Moving Charges and Magnetism Exemplar Solutions Class 12

Strategic angle. Rather than equating radius and pitch separately, use the gyration frequency _c = qB/m which alone fixes both shape and sense. This single scalar contains all the kinematic information about the helix and saves you two equations.

1. Identical helices ⇒ equal |_c|, hence equal |q/m|. (The magnitude of _c sets both the radius via v_⊥/_c and the pitch via 2π v_∥/_c.)
2. Opposite sense ⇒ _c values are opposite in sign, hence q/m values are opposite in sign. The sense of rotation about B is the sign of _c, full stop.
3. Combining: (q/m)₁ = -(q/m)₂, i.e. (q/m)₁ + (q/m)₂ = 0, which is option (D).
4. Alternative angle –- physical interpretation. A positive charge spirals one way and a negative charge of the same |q/m| spirals the other way at the same rate. The particle and ``mirror'' particle trace out identical helices in opposite senses; this is exactly the scenario in the problem.
5. Cross-check with options. (A) Equal z-momenta would require equal m v_∥, but m can differ between the two while m/|q| stays fixed. (B) Equal charges contradicts the opposite-sense requirement. (C) Particle–antiparticle pairs do satisfy (D), but the converse fails: an e^- and μ⁺ also satisfy (e/m)₁ + (e/m)₂ ≠ 0 in general but a proton and an electron with carefully chosen kinetic energies do not form a particle–antiparticle pair.

Why this matters. Helical motion is fully determined by q/m and the velocity; mass and charge separately are not needed. This is why mass spectrometers measure m/q, not m.

Option (D).

Quick reading. The Biot–Savart formula contains a cross product; cross products always yield a vector perpendicular to both factors. That single structural fact answers the entire question without any geometry.

1. B ∝ v × r is the entire content of the law for a point charge.
2. Perpendicularity is a property of the cross product, not of any special geometry. So (A) holds in every configuration.
3. Magnitude scales with |v|sinθ / r², ruling out (C) (inverse cube) and (D) (along r).
4. Alternative method –- direct vector computation. Pick an electron moving along +x with r along +y; then v× r ∝ z. The field at that observation point is along ± z, which is perpendicular to both v (x) and r (y). The construction generalises: any other choice of r still gives a field perpendicular to v.
5. Unit/structure check. [₀] = T m/A, [qv/r²] = A m^-1; product gives Tesla. Correct.

Why this matters. The perpendicularity of B to v is what causes magnetic field lines to form closed loops around a current; if B were along v they would diverge from the charge like an electric field. This is why no magnetic monopoles are observed –- the very algebra of Biot–Savart forbids a radial B.

Option (A).

Picture-first. Two perpendicular half-discs of equal area behave like the legs of a right triangle: the vector sum is 2 times one leg, not 2 times. The bend has split a single z-aligned moment of magnitude M₀ into two halved moments along z and x, and their orthogonal addition yields M₀/2.

1. Each half-disc carries half the original area. Treating each as a half-loop with the wire current I, the magnitude of each moment is |M₁| = |M₂| = Iπ R²/2 = M₀/2.
2. These two moments are perpendicular (z and x). Their resultant has magnitude √(M₀/2)² + (M₀/2)² = M₀/2 ≈ 0.707 M₀.
3. M₀/2 ≈ 0.71 M₀ < M₀. So the moment shrinks.
4. Alternative method –- continuous deformation. Imagine slowly bending the right half through angles 0,π/6,π/3,π/2. The z-component of the right half's moment falls as cosφ while a new x-component grows as sinφ. At φ = π/2 the z-component of the right half is 0, the x-component is M₀/2, and combined with the unchanged left half (M₀/2 along z) the resultant magnitude is M₀/2. Same answer, by tracking the geometry.
5. Numerical sanity check. M₀/2 = 0.707 M₀, so the on-axis dipole field at large z falls by the same factor 1/2 ≈ 0.71 (since dipole field ∝ M). The field does decrease, confirming options (C), (D) are wrong.

Why this matters. For a vector quantity, geometry matters: splitting along orthogonal directions does not preserve the magnitude. This is the same effect that makes the magnetic moment of a folded coil less than that of the flat one –- a fact exploited in magnetic-shielding design.

Option (A).

Strategic angle. The cross product vanishes when the two vectors are parallel. Spot this and the answer is immediate; no calculation required.

1. Velocity is along axis; field is along axis. Parallel.
2. v × B = 0, so F_mag = 0.
3. No force, no acceleration. Uniform motion continues.
4. Alternative method –- decompose the velocity. Write v = v_∥n + v_⊥. The general motion in a solenoid would be helical: the parallel piece slides along n unaffected, and the perpendicular piece circles at _c = eB/m. Here v_⊥ = 0, so the circle collapses to a point and only the parallel slide remains –- straight-line uniform motion.
5. Concept linkage. This is exactly the principle of magnetic confinement in fusion devices: charged particles spiral tightly around field lines but slide freely along them, just like the electron here.

Why this matters. A solenoid acts as a velocity filter: charges moving along its axis are unaffected, while charges with a transverse component spiral. The same physics underlies charged-particle beams in mass spectrometers and electron-microscope columns.

Option (D).

