Magnetism block: Moving Charges and Magnetism, Magnetism and Matter, Electromagnetic Induction. Twelve questions went wrong. Each one is rebuilt below — correct answer, the concept behind it, the exact trap that caught it, and the one line to lock in.
The 12 errors are not spread evenly. Two of the three chapters are holding up; one has collapsed completely.
| Chapter | Asked | Wrong | Question numbers missed | Accuracy |
|---|---|---|---|---|
| Moving Charges and Magnetism | 30 | 6 | 11, 16, 17, 20, 22, 35 | 80.0% |
| Magnetism and Matter | 13 | 4 | 6, 27, 28, 29 | 69.2% |
| Electromagnetic Induction | 2 | 2 | 19, 42 | 0.0% |
| Type of error | Count | Questions | What it means |
|---|---|---|---|
| Pure theory / recall — no calculation involved | 7 | 16, 17, 19, 27, 28, 29, 35 | The dominant leak. These are one-line facts, not problems. Zero working time, zero risk — they should be automatic. |
| Formula applied to the wrong quantity | 3 | 6, 20, 42 | Formula known, but the wrong length / wrong unit / wrong final quantity went into it. |
| Graph and derivative reading | 1 | 11 | Slope confused with curvature. |
| Multi-step reasoning | 1 | 22 | Two-force separation missed in the second stage. |
Seven of twelve errors carry no arithmetic at all. That is the single most encouraging fact in this analysis — it means most of the lost 60 marks are recoverable by memory work, not by getting faster at problems.
Bar magnet of effective length 31.4 cm, pole strength 0.8 A·m, bent into a semicircle. Find the new magnetic moment.
Pole strength never changes when a magnet is bent. Only the straight-line distance between the two poles changes, and that is the length the moment uses.
B–X graph for a circular coil. Points A and A′ are marked on the two shoulders. Which statement is false?
They are the points of inflection of the axial field curve, sitting at X = ± r/2. At an inflection point the second derivative vanishes and the curve is locally straight.
At A, A′ : d²B/dX² = 0 but dB/dX ≠ 0 (it is maximum here) At X = 0 (centre) : dB/dX = 0 but d²B/dX² ≠ 0So (a), (b) and (d) are all true descriptions of an inflection point. Only (c) is false — the slope there is at its steepest, not zero.
A cyclotron is used to accelerate —
A cyclotron needs two things from the particle: charge for the electric kick across the dee gap, and charge for the magnetic force that bends it back. The sign of the charge only decides which way round the particle circulates and the phase of the applied RF voltage — it does not decide whether the machine works.
A neutron has no charge, so neither force acts. That is the only exclusion in the options.
Two parallel beams of positrons moving in the same direction will —
Both beams are positively charged, so there is Coulomb repulsion. Both are also moving charges — parallel currents — so there is magnetic attraction. Which one wins:
F_magnetic / F_electric = v² / c²Since v < c always, this ratio is always less than 1. Electric repulsion wins at every possible speed. Net result: the beams push apart.
A circular loop sits in a uniform magnetic field with its plane perpendicular to the field. An emf will be induced if —
An emf exists only if this product changes. Three things can change it: B, A, or θ.
(a) Translated parallel to itself — the field is uniform, so moving to a new spot gives the same B. Nothing changes. No emf.
(b) Rotated about a diameter — the plane tilts, θ changes, cos θ changes. Flux changes. Emf induced.
(c) Rotated about its own axis parallel to the field — the loop spins onto itself. Area, orientation and field all unchanged. No emf.
Solenoid of length 0.4 m, radius 1 cm, 400 turns, current 5 A. Find B inside.
The radius 1 cm is a deliberate distractor. Inside an ideal solenoid the field does not depend on the radius at all.
A proton released from rest accelerates west at a₀. Projected north at v₀, it accelerates west at 3a₀. Find E and B in the room.
The proton is positive, so the force is along E. Force is west, so E is west.
The electric force is still ma₀ west. Total is 3a₀ west. So the magnetic force supplies the difference:
F_magnetic = 3ma₀ − ma₀ = 2ma₀, west e v₀ B = 2ma₀ → B = 2ma₀ / (e v₀)v is north, force must be west. Right hand: fingers point north, curl them downward, thumb points west. So B is directed downward.
A compass needle free to move in a horizontal plane is taken to a geomagnetic pole. It will —
At a geomagnetic pole the earth's field points straight down (or straight up) — the dip angle is 90°. That means the horizontal component BH is zero. A needle confined to the horizontal plane has nothing to align to, so no restoring torque acts and it rests wherever it is left.
N-pole of a bar magnet points south and S-pole points north. The null points lie on —
A null point is where the magnet's own field exactly cancels the earth's horizontal field BH, which always points north. So the null points sit wherever the magnet's field points south with the right magnitude.
With the N-pole facing south, the magnet's axial field on either end points south — so the cancellation happens on the axial line, i.e. along the magnetic axis.
Because of earth's magnetic field, charged cosmic-ray particles —
Cosmic rays arrive radially, pointing at the earth's centre. What matters is the angle between that velocity and the local field, because F = qvB sin θ.
At the equator the field lines run horizontally, i.e. perpendicular to the incoming particle. sin θ = 1, maximum deflection — only very energetic particles punch through.
At the poles the field lines are vertical, i.e. parallel to the incoming particle. sin θ = 0, no deflection at all — even low-energy particles get in freely. This is why aurorae happen at the poles.
A charged particle experiences a magnetic force. Which statement is correct?
Three conditions must all hold for a non-zero force: charge present, velocity non-zero, and θ ≠ 0 or 180°. Check the options: (c) and (d) both say stationary, so v = 0 and F = 0 immediately. (b) says parallel, so sin 0° = 0 and F = 0. Only (a) satisfies everything, and at θ = 90° the force is at its maximum.
A rod of length l cuts across a uniform field B with velocity v. Circuit resistance is r. Find the force required to move the rod.
The induced current sits in the same field, so the rod feels an opposing force. To keep it moving at constant velocity, an equal applied force is needed — that is what the question asks for.
Every one of the twelve errors reduces to one of these. Recite them cold, then re-attempt the twelve questions without looking.