Moving Charges & Magnetism

Q01

Biot–Savart law — direction of B due to a moving electron

NCERT Exemplar

Given

An electron moves with velocity v. It produces a magnetic field B at a point of observation.

Asked

Which statement about B is correct.

Concept

A moving charge is an element of current, so Biot–Savart applies to it directly.
The field involves a cross product, so it is perpendicular to both the velocity and the position vector.
Biot–Savart is an inverse-square law, never inverse-cube.

Formula to be used

B = μ₀4π · q (v × r̂)r²

Diagram

Steps involved

The field comes from v × r̂, so B is perpendicular to v and also to the line joining the electron to the point. Options (b) and (d) fall.
The denominator carries r² — an inverse-square law, not inverse-cube. Option (c) falls.
Perpendicularity to v holds for every observation point, so (a) is universally true.

ShortcutAny cross product ⇒ the result is perpendicular to both vectors. And Biot–Savart is always 1/r².

Answer(a) B is perpendicular to v

Q02

Field above a wire carrying current east → west

Given

A straight conductor carries current from east to west. The observation point is directly above the wire.

Asked

The direction of the magnetic field at that point.

Concept

Right-hand grip rule: thumb along the current, curled fingers give the sense of B.
Equivalently, B ∝ I × r̂, with r̂ pointing from the wire to the field point.

Formula to be used

B = μ₀I2πd (direction from I × r̂)

Diagram

Steps involved

Take x̂ = east, ŷ = north, ẑ = up (a right-handed set).
Current direction = west = −x̂; the point above gives r̂ = +ẑ.
B ∝ (−x̂) × (ẑ) = −(x̂ × ẑ) = −(−ŷ) = +ŷ.
+ŷ is north.

ShortcutPoint the right thumb west and grip the wire — above the wire the fingers emerge towards the north, below it towards the south. Memory hook: current east ⇒ field above is south; reverse the current, reverse the field.

Answer(a) towards north

Q03

Cyclotron — behaviour of the charged particle

NCERT Exemplar

Given

A charged particle inside a cyclotron: two dees, a perpendicular magnetic field, an alternating electric field across the gap.

Asked

Which statement is true.

Concept

Inside a dee only B acts. The force is perpendicular to v, so the speed is constant but there is centripetal acceleration.
In the gap only E acts, so the particle gains speed there.
Constant speed is not the same thing as zero acceleration.

Formula to be used

r = mvqB T = 2πmqB (independent of v)

a_centripetal = v²r inside a dee

Diagram

Steps involved

In a dee the particle moves on a circle — the direction of v changes, so acceleration is present even though the speed is not.
In the gap the electric field does work, so the particle speeds up — acceleration again.
Acceleration therefore exists at every instant ⇒ (a).
(b) is wrong: the speed-up in the gap is due to the electric field. (c) is wrong: speed is constant inside a dee. (d) is wrong on both counts.

ShortcutA magnetic force can never change speed, only direction — but “no change in speed” ≠ “no acceleration”. Circular motion is accelerated motion.

Answer(a) undergoes acceleration all the time

Q04

Electron projected along the axis of a solenoid

NCERT Exemplar

Given

A long current-carrying solenoid. An electron is projected with uniform velocity along its axis.

Asked

What happens to the electron.

Concept

Inside a long solenoid the field is uniform and directed along the axis.
The magnetic force vanishes when velocity and field are parallel.

Formula to be used

F = q(v × B), F = qvB sin θ

B_inside = μ₀nI (along the axis)

Diagram

Steps involved

B lies along the axis; v also lies along the axis, so θ = 0° (or 180°).
sin 0° = 0 ⇒ F = 0.
No magnetic force ⇒ no acceleration, no bending, no helix.
The electron sails straight through with unchanged velocity.

ShortcutThree cases cover nearly every question of this type: v ∥ B ⇒ F = 0 (straight line); v ⊥ B ⇒ circle; any other angle ⇒ helix.

Answer(d) continues to move with uniform velocity along the axis

Q05

Deflection of a proton in a perpendicular field

Given

Proton (charge +q) moving along the negative X-axis: v = −v x̂. Field along the positive Y-axis: B = B ŷ.

Asked

Along the negative direction of which axis is the proton deflected.

Concept

The magnetic force is perpendicular to both v and B, so with v along X and B along Y the force must lie along Z.
Only the sign then needs to be settled.

Formula to be used

F = q(v × B)

x̂ × ŷ = ẑ, ŷ × ẑ = x̂, ẑ × x̂ = ŷ

Diagram

Steps involved

F = q(−v x̂) × (B ŷ) = −qvB (x̂ × ŷ).
x̂ × ŷ = ẑ, so F = −qvB ẑ.
The charge is positive, so the force points along −ẑ — the negative Z-axis.
Consistency check: F must be perpendicular to both X and Y, so only a Z-answer is possible; (a) and (b) die on inspection.

ShortcutOptions (a) and (b) are impossible before any algebra. Right-hand rule: fingers along −x̂, curl towards +ŷ, thumb gives −ẑ. Reverse the thumb for a negative charge.

Answer(c) Z-axis

Q06

Dimensional formulae of ε₀ and μ₀

Given

ε₀ = permittivity of vacuum, μ₀ = permeability of vacuum; M = mass, L = length, T = time, I = current.

Asked

Which dimensional formula is correct.

Concept

Get ε₀ from Coulomb's law with [q] = [IT].
Get μ₀ from the force between parallel wires, or from c = 1/√(μ₀ε₀).
This is an Exemplar multiple-correct item — more than one option can be right.