Strategic angle. Acceleration includes change in direction, not just change in speed. Once that is clear, (A) is the only choice; the magnetic-field dee region and the gap each provide their own non-zero a.

1. Centripetal acceleration in the dees: speed fixed, direction rotates, a_c = v²/r pointed toward the centre of the arc, a ≠ 0.
2. Tangential acceleration in the gap: speed grows under F = qE, magnitude a_t = qE/m, a ≠ 0.
3. Net: acceleration is non-zero throughout the orbit, so (A).
4. Quantitative check. For protons in a 1.5 T cyclotron at radius r = 0.5 m: v = qBr/m = (1.610^-19)(1.5)(0.5)/(1.67× 10^-27) ≈ 7.2× 10⁷ m/s, so a_c = v²/r ≈ 1.0× 10¹⁶ m/s². Even inside the dee the acceleration is enormous –- option (A) is unambiguously right.
5. Alternative angle –- energy ledger. In the dee, the magnetic force does no work (F⊥ v), so kinetic energy stays constant. In the gap, the electric force does positive work each time, and the kinetic energy grows. Both regions feature non-zero a, even though the energy story is different in each.

Why this matters. The cyclotron is the textbook demonstration that constant-magnitude circular motion is still accelerated motion –- a fact that students often miss when they conflate ``speed'' with ``velocity''. Real cyclotrons exploit this duality: the dees provide the geometric (centripetal) turning, the gap provides the energy injection.

Option (A).

Strategic angle. Identify the rotation axis first; the behaviour of M follows. The axis-direction is doing all the work in this problem.

1. Loop plane ⊥ M, so the rotation axis (perpendicular to plane) is parallel to M.
2. A vector rotated about its own direction is unchanged: the rotation matrix about M fixes M.
3. Energy depends only on M· B. With M unchanged, U unchanged, W = 0.
4. Alternative method –- torque integral. Work done by an external agent rotating the loop by dθ at angle θ against the field torque is dW_ext = -_ext· dθ = MBsinθ dθ, but here dθ is along M and τ = M×B is perpendicular to M; their dot product is zero for every dθ. So ∫ dW = 0 irrespective of the 30^∘. Same answer.
5. Concept linkage. A free coil in a magnetic field precesses about its own axis (like a top about gravity) –- the precession does no work against the field for exactly this reason. Galvanometer suspension wires exploit this to allow rotation without energy loss into the magnetic background.

Why this matters. It explains why a freely-rotating coil about its own axis in a uniform field does not exchange energy with the field; only the alignment with B matters.

Option (D).

Structural observation. The cancellation of r and v in the ratio M/L is the key. Once that is seen, the orbit dependence disappears and the answer follows from the sign of q alone.

1. Magnetic moment of orbit: M = qvr/2 (treat orbit as current loop with effective current I = q/T = qv/(2π r), area π r²).
2. Orbital angular momentum: L = mvr.
3. Ratio: γ = q/(2m), no r or v or n.
4. Sign of γ = sign of q. Electron: q = -e, so γ = -e/(2m_e) < 0.
5. Alternative method –- Bohr-radius formulation. In Bohr's model r_n = n² a₀, v_n = v₁/n. So M⁽ⁿ⁾ = qv_nr_n/2 = (q/2)(v₁/n)(n² a₀) = (nq v₁ a₀)/2 and L_n = nm v₁ a₀. Their ratio M⁽ⁿ⁾/L_n = q/(2m), with all the n-dependence cancelling. Same answer.
6. Concept linkage. The Bohr magneton _B = e/(2m_e) pops out from this when L = is substituted. So γ is the constant of proportionality between angular momentum and the magnetic dipole moment –- a recurring theme in atomic and nuclear physics.

Why this matters. The same logic gives the electron's spin g-factor close to -1 for orbital motion; spin requires the relativistic correction g ≈ -2. The orbital gyro-magnetic ratio is the simplest manifestation of a deep symmetry between angular momentum and magnetic moment.

Options (A), (B).

Strategic angle. Separate force on charges (drift electrons) from force on the bulk (ions). Magnetic forces do no work on the carrier they push, but the carriers can transfer momentum via internal fields.

1. Drift electrons feel transverse F; they pile up on the wire surface. (B) follows.
2. For a wire moving under B: ions move with the wire. F on ions is ⊥ u, so no work on ions. (D) follows.
3. (A) and (C) are too sweeping: (A) ignores Hall accumulation; (C) confuses ``no work on the charges'' with ``no work on the wire''.

Why this matters. This is the principle behind every electric motor: magnetic forces do no work on individual carriers, yet a motor delivers mechanical power because the carriers transmit the force to the lattice. The energy ledger ties back to the battery maintaining the current, not to the magnetic field as a source.

Alternative angle –- microscopic check. The instantaneous power on a single drift electron is F· v = q(v× B)· v = 0 identically. Sum over ∼ 10²⁸ m^-3 electrons and the integrated power is still zero per electron. The macroscopic power comes from the Hall electric field, which is a real, work-doing E even though it originated from a magnetic field.

Options (B), (D).

Strategic angle. Distinguish ``integral = 0'' from ``integrand = 0''. The first holds because currents cancel; the second does not. The line integral is a global property; the integrand is a local one.

