A charged particle enters a uniform magnetic field perpendicular to it. The field does:

No work, so the speed is unchanged and only the direction turns

Positive work, increasing the particle’s speed

No work, so the speed is unchanged and only the direction turns

Negative work, bringing the particle to rest

Work equal to qvB times the distance travelled

A bar magnet is dropped with its north pole downward through a vertical copper pipe. It falls:

Slower than in free fall, because the induced currents oppose its motion

Faster than in free fall, because copper is non-magnetic

Slower than in free fall, because the induced currents oppose its motion

At exactly the acceleration due to gravity, since copper carries no permanent magnetism

Slower only if the pipe is made of iron instead

A transformer will not work on a direct-current supply because:

A steady current gives a steady flux, and only a CHANGING flux induces an emf

Direct current cannot flow through the primary winding

A steady current gives a steady flux, and only a CHANGING flux induces an emf

The iron core cannot be magnetised by direct current

Direct current has no rms value to step up

Magnetism and Electromagnetic Induction — Fields, Forces, Flux and AC

This block is four NCERT chapters and one habit of mind: a current makes a field, a field pushes a current, and a changing flux makes an emf that opposes the change. Everything CUET asks here is one of those three read in some direction. The questions are standard results and directions rather than derivations — the field at the centre of a circular loop, the force on a wire, why the induced current opposes the motion, what a transformer conserves and what it changes. Two things are worth fixing before the formulas: a magnetic force never does work on a moving charge, because it is always perpendicular to the velocity, and Lenz's law is conservation of energy, not a separate rule to memorise.

Fields From Currents, and Forces On Them

The standard results, and where each applies
Situation	Result	Note
Long straight wire	`B = μ₀I/(2πr)`	`μ₀ = 4π × 10⁻⁷ T m/A`. Direction by the right-hand thumb rule: thumb along I, fingers curl along B.
Centre of a circular loop	`B = μ₀I/(2R)`, and `μ₀NI/(2R)` for N turns	Note the 2R, not 2πR — the π cancels for a loop and survives for a straight wire.
Inside a long solenoid	`B = μ₀nI`, with n turns per unit length	Uniform along the axis and independent of the radius; halves at either end.
Charge moving in a field	`F = qvB sin θ`; radius `r = mv/(qB)`	Zero when v is parallel to B. The force does NO work, so the speed never changes — only the direction.
Current-carrying wire in a field	`F = BIL sin θ`	Maximum when the wire is perpendicular to B, zero when along it.
Current loop as a magnet	Moment `m = NIA`; torque `τ = mB sin θ`	The same form as an electric dipole in a field, which is the analogy worth carrying across.

⚠️ A magnetic force cannot speed a particle up

Because F = qv × B is always perpendicular to v, it does zero work: the kinetic energy and the speed are constant, and only the direction changes — which is why a charge entering a uniform field perpendicular to it travels a circle of radius mv/qB. Any option claiming that the field increases the particle's energy or speed is wrong on this ground alone. A cyclotron does speed particles up, but the work there is done by the alternating ELECTRIC field between the dees, with the magnetic field only bending the path back for the next kick.

Induction, and the Alternating Current It Feeds

Faraday: the induced emf is ε = −dφ/dt, with φ = BA cos θ — so flux changes if B changes, if the area changes, or if the coil turns. The minus sign is Lenz's law and it is conservation of energy: the induced current always opposes the change that produced it, which is why a magnet dropped through a copper pipe falls slowly and why work must be done to keep a generator turning. A rod of length L moving with speed v across a field gives the motional emf ε = BLv. Self-inductance stores energy as U = ½LI², the magnetic twin of ½CV².

RMS values are what a meter reads: I_rms = I₀/√2 and V_rms = V₀/√2, so 220 V mains has a peak of about 311 V. Average power is V_rms I_rms cos φ, and cos φ is the power factor.
Reactances: inductive X_L = 2πfL rises with frequency, capacitive X_C = 1/(2πfC) falls with it. So a capacitor blocks DC and passes high frequencies; an inductor does the opposite.
Resonance in a series LCR circuit is X_L = X_C, so f = 1/(2π√(LC)): the impedance is then just R, the current is maximum, and the power factor is 1.
A pure inductor or a pure capacitor consumes no average power, because the phase difference is 90° and cos 90° = 0. Only resistance dissipates.

🧠 A transformer changes voltage, not power

V_s/V_p = N_s/N_p and, for an ideal transformer, V_p I_p = V_s I_s — so stepping the voltage UP steps the current DOWN in the same ratio, and the power is unchanged. Two consequences the paper asks: a transformer cannot work on DC, because a steady current produces no changing flux; and its losses are named — copper loss in the windings, eddy currents (reduced by laminating the core), hysteresis (reduced by choosing soft iron) and flux leakage.