Formula to be used

F = 14πε₀q₁q₂r² c = 1√(μ₀ε₀)

Fℓ = μ₀I₁I₂2πd

Steps involved

ε₀ = q₁q₂ / (4πFr²), so [ε₀] = [I²T²] / ([MLT⁻²][L²]).
[ε₀] = [M⁻¹L⁻³T⁴I²] — this is option (b).
μ₀ = 1/(ε₀c²) = [ML³T⁻⁴I⁻²] / [L²T⁻²].
[μ₀] = [MLT⁻²I⁻²] — this is option (c).

ShortcutFastest route to μ₀ is the parallel-wire law: [μ₀] = [MT⁻²][L]/[I²]. Then ε₀ = 1/(μ₀c²). Verification trick: the two dimensions must multiply to [L⁻²T²] (= 1/c²). Check: M⁻¹L⁻³T⁴I² × MLT⁻²I⁻² = L⁻²T² ✓

Answer(b) and (c) — both are correct (multiple-correct item)

Q07

Coaxial cable — where is B zero?

Given

A coaxial cable: the central conductor and the outer conductor carry equal currents in opposite directions.

Asked

The region in which the magnetic field is zero.

Concept

Ampère's law depends only on the current enclosed by the chosen loop.
Outside the cable the go and return currents cancel exactly.

Formula to be used

∮ B·dl = μ₀I_enc ⇒ B = μ₀I_enc2πr

Diagram

Steps involved

Inside the inner conductor (r < a): I_enc = I·r²/a² ≠ 0 ⇒ B ≠ 0.
Between the conductors (a < r < b): I_enc = I ⇒ B = μ₀I/2πr, the strongest region.
Inside the outer conductor: part of the return current is enclosed, so B falls smoothly but is not zero until the outer surface.
Outside the cable (r > b): I_enc = I − I = 0 ⇒ B = 0.

ShortcutNet enclosed current = 0 ⇒ B = 0. That is the whole point of a coaxial cable: the return current cancels the external field so the cable neither radiates nor picks up interference.

Answer(a) outside the cable

Q08

Half of a circular loop bent into the perpendicular plane

NCERT Exemplar

Given

Circular loop of radius R, current I, in the xy-plane centred at the origin. The half with x > 0 is bent so that it now lies in the yz-plane.

Asked

Which statement is true.

Concept

Magnetic moment is a vector: the two halves must be added vectorially, not arithmetically.
For a fixed current and wire, the moment is largest when the loop is planar.

Formula to be used

m = IA, m_initial = IπR²

B_axial = μ₀4π2mz³, B_equatorial = μ₀4πmz³

Diagram

Steps involved

Each semicircular half carries a moment of magnitude m/2 = ½IπR².
Remaining half (xy-plane): m₁ = (m/2)ẑ. Bent half (yz-plane): m₂ = (m/2)x̂.
They are perpendicular, so m_net = √[(m/2)² + (m/2)²] = m/√2 ≈ 0.707 m — it diminishes.
Field check at (0,0,z), z ≫ R: the xy-half gives an axial field μ₀m/4πz³, the yz-half puts that point in its equatorial position giving μ₀m/8πz³ at right angles. Net ≈ 1.12 μ₀m/4πz³ against the original 2 μ₀m/4πz³, so B decreases — (c) and (d) both fail.

ShortcutBending any part of a loop out of plane can only shrink the vector sum — the moment is maximum when the loop is planar. Two equal perpendicular halves give the factor 1/√2 immediately.

Answer(a) the magnitude of the magnetic moment diminishes (m → m/√2)

Q09

Ratio of magnetic moment to angular momentum

Given

A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω.

Asked

What the ratio M/L depends on.

Concept

An orbiting charge is an equivalent current loop.
Both M and L carry the same factors of ω and r², so those cancel.

Formula to be used

i = qT = qω2π, M = iπr², L = mr²ω

Steps involved

M = (qω/2π)·πr² = qωr²/2.
L = mωr².
M/L = (qωr²/2)/(mωr²) = q/2m.
Both ω and r cancel — only q and m survive.

ShortcutMemorise the gyromagnetic ratio M/L = q/2m (for an electron, e/2m_e = 8.8 × 10¹⁰ C/kg). It is independent of orbit size and speed, which is why it is a universal constant of the particle.

Answer(c) q and m

Q10

Maximum proton energy from the same cyclotron

Given

A cyclotron gives a deuteron (q = e, m = 2m_p) a maximum energy of 20 MeV. The same machine ⇒ same B and same dee radius R.

Asked

Maximum energy obtainable for a proton.

Concept

The particle leaves at the outermost radius, so R fixes the maximum speed.
With B and R fixed, K_max ∝ q²/m.

Formula to be used

v_max = qBRm, K_max = ½mv_max² = q²B²R²2m

Steps involved

With B and R fixed, K_max ∝ q²/m.
Both particles carry charge e, so K_max ∝ 1/m.
K_p/K_d = m_d/m_p = 2m_p/m_p = 2.
K_p = 2 × 20 = 40 MeV.

ShortcutA deuteron is a proton + neutron, so m_d ≈ 2m_p — that factor of 2 is the entire question. Same charge, half the mass ⇒ double the energy. Retuning the oscillator frequency f = qB/2πm does not affect K_max.

Answer(d) 40 MeV

Q11

B-r graph for a long thin hollow cylinder

Given

A long thin hollow metallic cylinder of radius R carries current i. The current flows entirely on the surface.

Asked

Which graph correctly shows B against the distance r from the axis.