1. I_enc = I - I = 0 ⇒ integral = 0.
2. A scalar that equals 0 has the same value under reversal of the integration sense. (B) follows.
3. Local B ⊥ d on parts of C is consistent with non-zero B. (C) follows.
4. Field-cancellation everywhere on C (option D) would require B = 0 globally, contradicted by Biot–Savart for each loop (each loop alone produces a non-zero field everywhere except on its own axis).
5. Alternative method –- superposition. Apply Amp`ere's law to each loop separately. For loop 1 alone: _C B₁ · d= +₀ I. For loop 2 alone: _C B₂ · d= -₀ I (opposite sense). Their sum is zero. The local field B = B₁ + B₂ is non-zero at any generic point.
6. Concept linkage. This is the principle behind anti-Helmholtz coils used in magneto-optical traps: opposing currents give zero net field at the centre but a strong gradient that confines cold atoms.

Why this matters. Helmholtz coils and anti-Helmholtz coils exploit exactly this difference between net enclosed current and local field. Modern ion-trap experiments depend on getting this distinction right.

Options (B), (C).

Strategic angle. Track how each force depends on the sign of charge and the sign of velocity. Electric: only on q. Magnetic: on q v, so on the product. This separation immediately sorts the options.

1. Electric force flips between e^- and e⁺ (different q). Accelerations are equal in magnitude but opposite. Rule out (A).
2. Magnetic force is the same on both because both q and v flip and their product is unchanged. (B), and consequently (D) for the CM.
3. Power F_E·v flips twice (in q and in v), so it is the same number. (C).
4. Alternative method –- CM coordinates. The CM of the electron–positron system moves at (mv + m(-v))/(2m) = 0 at t = 0. The net external electric force is -eE + (+eE) = 0, so E never accelerates the CM. The net magnetic force is 2F_B^(e-) ≠ 0 in general, so it alone drives the CM. (D) follows from Newton's-second-law for the CM.
5. Concept linkage. Annihilation experiments (such as positron-emission tomography, PET) rely on the symmetric kinematics of an e⁺e^- pair: the two photons emitted in opposite directions can be tracked precisely because the magnetic environment treats both particles identically.

Why this matters. This is exactly the symmetry that makes e⁺ e^- pairs travel along the same trajectory inside a uniform B but opposite to each other inside a uniform E. The electric–magnetic split is central to particle-physics detector design.

Options (B), (C), (D).

Strategic angle. List all the ways to make the Lorentz force vanish; that gives the answer set immediately. The Lorentz force has exactly two pieces (electric, magnetic), and each can be killed in a specific way.

1. No fields: no force. Always works (D).
2. Magnetic only: force vanishes when v ∥ B (so v×B = 0). (A) works.
3. Both fields: balance them out, E + v×B = 0. Equivalently E = -v×B, a velocity-selector geometry. (B) works.
4. Electric only: F = qE≠ 0 always (for q≠ 0). Nothing can cancel an electric force on a stationary or moving charge in pure E. (C) fails.
5. Alternative method –- think about it kinematically. Constant velocity ⇔ a = 0 ⇔ F = 0. In each option, find conditions on v that make q(E + v× B) = 0. Only (C) admits no such v for a non-zero charge.
6. Numerical sanity for (B). In a velocity selector with E = 10⁴ V/m and B = 10^-2 T (crossed), the selected speed is v = E/B = 10⁶ m/s –- a typical thermionic-electron speed. The configuration is physically accessible, not pathological.

Why this matters. The crossed-fields configuration of (B) is the J. J. Thomson velocity selector that isolated a specific v from the cathode beam. It is also the working principle of the Wien filter used to purify ion beams in modern mass spectrometers.

Options (A), (B), (D).

Quick check. Read off each dimension from the SI base units and multiply. Dimensional analysis is a sanity tool, not just a verification trick –- if the dimensions disagree, a numerical calculation cannot save you.

1. [e] = A s, [B] = kg A^-1 s^-2 (from F = qvB so [B] = [F]/[qv] = kg m s^-2/(A ss^-1) = kg A^-1 s^-2).
2. [eBm] = (A s)(kg A^-1 s^-2)kg = s^-1. Verified.
3. Alternative method –- track the units of motion. The cyclotron period is T = 2π m/(qB), with [T] = s. Hence [qB/m] = s^-1. The factor of 2π is dimensionless, so ω = qB/m and ω = 2π/T have the same dimensions automatically.
4. Numerical sanity. For an electron in B = 1 T: _c = (1.6× 10^-19)(1)/(9.1× 10^-31) ≈ 1.76× 10¹¹ rad/s, i.e. ∼ 28 GHz –- microwave range, which is exactly where electron paramagnetic resonance (EPR) operates. Order of magnitude consistent.

Why this matters. The dimension T^-1 is the universal signature of frequency or rate. Every classical-mechanics formula yielding a frequency must reduce to a T^-1 combination of input parameters –- there is no other route.

Dimension T^-1.

Logical angle. Use proof by contradiction. This is a short question, but it admits a few flavours of proof; the contradiction flavour is the cleanest.

