vacuum condition must also be respected.

Explanation: \(\textbf{Determine the gravitational force per unit length:}\) The upper wire's mass per unit length is \[ \lambda=2.0\times10^{-3}\,\text{kg m}^{-1} \] Its weight per unit length is \[ w=\lambda g \] Substituting, \[ w=(2.0\times10^{-3})(10) \] \[ w=2.0\times10^{-2}\,\text{N m}^{-1} \] This force acts vertically downward. \(\textbf{Calculate the initial magnetic interaction:}\) The force per unit length between two long parallel wires is \[ f=\frac{\mu_0I_1I_2}{2\pi d} \] The current magnitudes are \[ I_1=10\,\text{A},\qquad I_2=20\,\text{A} \] Their separation is \[ d=2.0\times10^{-3}\,\text{m} \] Therefore, \[ f=(2.0\times10^{-7})\frac{(10)(20)}{2.0\times10^{-3}} \] \[ f=(2.0\times10^{-7})(1.0\times10^5) \] \[ f=2.0\times10^{-2}\,\text{N m}^{-1} \] This equals the upper wire's weight per unit length. \(\textbf{Infer the initial current directions:}\) To support the upper wire, the magnetic interaction must act upward. Since the lower wire lies below it, an upward interaction requires repulsion. Parallel wires repel when their currents flow in opposite directions. Therefore, the initial currents must be oppositely directed. \(\textbf{Apply the current reversal:}\) Reversing the lower current makes the two currents flow in the same direction. The interaction becomes attractive. The magnetic force on the upper wire therefore points downward, towards the lower wire. Its magnitude per unit length remains equal to \(f\). Immediately after reversal, the separation is still \(d\). \(\textbf{Determine the instantaneous net force per unit length:}\) Gravity and magnetic attraction both act downward: \[ f_{\mathrm{net}}=w+f \] \[ f_{\mathrm{net}}=0.020+0.020 \] \[ f_{\mathrm{net}}=0.040\,\text{N m}^{-1} \] \(\textbf{Calculate the acceleration:}\) Using force per unit length divided by mass per unit length, \[ a=\frac{f_{\mathrm{net}}}{\lambda} \] \[ a=\frac{0.040}{2.0\times10^{-3}} \] \[ a=20\,\text{m s}^{-2} \] \(\textbf{Final answer:}\) \[ \boxed{a=20\,\text{m s}^{-2}\ \text{downward}} \] The initial currents must be opposite because repulsion is required to balance gravity. After reversal, the magnetic interaction becomes attractive. Gravity and magnetic attraction then act in the same downward direction. The acceleration is evaluated immediately after reversal, before the wire separation changes appreciably. The force per unit length is the appropriate quantity because the wire's mass is specified per unit length.

Explanation: \(\textbf{Express the magnetic interaction using the variable current split:}\) For two long parallel wires, \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] The total current is fixed: \[ I_1+I_2=8.0\,\text{A} \] Let the current in the first branch be \(I_1=x\). The other current is then \[ I_2=8.0-x \] Therefore, \[ \frac FL=\frac{\mu_0}{2\pi d}x(8.0-x) \] The separation and permeability are fixed. Maximising the magnetic interaction is therefore equivalent to maximising the current product. \(\textbf{Determine the current distribution giving the maximum:}\) The product is \[ P=x(8.0-x) \] Expand and complete the square: \[ P=8.0x-x^2 \] \[ P=16-(x-4.0)^2 \] Because a square is nonnegative, \[ P\leq16\,\text{A}^2 \] Equality occurs when \[ x=4.0\,\text{A} \] Thus, the interaction is largest when the currents are equal: \[ I_1=I_2=4.0\,\text{A} \] \(\textbf{Calculate the maximum force per unit length:}\) \[ f_{\max}=(2.0\times10^{-7})\frac{(4.0)(4.0)}{0.040} \] \[ f_{\max}=(2.0\times10^{-7})(400) \] \[ f_{\max}=8.0\times10^{-5}\,\text{N m}^{-1} \] \(\textbf{Calculate the initial interaction:}\) Initially, \[ I_1=2.0\,\text{A},\qquad I_2=6.0\,\text{A} \] Therefore, \[ f_{\mathrm{initial}} =(2.0\times10^{-7})\frac{(2.0)(6.0)}{0.040} \] \[ f_{\mathrm{initial}}=6.0\times10^{-5}\,\text{N m}^{-1} \] \(\textbf{Compare the two forces:}\) \[ \frac{f_{\max}}{f_{\mathrm{initial}}} =\frac{8.0\times10^{-5}}{6.0\times10^{-5}} \] \[ \frac{f_{\max}}{f_{\mathrm{initial}}}=\frac43 \] \(\textbf{Final answer:}\) \[ \boxed{f_{\max}=8.0\times10^{-5}\,\text{N m}^{-1},\qquad \frac{f_{\max}}{f_{\mathrm{initial}}}=\frac43} \] The interaction depends on the product of the branch currents, not merely on their sum. For a fixed positive sum, the product is maximised when the currents are equal. The currents still flow in the same direction, so the interaction remains attractive. Redistributing current can therefore change the attraction even though the total supplied current remains fixed. The force increase is determined by the change in current product at unchanged separation.