Concept

Ampere's law with a coaxial circular loop.
For a thin shell, no current is enclosed anywhere inside, so the interior field is exactly zero.
The hollow case and the solid case give two different interior shapes - examiners pair them deliberately.

Formula to be used

B(2πr) = μ₀I_enc ⇒ B = μ₀I_enc2πr

Solid conductor, r < R: B = μ₀i r2πR² ∝ r

Diagram

Steps involved

Inside the shell (r < R): all of i lies on the surface at radius R, so I_enc = 0 ⇒ B = 0. The graph runs flat along the r-axis.
At the surface (r = R): B jumps discontinuously to B_max = μ₀i/2πR.
Outside (r > R): I_enc = i ⇒ B = μ₀i/2πr, a hyperbolic 1/r decay.
So the curve is: zero up to R → sharp vertical jump → 1/r fall-off.

ShortcutTwo shapes cover almost every cylinder question. Hollow / thin shell: 0 inside, jump, then 1/r. Solid conductor: straight line ∝ r inside, peak at the surface, then 1/r. When matching by eye, check that the flat portion sits on the r-axis, not at some finite height - that is what separates (b) from (c).

Answer(b) flat at zero up to R, then a 1/r decay

Q12

Magnetic field at the centre of the hydrogen orbit

Given

r = 0.53 Å = 0.53 × 10⁻¹⁰ m, revolution frequency f = 6.6 × 10¹⁵ rot/s, e = 1.6 × 10⁻¹⁹ C.

Asked

B at the centre of the orbit.

Concept

An orbiting electron is equivalent to a circular current loop of current i = ef.
Then apply the standard centre-of-loop formula.

Formula to be used

i = ef = eT, B = μ₀i2r = μ₀ef2r

Diagram

Steps involved

i = (1.6 × 10⁻¹⁹)(6.6 × 10¹⁵) = 1.056 × 10⁻³ A ≈ 1.06 mA.
B = μ₀i / 2r = (4π × 10⁻⁷)(1.056 × 10⁻³) / (2 × 0.53 × 10⁻¹⁰).
Numerator = 1.327 × 10⁻⁹; denominator = 1.06 × 10⁻¹⁰.
B = 12.5 Wb/m².

ShortcutTrack powers of ten only: (10⁻⁶ × 10⁻¹⁹ × 10¹⁵)/10⁻¹⁰ = 10⁰, so the answer is of order 10 T - only one option is in that range. On a timed paper the exponent alone settles it.

Answer(c) 12.5 Wb/m²

Q13

Undeflected passage through crossed E and B (velocity selector)

Given

A charge q with velocity v passes undeflected through a region with non-zero E and B.

Asked

Under which changes does the undeflected condition still hold.

Concept

The balance depends on the ratio E/B, not the product EB.
Both force terms carry a common factor of q, so reversing q alone changes nothing.
Another Exemplar multiple-correct item.

Formula to be used

qE + q(v × B) = 0 ⇒ E = vB ⇒ v = EB

Diagram

Steps involved

(a) Signs of q and E both reversed: the electric force (−q)(−E) is unchanged, but the magnetic force (−q)(v×B) reverses. They now add ⇒ deflection. Fails.
(b) Signs of q and B both reversed: electric force reverses; magnetic force (−q)v×(−B) = +q(v×B) also reverses. Both flip together, so they still cancel. Holds.
(c) E and B changed keeping EB fixed: the condition needs E/B constant. E → 2E, B → B/2 keeps the product but quadruples the ratio. Fails.
(d) Both doubled: 2E/2B = E/B = v. Holds.

ShortcutWrite the condition as v = E/B and ask two questions only: is the ratio E/B preserved, and do the two forces flip together? Reversing q alone never breaks the balance - q is a common factor; it is reversing q with only one of E or B that breaks it.

Answer(b) and (d) - both correct (multiple-correct item)

Q14

Ratio of fields at the centres of two coils

Given

Coil 1: radius r, current I, centre field B₁. Coil 2: radius 2r, same current I, centre field B₂.

Asked

The ratio B₁/B₂.

Concept

At the centre of a coil, with the current fixed, B ∝ 1/R.
Do not confuse this with the axial far-field (1/z³) or the straight-wire law (1/d from the wire).

Formula to be used

B = μ₀I2R

Steps involved

B₁ = μ₀I / 2r.
B₂ = μ₀I / 2(2r) = μ₀I / 4r.
B₁/B₂ = 4r / 2r = 2.

ShortcutWith I fixed, B ∝ 1/R. Double the radius ⇒ half the field ⇒ ratio 2 : 1. No numbers needed.

Answer(c) 2

Q15

Resultant force on the middle wire B

Given

Three long parallel wires A, B, C carrying 1 A, 2 A and 3 A, all in the same direction. Spacings A-B = d and B-C = d.

Asked

The direction of the resultant force on B.

Concept

Parallel currents attract; antiparallel currents repel.
Forces between parallel wires always lie in the plane of the wires - never out of the page, so two options die immediately.

Formula to be used

Fℓ = μ₀I₁I₂2πd

Diagram

Steps involved

Force on B due to A - same direction ⇒ attraction, pulling B towards A: F/ℓ = μ₀(1)(2)/2πd.
Force on B due to C - same direction ⇒ attraction, pulling B towards C: F/ℓ = μ₀(2)(3)/2πd.
The two pulls are collinear and opposite. Net = (6 − 2)μ₀/2πd = 2μ₀/πd.
F_BC > F_BA, so the resultant is towards C.