1. Assume F does no work and is velocity-independent.
2. Pick a v along F (possible since F≠0). Then F·v = |F||v|≠ 0. Contradiction with ``no work''.
3. Hence F must depend on v.
4. Alternative method –- geometric. A non-zero F defines a fixed direction in space. For F· v = 0 to hold over a trajectory, v must always stay in the plane perpendicular to F. This severely constrains the motion and is generally impossible for arbitrary initial conditions –- unless F rotates with v, i.e. depends on v.
5. Concept linkage. Constraint forces (e.g. normal reaction on a particle constrained to a surface) also do no work; they too depend on the motion (specifically on the velocity direction at every instant). The magnetic force and constraint forces share this defining property.

Why this matters. The fact that magnetic forces do no work forces all the energy book-keeping in a generator/motor to be done by the electric field. This is the cornerstone of electromechanical energy conversion –- magnetic fields mediate the coupling but never sit on either side of the energy ledger.

No-work non-zero forces are necessarily velocity-dependent.

Strategic angle. Field transformations rescue Newton's second law from apparent frame-dependence. The slick way to see this: E and B are not separately physical objects but two faces of a single object (the electromagnetic field tensor), and the transformation of that single object preserves the force.

1. In frame S: F = q(E + v×B).
2. In frame S' moving with velocity u: E' = E + u×B, B' = B (non-relativistic limit). Plugging in v' = v - u gives the same F.
3. Mass is invariant; a is the same. Newton's laws hold in every inertial frame, and they say the same thing.
4. Alternative method –- worked example. A wire at rest with current I in lab frame S produces a magnetic field but no electric field. A free charge moving along the wire feels a pure magnetic force in S. Now look at the rest frame S' of the charge –- in S', the wire's positive ions move backward and the electrons either stay put or move at a different speed. Length contraction makes the wire slightly charged in S', producing an electric field in S' that exerts the same force on the (now-stationary) charge. Magnetism in S has become electricity in S'; the force is the same.
5. Concept linkage. This thought experiment is the starting point of Einstein's 1905 paper on special relativity. The frame-mixing of E and B forced him to rebuild kinematics from scratch.

Why this matters. The frame-mixing of E and B is the seed of special relativity's full Lorentz transformation of the field tensor.

Magnetic part of the force varies, but a does not.

Strategic angle. Compare the oscillation period of the rf to the half-revolution time inside a dee. Once you have the ratio of periods, the qualitative answer drops out.

1. Half-revolution time: T_c/2 = π m/(qB) (one dee transit).
2. Rf half-period: T_rf/2 = 1/(2 f_rf). Doubling f_rf halves T_rf, so T_rf/2 = T_c/4 –- one quarter of a dee transit.
3. The rf reverses twice during one dee-transit: accelerate then decelerate. Net energy gain ≈ 0 per orbit, so the particle's kinetic energy plateaus.
4. Alternative method –- phase diagram. Plot the oscillating gap voltage V(t) = V₀sin(2π f_rf t) and the particle's gap-crossing times t_n = n T_c/2. At resonance, V(t_n) = ± V₀ alternately, always with the sign that accelerates. At f_rf = 2f_c, V(t_n) = V₀sin(2π n) = 0 every crossing –- the gap voltage is zero each time the particle arrives, so no energy transfer.
5. Numerical check (proton in B=1 T). f_c = qB/(2π m) = (1.610^-19)/(2π· 1.6710^-27) ≈ 15.3 MHz. Doubling to 30.6 MHz, the rf cycles in 33 ns versus the dee transit time of 65 ns: rf reverses twice per transit, exactly as above.

Why this matters. Cyclotrons are tuned by adjusting the rf to match the cyclotron frequency; for relativistic particles the synchrocyclotron lowers f_rf slowly as the particle's mass grows. Modern synchrotrons keep a fixed circular path by ramping both B and f_rf together –- both are tuned to maintain the resonance condition.

Resonance broken; particle circulates at fixed radius.

Picture-first. Look along -x (i.e. down wire 1). The B-field circles the x-axis; at O₂ (above the axis) it points along -y. Wire 2's current direction at O₂ is also along y. Parallel current and field ⇒ zero force.

1. Direction of B at O₂ from wire 1: -y (right-hand rule: thumb along x = direction of I₁, fingers curl from above the wire toward -y at z = d).
2. Direction of I₂ at O₂: +y.
3. Cross product of parallel/antiparallel vectors vanishes: y×(-y) = 0. Hence F = 0 at O₂.
4. Alternative method –- parallel-current rule. The standard ``force per unit length on parallel wires'' formula ₀ I₁ I₂/(2π d) applies only when the wires are parallel to each other. Here, wires 1 and 2 are perpendicular, so the simple formula does not apply; you must use F = I× B directly. At the specific point O₂, the parallel-ness of B(O₂) and I₂ makes the integrand zero.
5. What about elsewhere on wire 2? Away from O₂, wire 2 is no longer at the closest approach; B from wire 1 there is no longer purely along -y. The local force per length there is non-zero. The question only asks about O₂, where the geometry forces zero.

Why this matters. The local force on a current element can vanish even when the two wires interact strongly elsewhere. Total force and torque on the wires require integrating along their lengths, which is what gives the standard parallel-wire result for two infinite parallel wires.

F = 0 at O₂.

Strategic angle. Each quarter contributes a perpendicular field component along one axis. Sum them as a 3D vector. The symmetry of the configuration forces the result to lie along the body diagonal (1,1,1)/3 of the positive octant.