Key takeaways

B = μ₀I/2πr for a long wire, μ₀I/2R at the centre of a loop (2R, not 2πR), and μ₀nI inside a solenoid. A charge in a field moves in a circle of radius mv/qB.
A magnetic force is always perpendicular to the velocity, so it does no work: speed and kinetic energy stay constant and only direction changes. In a cyclotron the electric field does the accelerating.
ε = −dφ/dt with φ = BA cos θ, and the minus sign is Lenz's law — conservation of energy, not a separate rule. A rod moving across a field gives ε = BLv, and an inductor stores ½LI².
X_L = 2πfL rises with frequency and X_C = 1/2πfC falls, resonance is X_L = X_C at f = 1/2π√(LC), and a pure L or C consumes no average power. A transformer changes voltage and current inversely, leaving power unchanged, and cannot work on DC.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The magnetic field at the centre of a circular coil of N turns, radius R, carrying current I is:
1. μ₀NI/(2πR)
2. μ₀NI/(2R)
3. μ₀NI/R
4. μ₀NI/(4πR²)
Show answer
Answer: B — μ₀NI/(2R)
At the centre of a loop the Biot–Savart contributions all add along the axis and the π cancels, leaving B = μ₀NI/2R. The version with 2πR is the LONG STRAIGHT WIRE formula with N attached, and it is the single commonest confusion in this topic because the two look alike on the page — the loop keeps 2R and the wire keeps 2πr. μ₀NI/R is the loop result with the factor of 2 dropped. And the last option has an inverse-square dependence, which no magnetic field of a steady current in this syllabus has.
A charged particle enters a uniform magnetic field perpendicular to it. The field does:
1. Positive work, increasing the particle’s speed
2. No work, so the speed is unchanged and only the direction turns
3. Negative work, bringing the particle to rest
4. Work equal to qvB times the distance travelled
Show answer
Answer: B — No work, so the speed is unchanged and only the direction turns
The magnetic force qv × B is perpendicular to the velocity at every instant, and a force perpendicular to the displacement does zero work — so the kinetic energy and the speed are constant while the direction changes continuously, giving a circular path of radius mv/qB. Positive work increasing the speed is what an electric field would do, and importing it here is the error the question is set on. Bringing the particle to rest would need a force with a component against v. And the last option computes force times distance without noticing that the angle between them is 90°, which is precisely why the product vanishes.
A bar magnet is dropped with its north pole downward through a vertical copper pipe. It falls:
1. Faster than in free fall, because copper is non-magnetic
2. Slower than in free fall, because the induced currents oppose its motion
3. At exactly the acceleration due to gravity, since copper carries no permanent magnetism
4. Slower only if the pipe is made of iron instead
Show answer
Answer: B — Slower than in free fall, because the induced currents oppose its motion
The falling magnet changes the flux through each ring of the pipe, inducing eddy currents whose own field, by Lenz's law, opposes the change — attracting the magnet from above and repelling it from below, which is a retarding force. This is conservation of energy in visible form: the lost gravitational energy appears as heat in the copper. Copper being non-magnetic is true and irrelevant, since what matters is that it conducts; that mistaken step produces both the "faster" and the "exactly g" options. And iron would add attraction but is not needed for the effect — a good conductor is what the effect requires.
An ideal transformer steps 220 V up to 2200 V. If the primary draws 5 A, the secondary current is:
1. 50 A
2. 0.5 A
3. 5 A
4. 10 A
Show answer
Answer: B — 0.5 A
An ideal transformer conserves power, so V_p I_p = V_s I_s: 220 × 5 = 1100 W, and 1100/2200 = 0.5 A. Stepping the voltage up by ten therefore steps the current down by ten — the two move inversely, and 50 A is the answer given by moving them together, which would create power from nothing. 5 A assumes the current is unchanged, which is true of neither an ideal nor a real transformer. And 10 A corresponds to the turns ratio itself rather than to any current in the problem.
The peak value of a 220 V (rms) alternating supply is about:
1. 220 V
2. 311 V
3. 156 V
4. 440 V
Show answer
Answer: B — 311 V
V₀ = √2 × V_rms = 1.414 × 220 ≈ 311 V, so the peak is higher than the rms value — which is the point of the rms definition, since a meter reads the equivalent steady value rather than the maximum. 156 V is 220/√2 and applies the factor the wrong way round, which is the commonest error here. 220 V would be the answer if rms and peak were the same, which they are only for a steady DC. And 440 V is twice the rms, corresponding to the peak-to-peak value being confused with the peak.
At resonance in a series LCR circuit, the impedance and the power factor are respectively:
1. Maximum and zero
2. Equal to R and one
3. Zero and one
4. Equal to X_L and zero
Show answer
Answer: B — Equal to R and one
Resonance is the condition X_L = X_C, so the two reactances cancel in the impedance Z = √(R² + (X_L − X_C)²), leaving Z = R — its minimum value — with the current maximum and the voltage in phase with it, so cos φ = 1. "Maximum and zero" describes the series circuit far from resonance and is also the correct description of a PARALLEL LCR circuit at resonance, which is why it is the tempting option. Zero impedance would require R itself to vanish, which resonance does not do. And Z equal to X_L ignores the cancellation that defines resonance.
A transformer will not work on a direct-current supply because:
1. Direct current cannot flow through the primary winding
2. A steady current gives a steady flux, and only a CHANGING flux induces an emf
3. The iron core cannot be magnetised by direct current
4. Direct current has no rms value to step up
Show answer
Answer: B — A steady current gives a steady flux, and only a CHANGING flux induces an emf
A transformer works by mutual induction, and Faraday's law gives an emf only from dφ/dt. A steady current makes a steady flux, dφ/dt is zero, and the secondary reads nothing — worse, with no back-emf the primary behaves as a low resistance and can burn out. Direct current flows through the primary perfectly well, so the first option is factually wrong. Direct current magnetises iron very effectively, which is how an electromagnet works. And direct current does have an rms value, equal to itself; the missing ingredient is change, not a value.
A conducting rod of length 0.5 m moves at 4 m/s perpendicular to a 0.2 T field. The emf induced across its ends is:
1. 0.1 V
2. 0.4 V
3. 4 V
4. 1.6 V
Show answer
Answer: B — 0.4 V
Motional emf is ε = BLv = 0.2 × 0.5 × 4 = 0.4 V, and the formula holds in this simple form only when B, L and v are mutually perpendicular, which the question states. 0.1 V divides by 4 instead of multiplying, the arithmetic slip the small numbers invite. 4 V is the speed carried through as a voltage, a unit error worth catching by asking whether the answer can be larger than BLv with all three factors less than or equal to 4. And 1.6 V multiplies L by v twice over, in effect using v² — a dimensional check rules it out immediately.