ShortcutWith equal distances, only the currents decide: compare the products 1 × 2 = 2 against 2 × 3 = 6. The larger product wins ⇒ towards C.

Answer(b) towards C

Q16

Magnetic moment of a wire bent into a circle

Given

Wire length L = 2 m, current I = 1 A, bent into a single circular turn.

Asked

The magnetic moment M in A·m².

Concept

The perimeter fixes the radius, which fixes the area, which fixes the moment.

Formula to be used

M = IA = Iπr², L = 2πr

General: M = L²I4πN for N turns from a wire of length L

Steps involved

r = L/2π = 2/2π = 1/π m.
A = πr² = π(1/π²) = 1/π m².
M = (1)(1/π) = 1/π A·m².

ShortcutUse M = L²I / 4πN directly: (2)²(1)/(4π) = 1/π. Note M ∝ 1/N - a single turn maximises the moment.

Answer(d) 1/π

Q17

Same wire: one turn versus two turns

Given

A conductor of length ℓ carrying current I is bent first into one turn (field B), then into two turns (field B′).

Asked

The relation between B′ and B.

Concept

For a fixed length of wire, adding turns also shrinks the radius - both effects raise the field.
Result: B ∝ N² (while the moment goes as 1/N).

Formula to be used

B = μ₀NI2r, ℓ = N(2πr) ⇒ r = ℓ2πN

⇒ B = μ₀πN²Iℓ ∝ N²

Steps involved

Substituting r into the centre-field formula gives B = μ₀πN²I/ℓ.
So for fixed wire length, B ∝ N².
B′/B = (2/1)² = 4 ⇒ B′ = 4B.

ShortcutFixed wire length ⇒ B ∝ N² and M ∝ 1/N. Two turns: field × 4, moment ÷ 2. The trap is answering 'twice' by counting turns and forgetting that the radius halves.

Answer(a) B′ = 4B

Q18

Identical helical paths traversed in opposite sense

NCERT Exemplar

Given

Two charged particles trace identical helical paths in a completely opposite sense in a uniform field B = B₀k̂.

Asked

Which condition necessarily holds.

Concept

Helical geometry (radius and pitch) is governed entirely by the ratio q/m.
The sense of winding is set by the sign of the charge.

Formula to be used

r = mv_⊥qB, pitch p = 2πm v_∥qB

Steps involved

Identical geometry (same r, same pitch) ⇒ |q/m|₁ = |q/m|₂.
Opposite sense of traversal ⇒ the charges have opposite sign.
Combining: (q/m)₁ = −(q/m)₂ ⇒ (e/m)₁ + (e/m)₂ = 0.
Why the others fail - (a): the z-components could be opposite in sign, not equal. (b): the charges are equal in magnitude but opposite in sign. (c): a particle/antiparticle pair is one possibility, but any pair with matching |q/m| works, so it is not necessary.

Shortcut'Identical shape' ⇒ equal magnitude of q/m; 'opposite sense' ⇒ opposite sign. Put together, that is option (d).

Answer(d) (e/m)₁ + (e/m)₂ = 0

Q19

One loop versus three loops from the same wire

Given

A fixed length of wire is turned first into one circular loop (field B₁), then into three loops; the current is the same.

Asked

The new field at the centre.

Concept

Same principle as Q17: for fixed wire length, B ∝ N².

Formula to be used

B = μ₀πN²Iℓ ∝ N²

Steps involved

ℓ = 2πr₁ = 3(2πr₂) ⇒ r₂ = r₁/3.
B₂ = μ₀(3)I / 2r₂ = μ₀(3)(3)I / 2r₁.
B₂ = 9 B₁.

ShortcutB ∝ N² ⇒ 3² = 9. Option (c) 3B₁ is the trap for anyone who counts only the turns.

Answer(b) 9B₁

Q20

Charge released from the top of a vertical solenoid

Given

A long solenoid with its axis vertical carries current I. A particle of mass m and charge q is released from rest at the top, on the axis.

Asked

Its acceleration.

Concept

Released from rest ⇒ no magnetic force initially.
It then falls along the axis, so v stays parallel to B and the magnetic force stays zero.

Formula to be used

F = qvB sin θ, B_inside = μ₀nI along the axis

Diagram

Steps involved

At release v = 0 ⇒ magnetic force = 0; only gravity acts.
It then falls vertically along the axis, so v is parallel (anti-parallel) to B: θ = 180°, sin θ = 0.
The magnetic force stays zero throughout the fall.
Net force = mg ⇒ acceleration = g.

ShortcutSame rule as Q4: v ∥ B ⇒ F = 0. The solenoid is decoration - the particle simply free-falls.

Answer(c) equal to g

Q21

Distance advanced along B after two complete circles

Given

v = 2 × 10⁶ m/s, B = 0.05 T, θ = 30° to B; proton (m ≈ 1.6 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C). Two revolutions.

Asked

Distance moved along the direction of B.

Concept

Resolve the velocity: v_⊥ makes the circle, v_∥ drives the drift along B.
The period depends only on m, q, B - not on speed or angle.

Formula to be used

T = 2πmqB, pitch p = v_∥T = 2πm v cos θqB

Diagram

Steps involved

T = 2π(1.6 × 10⁻²⁷) / [(1.6 × 10⁻¹⁹)(0.05)] = 1.257 × 10⁻⁶ s.
v_∥ = 2 × 10⁶ cos 30° = 1.732 × 10⁶ m/s.
Pitch p = (1.732 × 10⁶)(1.257 × 10⁻⁶) = 2.177 m.
Two revolutions: 2p = 4.35 m.