1. Quarter loop field magnitude = 1/4 full loop = ₀ I/(8R). The factor of 1/4 follows directly from Biot–Savart applied to a circular arc subtending an angle of π/2 instead of 2π.
2. Three orthogonal contributions, equal magnitude, along x,y,z. By right-hand rule, the x-y-plane quarter contributes along +z (current sense determines sign), the y-z-plane quarter along +x, the z-x-plane quarter along +y.
3. Resultant magnitude 3 times one component; direction (1,1,1)/3 (the body diagonal of the unit cube in the positive octant).
4. Alternative method –- Biot–Savart from scratch. For the x-y-plane quarter: d= R dφ (-sinφ,cosφ,0), r from element to origin is -r_element. Integrating ₀^π/2d×r/r² = π/(2R²)z and prepending ₀ I/(4π) gives ₀ I/(8R)z. The full computation is tedious; the recipe via fraction of a full circle is cleaner.
5. Cross-check the magnitude. If all three quarters were in the same plane, total moment along z would be 3× ₀ I/(8R) = 3₀ I/(8R). Spreading them orthogonally reduces this to 3 ₀ I/(8R), a factor of 3/3 = 1/3 smaller. Same effect as in Q4.3.

Why this matters. The 3D symmetry (three mutually perpendicular quarter-arcs) forces the field to lie along the body diagonal of the positive octant. This is the simplest geometry in which a vector field acquires equal components along three Cartesian axes by construction, not by accident.

B = ₀ I8R(x+y+z).

Strategic angle. Use the cyclotron frequency as the canonical T^-1 scale; build dimensionless ratios from E/B and a velocity. Both constructs have natural physical meanings.

1. eB/m has dimension s^-1 (cyclotron frequency). This is the rate at which a charge gyrates around B; the cancellation of r and v in the dimensional check is the same cancellation that made the cyclotron a useful accelerator.
2. E/B has dimension of m/s; dividing by c gives a pure number E/(cB). Physically, this is the ratio of the characteristic ``drift speed'' E/B to the speed of light –- small when the system is non-relativistic.
3. Either result also follows from inserting SI base units directly: [E/B] = V m^-1/T = V/(T m) = m s^-1, then dividing by c in m/s gives dimensionless.
4. Alternative method –- velocity selector reading. In a velocity selector, the condition F = 0 gives E = vB, so v = E/B. The ratio E/(cB) = v/c is just β, the relativistic speed parameter. So E/(cB) is a physically meaningful pure number, not a mathematical oddity.
5. Alternative inverse-time quantity. If the problem allowed a length scale , then ω = eE/(m) does not have time dimensions; instead √eE/(m) does. Without a length, eB/m is the only natural rate.

Why this matters. Dimensional analysis classifies which quantities can appear together in a physically meaningful combination. The same kind of analysis is what guides theorists in building new physical theories: any new law must have consistent dimensions on both sides, and dimensionless combinations correspond to physically meaningful ratios (think of fine-structure constant, Mach number, Reynolds number).

_c = eB/m ([T]^-1);   E/(cB) (dimensionless).

Strategic angle. Separate the roles: B produces the circle; E pumps energy in to grow it. The radius r = mv/(eB) grows as v grows because E does positive work.

1. B must be perpendicular to the plane of motion. Plane is x-y ⇒ B ∥ z. Take B = B₀z with B₀ > 0.
2. For energy gain, E·v > 0 on average. With initial v = v₀x, take E along -x so that the force -eE on the electron is along +x, parallel to v. (Recall q = -e for the electron, so a -x field accelerates it along +x.)
3. Result: circular motion radius grows with time, producing the outward spiral. The kinetic energy grows linearly in time (since dK/dt = F_E·v = eE₀ v and v increases), and r = mv/(eB₀) tracks v.
4. Alternative configuration. A radial inward E (in the spiral plane) would also work –- the inward component continuously decelerates the radial drift while accelerating tangentially. The uniform-E scenario chosen here is the simplest, and the one most commonly sketched in NCERT-Exemplar diagrams.
5. Why ``spiral down'' is a slight subtlety. The problem says the electron spirals down (i.e. in a plane parallel to xy). The spiral itself is in xy with growing radius; a small E<sub>z</sub> would tilt the motion toward -z, but the dominant motion stays in the xy plane. The statement of the question is consistent with the ``radius-growing'' spiral interpretation.

Why this matters. Crossed E-B configurations are the basis of magnetrons (in microwave ovens), cycloid drives for cathode-ray oscilloscopes, and ion-trap diagnostics. The energy-gain mechanism in each case follows the same logic: E does the work, B provides the geometry.

B = B₀z and E = -E₀x.

Strategic angle. Compute F₁₂ and F₂₁ explicitly; observe asymmetry. The whole content of the question is in this asymmetry.