Explanation: \(\textbf{Interaction between wires }P\textbf{ and }Q\textbf{:}\) The current in \(P\) is along positive \(z\), while that in \(Q\) is along negative \(z\). Oppositely directed parallel currents repel. Since \(Q\) lies to the right of \(P\), the force on \(P\) due to \(Q\) points along negative \(x\). Their separation is \(a\). Therefore, \[ f_{PQ}=\frac{\mu_0(I)(2I)}{2\pi a} \] \[ \vec f_{PQ}=-2f_0\hat i \] \(\textbf{Interaction between wires }P\textbf{ and }R\textbf{:}\) Both currents are directed along positive \(z\). The wires therefore attract. Since \(R\) lies above \(P\), the force on \(P\) points along positive \(y\). Thus, \[ f_{PR}=\frac{\mu_0(I)(3I)}{2\pi a} \] \[ \vec f_{PR}=+3f_0\hat j \] \(\textbf{Interaction with the diagonally placed wire:}\) Wire \(S\) carries current opposite to that in \(P\). The force is repulsive and points away from \(S\), along the diagonal towards negative \(x\) and negative \(y\). Their separation is \[ d_{PS}=\sqrt{a^2+a^2}=\sqrt2a \] The force magnitude per unit length is \[ f_{PS}=\frac{\mu_0(I)(4I)}{2\pi\sqrt2a} \] \[ f_{PS}=2\sqrt2f_0 \] The negative diagonal direction has unit vector \[ \hat n=-\frac{\hat i+\hat j}{\sqrt2} \] Therefore, \[ \vec f_{PS}=2\sqrt2f_0\left(-\frac{\hat i+\hat j}{\sqrt2}\right) \] \[ \vec f_{PS}=-2f_0\hat i-2f_0\hat j \] \(\textbf{Add the three pairwise forces:}\) \[ \vec f_P=\vec f_{PQ}+\vec f_{PR}+\vec f_{PS} \] \[ \vec f_P=-2f_0\hat i+3f_0\hat j-2f_0\hat i-2f_0\hat j \] \[ \vec f_P=f_0(-4\hat i+\hat j) \] \(\textbf{Final answer:}\) \[ \boxed{\vec f_P=f_0(-4\hat i+\hat j)} \] The force is directed predominantly towards negative \(x\), with a smaller positive-\(y\) component. The diagonal wire contributes to both Cartesian components. It is incorrect to add the three pairwise force magnitudes directly because their directions differ. The net force on one wire must be determined from the magnetic fields generated by the other wires, not from its own field.

Explanation: \(\textbf{Determine the initial cancellation condition:}\) For a long straight wire, \[ B=\frac{\mu_0I}{2\pi r} \] Initially, both source currents flow along positive \(z\). Between the wires, their magnetic fields point in opposite directions. At \[ x_0=\frac{4a}{5} \] the distances from \(P\) and \(Q\) are \[ r_P=\frac{4a}{5} \] and \[ r_Q=4a-\frac{4a}{5}=\frac{16a}{5} \] The corresponding field magnitudes are \[ B_P=\frac{\mu_0I}{2\pi(4a/5)} \] \[ B_P=\frac54\frac{\mu_0I}{2\pi a} \] and \[ B_Q=\frac{\mu_0(4I)}{2\pi(16a/5)} \] \[ B_Q=\frac54\frac{\mu_0I}{2\pi a} \] They cancel because their directions are opposite. \(\textbf{Locate the new zero-field position:}\) After reversing \(Q\), the source currents are opposite. Their fields reinforce between the wires, so cancellation cannot occur there. The weaker current belongs to \(P\), so the possible cancellation point lies outside, to the left of \(P\). Let its coordinate be \(x=-s\), where \(s\gt 0\). Its distances from \(P\) and \(Q\) are \(s\) and \(4a+s\). For cancellation, \[ \frac{\mu_0I}{2\pi s} = \frac{\mu_0(4I)}{2\pi(4a+s)} \] Cancelling common factors, \[ \frac1s=\frac4{4a+s} \] \[ 4a+s=4s \] \[ s=\frac{4a}{3} \] Thus, \[ x=-\frac{4a}{3} \] \(\textbf{Force at the original test-wire position:}\) Define \[ b=\frac{\mu_0I}{2\pi a} \] At \(x=4a/5\), the field from \(P\) is \(+\frac54b\hat j\). After reversal, the field from \(Q\) also points along positive \(y\), with magnitude \(5b/4\). Hence, \[ \vec B_{\mathrm{net}}=\frac52b\hat j \] The test-wire current is \(2I\) along positive \(z\). Its force per unit length is \[ \vec f=(2I\hat k)\times\left(\frac52b\hat j\right) \] Since \(\hat k\times\hat j=-\hat i\), \[ \vec f=-5Ib\hat i \] Using \(Ib=f_0\), \[ \vec f=-5f_0\hat i \] \(\textbf{Final answer:}\) \[ \boxed{x_{\mathrm{zero}}=-\frac{4a}{3},\qquad \vec f_{\mathrm{original\ position}}=-5f_0\hat i} \] Reversing one source current moves the cancellation point from between the sources to the exterior region. At the original position, two previously opposing magnetic fields now reinforce. The force on the test wire consequently becomes nonzero and points towards negative \(x\). The field-zero location is determined by the source currents and geometry, independently of the nonzero current carried by the test wire.