ShortcutT is independent of speed and angle, so compute it first. Distance for n revolutions = n · v cos θ · (2πm/qB).

Answer(a) 4.35 m

Q22

Charge on the axis of a circular wire fed at diametrically opposite points

Given

Current i enters and leaves a uniform circular wire of radius a at diametrically opposite points. A charge q moves along the axis and passes the centre at speed v.

Asked

The magnitude of the magnetic force on the particle at the centre.

Concept

Two parallel arcs of equal length ⇒ equal resistance ⇒ equal currents.
The two arcs circulate oppositely about the centre, so their fields cancel there.

Formula to be used

B_arc = μ₀Iθ4πa, F = q(v × B)

Diagram

Steps involved

The two semicircles are identical in length and resistance, so each carries i/2.
Seen from the centre, one semicircle carries current clockwise and the other anticlockwise, so their fields at the centre are opposite.
Equal magnitudes, opposite directions ⇒ B_centre = 0.
F = q(v × 0) = 0.

ShortcutTwo watchwords make this instant: uniform wire and diametrically opposite feed points ⇒ equal split ⇒ B = 0 at the centre. General rule: B_centre = 0 whenever I₁θ₁ = I₂θ₂, which resistance-sharing guarantees for a uniform wire wherever you feed it.

Answer(d) zero

Q23

Minimum speed to cross a magnetic field slab

Given

Charge q, mass m, velocity v along +x. Uniform B along −ẑ occupies the slab from x = a to x = b.

Asked

The minimum v needed for the particle to just emerge at x > b.

Concept

Inside the slab the particle follows a circular arc.
The maximum horizontal depth it can penetrate equals the radius - beyond that point it curves back.

Formula to be used

r = mvqB, condition: r ≥ (b − a)

Diagram

Steps involved

Maximum penetration depth into the field region = r (at that point the velocity is purely along y; beyond it the particle turns back and exits through the x = a face).
To just cross: r ≥ (b − a).
mv/qB ≥ (b − a).
v ≥ q(b − a)B / m.

ShortcutSet radius = slab width - that is the whole problem. The width is (b − a), not b and not (b + a), which is exactly what the other options test.

Answer(b) q(b − a)B/m

Q24

Path of a proton projected at 60° to the X-axis

Given

m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C, v = 2 × 10⁶ m/s in the xy-plane at 60° to the X-axis; B = 0.14 ŷ T.

Asked

The shape of the path, its radius and time period.

Concept

The angle is quoted from the X-axis but B lies along Y - so resolve carefully.
Both velocity components are non-zero ⇒ helix.

Formula to be used

r = mv_⊥qB, T = 2πmqB

Steps involved

B is along ŷ, so measure from the Y-axis: velocity at 60° to X is at 30° to Y.
v_∥ = v cos 30° = 1.732 × 10⁶ m/s (along Y); v_⊥ = v cos 60° = 1 × 10⁶ m/s.
Both components non-zero ⇒ the path is a helix; options (a) and (b) fall.
r = (1.67 × 10⁻²⁷)(1 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.14) = 0.0745 ≈ 0.07 m.
T = 2π(1.67 × 10⁻²⁷) / (2.24 × 10⁻²⁰) = 4.68 × 10⁻⁷ ≈ 0.5 × 10⁻⁶ s.

ShortcutCompute T = 2πm/qB first - it needs no velocity resolution and instantly separates (c) from (d). The real trap is angle bookkeeping: since B is along Y, v_⊥ = v cos 60°, not v sin 60°.

Answer(c) helix of radius 0.07 m, period 0.5 × 10⁻⁶ s

Q25

Field at the centre of an equilateral triangle of uniform wire

Given

Equilateral triangle of side l made from wire of uniform resistance. Current I enters at vertex a and leaves at vertex c.

Asked

The magnitude of the magnetic field at the centroid O.

Concept

Current divides inversely as resistance, and resistance ∝ length.
By symmetry all three sides are the same perpendicular distance from O and subtend the same angles.
The two branches circulate oppositely about O.

Formula to be used

B_segment = μ₀i4πd(sin θ₁ + sin θ₂)

Diagram

Steps involved

Two parallel paths from a to c: the direct side ac (length l) and the route a→b→c (length 2l).
Resistances are in ratio 1 : 2, so currents split inversely: i_ac = 2I/3 and i_abc = I/3.
Each side contributes B = k·i with the same constant k, by symmetry.
Side ac gives k(2I/3) one way; sides ab + bc give k(I/3) + k(I/3) = k(2I/3) the other way.
Equal and opposite ⇒ B_O = 0.

ShortcutGeneral rule: for any uniform-resistance closed loop fed at two points, the field at the symmetric centre is zero, because current × path length matches on both branches ((2I/3)·l = (I/3)·2l) while the branches circulate oppositely. Triangle, square, circle - always zero. Spot 'uniform resistance' and stop calculating.

Answer(d) zero

Q26

Infinitely long conductor bent into a circle (tangent loop)

Given

An infinite straight wire carrying current I has a circular loop of radius R formed in it; the straight portions are tangential to the circle at the same point.

Asked

The magnetic induction at the centre O of the loop.

Concept

Split the arrangement into two standard pieces: a full circular loop and an infinite straight wire.
In the tangent geometry the two fields are parallel, so they add.