1. Field of element 1 at element 2: by Biot–Savart, dB₁₂ = ₀4π I di× jR² = ₀ I d4π R²k. Non-zero.
2. Force on element 2 due to this field: dF₁₂ = I dj× dB₁₂ = ₀ I²(d)²4π R²(j× k) = ₀ I²(d)²4π R²i.
3. Field of element 2 at element 1: ∝ j×(-j) = 0. Zero. So dF₂₁ = 0.
4. Forces follow; they are not equal and opposite. Newton-III violated: dF₁₂ + dF₂₁ = dF₁₂ ≠ 0.
5. Concept linkage –- field momentum. The ``missing'' momentum is carried by the electromagnetic field itself. The Poynting vector S = (E×B)/₀ encodes a momentum density S/c², and the rate of change of field momentum exactly compensates the mechanical imbalance. Closed loops integrate this out automatically.
6. Alternative angle –- complete-loop check. If we imagine wires 1 and 2 closed into full loops, the line integral ∮ d₁× (d₂× r) is symmetric under exchange (use the BAC-CAB identity). Newton-III is restored at the integrated level.

Why this matters. Field momentum closes the conservation gap; for full loops the loop integrals restore F₁₂=-F₂₁. This was the historical motivation for Maxwell to introduce the ``electromagnetic momentum'' in his 1873 Treatise –- without it, Newton's laws fail for radiating systems.

Newton's third law fails for isolated current elements.

Picture-first. The galvanometer plus R₁ alone gives the 2V scale. Adding R₂ in series raises the range to 20V; adding R₃ further raises to 200V. The structure is a ladder of series resistors with taps at each level.

1. 2V branch: R₁ = V₁/I_g - G = 2000 - 10 = 1990 Ω. (At full-scale, I_g = 1 mA flows through G + R₁, so the voltage across the combination is I_g(G+R₁) = 2 V.)
2. 20V branch: R₁ + R₂ = V₂/I_g - G = 20 000 - 10 = 19 990 Ω. Hence R₂ = 19 990 - 1990 = 18 000 Ω.
3. 200V branch: R₁ + R₂ + R₃ = V₃/I_g - G = 199 990 Ω. Hence R₃ = 199 990 - 19 990 = 180 000 Ω.
4. Alternative method –- ratio check. Each range is a factor of 10 larger than the previous. So the total series resistance is also a factor of 10 larger: 2000 : 20000 : 200000 = 1:10:100. The increments are R₁ = 1990, R₂ = 18000, R₃ = 180000 –- in the ratio 1:9:90, consistent with the powers-of-ten scaling.
5. Sanity check on impedance. A 200 V voltmeter built from this ladder has total resistance 200 kΩ, giving V/I_g = 200 V at full scale. Drawing only 1 mA from a 200 V source is the design intent: a voltmeter must draw negligibly small current to avoid disturbing the circuit.

Why this matters. A multi-range voltmeter is built by adding series resistance; a multi-range ammeter (next set) is built by adding shunts in parallel. The duality is exact and forms the design vocabulary of every analogue measurement instrument.

R₁ ≈ 1990 Ω, R₂ = 18 kΩ, R₃ = 180 kΩ.

Strategic angle. Set magnetic repulsion equal to weight; solve for h. The geometry (anti-parallel currents) guarantees repulsion, which then balances gravity at a unique equilibrium height.

1. Magnetic upward force per unit length: ₀ I²/(2π h) (anti-parallel, so repulsive, lifting PQ upward).
2. Weight per unit length: m g/L.
3. Equate: h = ₀ I²2π (m g/L) = ₀ I² L2π m g.
4. Plug numbers: h = (4π× 10^-7)(625)(1)2π(2.5× 10^-3)(9.8) = 2.5× 10^-44.9× 10^-2 ≈ 5.1× 10^-3 m.
5. Stability check. Is the equilibrium stable? If PQ rises above h, f decreases (∝ 1/h) while gravity stays constant; net force becomes downward, pulling PQ back. If PQ falls below h, f increases and pushes PQ back up. So h ≈ 5.1 mm is a stable equilibrium.
6. Order-of-magnitude check. 25 A is a sizeable current; 2.5 g is a light wire. The result h ∼ 5 mm is in the millimetre range –- reasonable for a tabletop demonstration. If the current were halved to 12.5 A, h would scale as I², falling to ∼ 1.3 mm; if doubled to 50 A, h rises to ∼ 20 mm. The scaling is consistent with the force-balance formula.

Why this matters. The same balance underlies every magnetic levitation demonstration with parallel wires, the operating principle of magnetic-track maglev trains, and the original SI definition of the ampere. The force formula ₀ I₁ I₂/(2π d) is the most-tested calculation in this chapter.

h ≈ 5 mm.

Picture-first. Only CD experiences a net magnetic force, because only CD is inside the field region. Compute that one force and balance it. The other three arms (AB, BC, AD) lie outside the field and contribute nothing.

1. Force per turn on CD: F₁ = B I = (0.2)(4.9)(0.01) = 9.8× 10^-3 N.
2. Total over 100 turns: F = 100 F₁ = 0.98 N.
3. Additional weight needed: m g = 0.98 ⇒ m = 0.1 kg = 100 g.
4. Direction check. The current in CD flows along (say) +x, and B acts inward along -y (B in xz-plane perpendicular to coil in xy-plane). Then F = I× B ∝ x× (-y) = -z. If the coil hangs with -z pointing downward, F is downward, requiring an additional mass on the balance arm to restore equilibrium. Consistent with the problem statement.
5. Alternative method –- torque-free balance. Set the problem up as a balance: pre-existing balance is undisturbed because the coil's weight is already balanced by 500 g. The only new force is the magnetic force on CD, which acts at the far end of the lever. To balance a downward F, an additional weight W = F must be added on the opposite pan (assuming equal lever arms). W = m g = F ⇒ m = F/g.
6. Numerical sanity. The 500 g pre-balance is a red herring –- it cancels with the coil's weight. The only relevant numbers are B, I, , N and g. Trusting the formula yields 100 g without ambiguity.

Why this matters. A current balance is a classic absolute measurement: a known B and converts a current into a mechanical force, allowing I to be calibrated from first principles. This was the foundation of the pre-2019 SI ampere definition (see the inlinerecall in Q4.22).

m = 100 g.

Strategic angle. The two wires carry different currents because their resistances differ; each contributes a torque about the rod-axis, with a cos 45^∘ factor from the field tilt. The asymmetry of the currents is what creates the net torque –- in a symmetric loop the two torques would cancel.

1. Currents: V₀/R and V₀/(2R), summing to 3V₀/(2R). The two branches are in parallel; the rods are equipotential connectors.
2. Force per wire: F = B I (wire ⊥ field component in the plane).
3. Lever arm about rod-axis: d/2. Tilt factor: cos 45^∘ = 1/2 enters because the force direction makes a 45^∘ angle with the perpendicular to the rod-axis.
4. Torque: τ = (I₁+I₂) B·(d/2)·(1/2) = 3 V₀ Bd42 R.
5. Alternative method –- magnetic-moment approach. The rectangular loop has effective magnetic moment M = NIAn where n is its normal. But here the ``effective I'' must average the two unequal currents. The result is the same as the direct force-balance computation, but the algebra is less transparent –- the explicit force-on-each-arm method is preferred for asymmetric circuits.
6. Numerical sanity for orders of magnitude. For V₀ = 10 V, R = 10 Ω, = 0.1 m, d = 0.1 m, B = 0.1 T: τ = 3· 10· 0.1· 0.1· 0.1/(42· 10) = 5.3× 10^-4 N·m. Small but measurable –- consistent with bench-scale magnetometry.

Why this matters. Unequal currents in opposite arms is the signature of a non-uniform loop in a magnetic field; this is the operating principle of asymmetric torque sensors and tilt-meters in industrial sensor design.

τ = 3 V₀ B d42 R.

Strategic angle. Two coplanar circles of equal radius R are non-intersecting iff the separation between their centres exceeds 2R. Choose momentum directions that displace centres far apart. The released points are 1.5R apart; we need to move the centres at least 0.5R further apart, which requires a sideways displacement of the centres relative to the line joining the start points.

1. Radius of each orbit: r = p/(eB) = R.
2. For electron at origin with momentum p ∥ k, the magnetic force is F = -ek× B₀i = -eB₀(k× i) = -eB₀j, so the centre lies at (0,-R,0).
3. For positron at (0,0,1.5R) with momentum ∥ k, the force is +eB₀j, so the centre is at (0,+R,1.5R).
4. Separation: √(2R)² + (1.5R)² = 2.5R > 2R. Circles do not intersect.
5. Geometric picture. The two starting points are on the z-axis at z = 0 and z = 1.5R. By choosing momenta along k, the magnetic forces push the centres to ±j (opposite signs because the charges differ). The centres end up at the diagonal corners of a 2R× 1.5R rectangle, and the diagonal length 2.5R exceeds 2R –- the circles are tangent-free.
6. Alternative configurations that work. Both momenta along -k also gives separation 2.5R (by symmetry). Momenta along j or -j keep the centres in the z-direction only, and separation falls below 2R as shown in the main solution. So the line-joining-the-start-points (k) is the optimal direction.

Why this matters. Orienting the momenta along the line joining the release points maximises the separation of the centres perpendicular to that line. This is the same idea behind storage-ring design: pair-creation events deposit electrons and positrons in specific orbits, and the geometry ensures the two beams stay clear of each other for many revolutions.

Both momenta along k (or both along -k).

Strategic angle. Same current, same total wire length; only NA changes with shape. Triangle has more turns but small area; hexagon has fewer turns but large area. The interplay between N and A is the heart of the question.

1. I = V₀/R throughout (the resistance is set by the total wire length, which is fixed at 12a).
2. Triangle: N=4, A=(3/4)a² ⇒ NA = 3 a².
3. Square: N=3, A=a² ⇒ NA = 3 a².
4. Hexagon: N=2, A=(33/2)a² ⇒ NA = 33 a².
5. M = N I A = NA· V₀/R. Listed in step (iv).
6. Numerical comparison. 3 ≈ 1.73, 3 ≈ 3.0, 33 ≈ 5.2. So hexagon's moment is about 3 times the triangle's, square's about 1.7 times. The order matches the isoperimetric ranking: more sides ⇒ more area-per-perimeter ⇒ larger M.
7. Circle limit. A circle of circumference 12a has radius r = 12a/(2π) = 6a/π and area π r² = 36a²/π. With N = 1 (single circular turn) and the same I = V₀/R: M_∘ = 36 a²/π · V₀/R ≈ 11.5 a² V₀/R. This is the limiting maximum, exceeding even the hexagon's value.

Why this matters. For the same length of wire, the magnetic moment is maximised by the shape with the largest area-per-unit-perimeter ratio: hexagon > square > triangle, with the limiting case being a circle. This is the isoperimetric inequality in electromagnetic disguise.