Formula to be used

B_loop = μ₀I2R, B_wire = μ₀I2πR

Diagram

Steps involved

Set the straight wire along the x-axis with current in +x̂; O sits at (0, R) and the tangent point at the origin.
Loop: at the tangent point the current runs in +x̂ along the bottom of the circle, so it traverses the circle anticlockwise ⇒ B₁ = μ₀I/2R, out of the page.
Straight wire: current +x̂, field point directly above ⇒ x̂ × ŷ = ẑ ⇒ B₂ = μ₀I/2πR, also out of the page.
Same direction ⇒ add: B = μ₀I/2R + μ₀I/2πR.
Factor out μ₀2I/4πR: B = (μ₀2I/4πR)(π + 1).

ShortcutIn the tangent-loop geometry the two fields are always parallel, giving the (π + 1) form. The (π − 1) option belongs to the other standard figure, where the straight wire crosses the loop as a diameter so the circulations oppose. One glance at 'tangent vs crossing' decides it.

Answer(a) (μ₀2I/4πR)(π + 1)

Q27

Field at O due to an infinite wire with a semicircular bend

Given

A wire comes in from infinity along the vertical, bends into a semicircle of radius R centred at O, then runs off horizontally to infinity. Current I.

Asked

The magnetic field at O.

Concept

Break the wire into three standard pieces and add their contributions vectorially.
A straight segment whose line passes through the field point contributes nothing.

Formula to be used

semicircle: μ₀I4R semi-infinite wire: μ₀I4πR collinear segment: 0

Diagram

Steps involved

Put O at the origin: the wire runs from y = +∞ down to (0, R), the semicircle sweeps left to (0, −R), then part 3 runs to x = +∞.
Part 1 lies along the line x = 0, which passes through O, so every element has dl ∥ r̂ ⇒ contributes 0.
Part 2 (semicircle): B = μ₀I/4R; tracing (0,R) → (−R,0) → (0,−R) is anticlockwise ⇒ out of the page.
Part 3 (semi-infinite, with O at the foot of the perpendicular at its end): B = μ₀I/4πR; current +x̂ with O above ⇒ also out of the page.
Same direction ⇒ B_O = μ₀I/4R + μ₀I/4πR.

ShortcutMemorise the ladder: full circle μ₀I/2R → semicircle μ₀I/4R → quarter μ₀I/8R; infinite wire μ₀I/2πd → semi-infinite μ₀I/4πd; collinear segment 0. Then only check whether the pieces add or subtract.

Answer(d) μ₀I/4R + μ₀I/4πR

Q28

Two identical coils with planes at right angles

Given

Two identical coils carrying equal currents share a common centre; their planes are mutually perpendicular. Each alone produces B₀ at the centre.

Asked

The ratio of the resultant field to the field of one coil.

Concept

A coil's field at its centre points along its own axis, i.e. perpendicular to its plane.
Perpendicular planes ⇒ perpendicular axes ⇒ perpendicular field vectors.

Formula to be used

B_net = √(B₁² + B₂²)

Diagram

Steps involved

B₁ = B₂ = B₀ (identical coils, equal currents).
The two field vectors are at 90°.
B_net = √(B₀² + B₀²) = √2 B₀.
Ratio = √2 : 1.

ShortcutPerpendicular planes ⇒ perpendicular B vectors ⇒ factor √2. Do not fall for 2 : 1, which would need the coils coplanar and coaxial. The key mental step: coil plane ⊥ coil field.

Answer(d) √2 : 1

Q29

Tension in a current-carrying flexible loop

Given

A flexible circular loop of radius r carries current I in a uniform field B perpendicular to the plane of the loop. The loop is under tension. B is then doubled.

Asked

What happens to the tension.

Concept

Each element Idl × B points radially outward, stretching the loop.
Balance the radial force on a small arc against the two tension components.

Formula to be used

2T sin(dθ/2) = BIr dθ ⇒ T = BIr

Diagram

Steps involved

The radial force per unit arc is BI, and balancing it against tension gives T = BIr.
Tension is directly proportional to B, with I and r fixed.
B → 2B ⇒ T → 2T.

ShortcutMemorise T = BIr - the magnetic twin of the hoop-stress formula. Everything appears to the first power, so doubling any one of B, I or r doubles the tension.

Answer(b) is doubled

Q30

Radius of an electron orbit from the field it produces

Given

An electron moves in a circular orbit of radius r at uniform speed v. It produces a field B at the centre of the circle.

Asked

r is proportional to what.

Concept

Read carefully: this is the field produced by the electron, not an applied field bending it. That distinction flips the answer.
Orbiting charge ⇒ equivalent current loop.

Formula to be used

i = eT = ev2πr, B = μ₀i2r = μ₀ev4πr²

Steps involved

Substitute the current: B = (μ₀/2r)(ev/2πr) = μ₀ev/4πr².
Rearrange: r² = μ₀ev/4πB.
r = √(μ₀e/4π) · √(v/B).
Hence r ∝ √(v/B).

ShortcutTwo cases, opposite answers. Field applied, electron bent by it: r = mv/eB ⇒ r ∝ v/B. Field produced at the centre: B ∝ v/r² ⇒ r ∝ √(v/B). Option (b) is planted precisely for the misreading.

Answer(c) √(v/B)

Q31

Magnetic moment - square versus circle from the same wire

Given

Two wires of the same length L, one bent into a square and the other into a circle. Both carry the same current I.

Asked

The ratio M_square : M_circle.

Concept

Same L and same I ⇒ the ratio of moments is simply the ratio of enclosed areas.
For a fixed perimeter the circle always encloses the largest area, so the ratio must be less than 1.

Formula to be used

M = IA; square: A = L²16; circle: A = L²4π

Steps involved

Square: side = L/4, so A_s = L²/16.
Circle: 2πr = L ⇒ r = L/2π, so A_c = πL²/4π² = L²/4π.
M_s/M_c = A_s/A_c = (L²/16)(4π/L²) = π/4.

ShortcutThe 'circle encloses more area' test kills every option greater than 1 (π/2 and 4/π) before any algebra. Handy set for fixed length L: square L²/16, circle L²/4π, equilateral triangle L²/12√3.

Answer(c) π : 4

Q32

Ratio of masses from radii after acceleration through the same potential

Given

Particles X and Y have equal charges, are accelerated through the same potential difference V, then enter the same field B describing radii R₁ and R₂.

Asked

The ratio of the mass of X to that of Y.

Concept

The accelerating stage fixes the speed via qV = ½mv².
Combine with r = mv/qB to get the mass-spectrometer relation.

Formula to be used

qV = ½mv² ⇒ v = √(2qV/m)

R = mvqB = 1B√(2Vm/q) ⇒ R ∝ √m

Steps involved

Substituting v into R gives R = (1/B)√(2Vm/q).
With V, q and B identical, R ∝ √m.
R₁/R₂ = √(m_X/m_Y).
m_X/m_Y = (R₁/R₂)².

Shortcut'Same V, same q' ⇒ R ∝ √m ⇒ m ∝ R². Contrast the same-speed case, where R ∝ m directly - that is what the (R₁/R₂) option is testing.

Answer(c) (R₁/R₂)²

Q33

Field at O due to two bent wires ABC and DEF

Given

Wire 1 runs horizontally A→B then turns down B→C; wire 2 comes down F→E then turns horizontally E→D. Both carry I. The bends B and E lie on the same horizontal line as O, each at perpendicular distance r.

Asked

The magnitude of the field at O.

Concept

Kill the collinear segments first - they contribute nothing.
The two surviving vertical semi-infinite pieces sit on opposite sides of O, so their fields oppose.

Formula to be used

collinear segment: 0 semi-infinite wire (foot at its end): μ₀I4πr

Diagram

Steps involved

Segment AB lies along the horizontal line that passes through O ⇒ contributes zero.
Segment ED likewise lies along that line ⇒ contributes zero.
Segment BC (vertical, downward, distance r to the left of O): B = μ₀I/4πr, direction (−ŷ) × (+x̂) = +ẑ, out of the page.
Segment FE (vertical, downward, distance r to the right of O): B = μ₀I/4πr, direction (−ŷ) × (−x̂) = −ẑ, into the page.
Equal magnitudes, opposite directions ⇒ they cancel.

ShortcutRemoving the collinear halves halves the work instantly. The two surviving vertical pieces both carry current downward but sit on opposite sides of O, so their fields must oppose; equal distances give exact cancellation.

Answer(d) zero

Q34

Circular conductor fed by a battery - field at the centre

Given

A circular conductor of uniform resistance per unit length, total resistance 4 Ω, is connected across a 4 V battery at two points.

Asked

The net field at the centre of the conductor.

Concept

Same principle as Q22 and Q25: uniform wire fed at two points.
Resistance ∝ arc length ∝ arc angle, so current splits inversely as the angle.

Formula to be used

B_arc = μ₀iθ4πa; parallel arcs give i₁θ₁ = i₂θ₂

Diagram

Steps involved

Let the two arcs subtend θ₁ and θ₂. Their resistances are in the ratio θ₁ : θ₂.
Being in parallel across the same battery, the currents split inversely: i₁θ₁ = i₂θ₂.
So the two field magnitudes μ₀iθ/4πa are equal.
The arcs circulate in opposite senses about the centre ⇒ the fields are antiparallel.
B_net = 0.

ShortcutThe 4 V and 4 Ω are pure decoration - they only fix the total current (1 A). Spot 'uniform resistance' plus two feed points and answer zero in five seconds.

Answer(d) zero

Q35

Question image not supplied

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Q36

Forces on wire A from wires B and C

Given

Three parallel conductors of equal length. Currents A = I, B = I, C = 2I, all in the same direction. Separations A-B = x and B-C = x, so A-C = 2x. F₁ = force on A due to B; F₂ = force on A due to C.

Asked

The relation between F₁ and F₂.

Concept

Force ∝ (other current) / (distance) - the doubled current and doubled distance cancel exactly.
All currents parallel ⇒ both forces are attractions in the same direction, so no minus sign appears.

Formula to be used

F = μ₀I₁I₂ ℓ2πd

Diagram

Steps involved

F₁ (from B, distance x) = μ₀(I)(I)ℓ / 2πx = μ₀I²ℓ / 2πx.
F₂ (from C, distance 2x) = μ₀(I)(2I)ℓ / 2π(2x) = μ₀I²ℓ / 2πx.
The magnitudes are equal.
All three currents are parallel, so both B and C attract A - the forces point the same way, not opposite. Option (d) with the minus sign fails.

ShortcutForce ∝ I_other/d. For B: I/x. For C: 2I/2x = I/x. The doubled current and doubled distance cancel - you can see the answer without writing a formula.

Answer(c) F₁ = F₂

Q37

Force on 10 cm of the middle wire Q

Given

Parallel wires R (20 A), Q (10 A), P (30 A). Spacings R-Q = 2 cm and Q-P = 10 cm. From the arrows, Q carries current opposite to both R and P. Length considered ℓ = 10 cm.

Asked

The force on Q, with direction.

Concept

Antiparallel currents repel, so each neighbour pushes Q away from itself.
Distance usually dominates: compare I/d for the two neighbours.