M:M:M_hex = 3 : 3 : 33.

Strategic angle. Use Amp`ere's law to dodge the z-integral in part (b); for part (c) do the integral directly. The contrast between the two approaches is itself the lesson –- one bypasses the field, the other works through it.

1. (L) = _-L^LB_z dz. B_z > 0 on the entire axis ⇒ (L) increases. Increment is (L+ε) - (L) = 2 B_z(L) ε > 0 for any ε > 0.
2. Close the axis with an arc at infinity to form a loop enclosing one turn of current I. Apply Amp`ere: (∞) = ₀ I. The closing arc at infinity contributes zero because the dipole field falls as 1/r³ while the arc length grows only as r.
3. Direct integral: substitute z = Rtanθ, dz = Rsec²θ dθ, (z²+R²)^3/2 = R³sec³θ. So _-∞^∞dz(z²+R²)^3/2 = _-π/2^π/2Rsec²θR³sec³θ dθ = 1R²_-π/2^π/2cosθ dθ = 2R², so (∞) = ₀ I R² · (1/R²) = ₀ I.
4. Square coil: same Amp`ere argument ⇒ same (∞) = ₀ I. Finite-L values differ in detail because the on-axis B_z(z) for a square coil has a different (closed-form) expression involving arctan, but the integral _-∞^∞B_z dz is shape-blind.
5. Concept linkage –- gauge of the result. The ₀ I is exactly the line integral that Amp`ere's law gives for any closed loop enclosing the current; the z-axis, extended to infinity and closed at infinity, is just one choice of such a loop.

Why this matters. The shape of the current loop drops out at infinity; only the total enclosed current matters. This is a preview of the more general statement that, asymptotically, only the multipole moments determine the field –- and the leading ``monopole'' term is what counts in the line integral here.

(∞) = ₀ I for both circular and square loops.

Strategic angle. Use the Ayrton-shunt simplification: at each range, I_g flows through G plus the shunts not in the parallel path, and the remainder I - I_g flows through the shunts in the parallel path. Equating voltages gives one equation per range, and three ranges give three equations in three unknowns.

1. 10 mA range: I_g(G + S₂ + S₃) = (10 mA - 1 mA)S₁ ⇒ G + S₂ + S₃ = 9 S₁.
2. 100 mA range: G + S₃ = 99(S₁ + S₂).
3. 1 A range: G = 999(S₁ + S₂ + S₃).
4. Solve: S₁ + S₂ + S₃ = G/999 = 10/999.
   S₂ + S₃ = (G/99) - S₁ - S₂ + ⋯ (algebra).
   Standard solution: S₁ = G/9, S₂ = G/99 - G/9, S₃ = G/999 - G/99 (Ayrton shunt).
5. Numerical sanity. Using the simple single-shunt formula S = I_g G/(I - I_g): S₁ = 10/9 ≈ 1.11 Ω, S₂ = 10/99 ≈ 0.101 Ω, S₃ = 10/999 ≈ 0.0100 Ω. These are the NCERT-Exemplar approved values: each range uses a single shunt selected by the rotary switch, giving the same overall current sensitivity in the galvanometer.
6. Why values are in Ω, not kΩ. Ammeters are low-impedance devices, drawing current with minimal voltage drop. The shunts are tens of mΩ to hundreds of mΩ –- never kΩ.

Why this matters. The Ayrton (universal) shunt keeps the parallel combination of G and shunt nearly constant across ranges, which is what makes the meter scale linearly. This is the preferred design for multi-range analogue ammeters because it preserves the deflection-vs-current calibration as you switch ranges.

S₁ ≈ 1.11 Ω, S₂ ≈ 0.101 Ω, S₃ ≈ 0.0100 Ω.

Strategic angle. Use the symmetry trick: the total of all five fields is zero. Switching A off subtracts B_A; reversing A subtracts 2B_A. The symmetric configuration provides a clean reference (zero), and any perturbation is then read off as the difference from that reference.

1. Total of five symmetric vectors = 0. The five contributions B_A, B_B, B_C, B_D, B_E each have magnitude ₀ I/(2π R) and are equally spaced at 72^∘; like the five edges of a closed pentagon, they sum to zero.
2. Remove A: net = 0 - B_A = -B_A. Magnitude ₀ I/(2π R).
3. Reverse A: A's contribution flips sign, so net change is -2B_A. Net field = 0 - 2B_A = -2B_A, magnitude ₀ I/(π R).
4. Direction sanity for (b) and (c). The field B_A at the axis is tangent to the circle of radius R centred on wire A, perpendicular to the line OA. After switching A off or reversing A, the resultant is along -B_A –- in the same plane, perpendicular to OA, and oppositely directed.
5. Alternative method –- direct vector sum (no symmetry). Add the four remaining vectors explicitly for part (b), each at 72^∘ from the previous. The sum collapses to -B_A by trigonometric identity _n=0⁴cos(72^∘ n) = 0, etc. The symmetric approach saves the algebra.

Why this matters. Many ``n-fold symmetric'' problems reduce to subtracting one term from a zero total –- this is the trick to use whenever the full symmetric set sums to zero. Phased-array antennas and multipole expansions both exploit this algebraic structure.

(a) 0;   (b) ₀ I/(2π R);   (c) ₀ I/(π R).