Formula to be used

Fℓ = μ₀I₁I₂2πd = 2 × 10⁻⁷ I₁I₂d

Diagram

Steps involved

Due to R (antiparallel ⇒ repulsion, pushing Q to the right): F = 2 × 10⁻⁷(20)(10)/0.02 × 0.1 = 2 × 10⁻⁴ N.
Due to P (antiparallel ⇒ repulsion, pushing Q to the left): F = 2 × 10⁻⁷(30)(10)/0.10 × 0.1 = 0.6 × 10⁻⁴ N.
Net = 2 × 10⁻⁴ − 0.6 × 10⁻⁴ = 1.4 × 10⁻⁴ N.
It points in the direction of the larger force, i.e. towards the right.

ShortcutCompare I/d: R gives 20/2 = 10, P gives 30/10 = 3. R wins comfortably despite the smaller current, because it is five times closer. Then the arrows alone set left vs right - if all three currents were parallel, the same magnitude would point left.

Answer(a) 1.4 × 10⁻⁴ N towards the right

Q38

Angle between B and I from the force

Given

I = 10 A, ℓ = 1.5 m, B = 2 T, F = 15 N.

Asked

The angle between B and the direction of the current.

Concept

BIℓ is the maximum possible force (at 90°); comparing F with it gives sin θ at once.

Formula to be used

F = BIℓ sin θ

Steps involved

15 = (2)(10)(1.5) sin θ.
15 = 30 sin θ.
sin θ = 0.5.
θ = 30°.

ShortcutCompute BIℓ = 30 N, the maximum force. The given force is exactly half, so sin θ = ½ ⇒ 30°. Half ⇒ 30°, 1/√2 ⇒ 45°, √3/2 ⇒ 60°.

Answer(a) 30°

Q39

Ionised gas in crossed E and B fields

Given

Positive and negative ions, initially at rest. E along +x, B along +z.

Asked

The direction in which each species deflects.

Concept

At t = 0 the magnetic force is zero, so E alone sets the ions moving - in opposite directions for the two signs.
Two sign reversals then cancel in the magnetic force.

Formula to be used

F = qE + q(v × B); x̂ × ẑ = −ŷ

Diagram

Steps involved

Positive ions: electric force +qE x̂ ⇒ they move along +x̂. Magnetic force = q(v x̂) × (B ẑ) = qvB(x̂ × ẑ) = −qvB ŷ ⇒ towards −y.
Negative ions: electric force (−q)E x̂ ⇒ they move along −x̂. Magnetic force = (−q)(−v x̂) × (B ẑ) = qvB(x̂ × ẑ) = −qvB ŷ ⇒ also towards −y.
Both species deflect the same way, along −y.

ShortcutTwo sign reversals cancel: the negative ion has its charge reversed and its velocity reversed, and (−q)(−v) = +qv. This is exactly why crossed E-B fields cannot separate positive from negative ions, though they do sort by q/m. Remember x̂ × ẑ = −ŷ - the one cross product that trips people up.

Answer(c) all ions deflect towards the −y-direction

Q40

Question image not supplied

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Q41

Torque about the Z-axis on a rectangular coil

Given

A rectangular coil of area A carrying current I lies in the plane of the page (the Y-Z plane, with Z up and Y to the right). The field B = B ŷ lies in the plane of the coil.

Asked

The torque about the Z-axis.

Concept

The coil's moment is normal to its plane, so here m lies along X while B lies along Y - they are perpendicular.
B in the plane of the coil ⇒ maximum torque; B perpendicular to the coil ⇒ zero torque.

Formula to be used

m = IA n̂, τ = m × B, τ = NIAB sin α

Diagram

Steps involved

The coil lies in the Y-Z plane, so its normal is along X: m = IA(±x̂).
B is along ŷ, so m ⊥ B, α = 90° and the magnitude is τ = IAB. This alone eliminates the options carrying a factor 2 (a factor 2 would need N = 2 turns; there is only one loop).
Sign: with the current flowing downward along the right arm (−ẑ there), ŷ × (−ẑ) = −x̂, so m = −IA x̂.
τ = (−IA x̂) × (B ŷ) = −IAB ẑ.
Torque of magnitude IAB directed along the negative Z-axis.

ShortcutTwo-step routine. Magnitude first: τ = NIAB sin α; B in the plane of the coil ⇒ sin α = 1 ⇒ maximum NIAB, which usually halves the options. Sign second: curl the right hand along the current for n̂, then use x̂ × ŷ = ẑ. Note the sign depends entirely on the arrowhead in the figure - if the current runs upward along the right arm, the answer flips to +Z. The magnitude IAB is certain either way.

Answer(a) IAB along the negative Z-axis

Q42

Magnetic moment of the orbiting electron in hydrogen

Given

f = 6.6 × 10¹⁵ rev/s, r = 0.528 Å = 0.528 × 10⁻¹⁰ m, e = 1.6 × 10⁻¹⁹ C.

Asked

The magnetic moment M in A·m².

Concept

Same physical picture as Q12: orbiting electron ⇒ current loop. There we found its field; here we want its moment.

Formula to be used

i = ef, M = iA = ef · πr²

Steps involved

i = (1.6 × 10⁻¹⁹)(6.6 × 10¹⁵) = 1.056 × 10⁻³ A.
A = π(0.528 × 10⁻¹⁰)² = 8.756 × 10⁻²¹ m².
M = (1.056 × 10⁻³)(8.756 × 10⁻²¹) = 9.25 × 10⁻²⁴ ≈ 1 × 10⁻²³ A·m².

ShortcutTrack exponents only: 10⁻¹⁹ × 10¹⁵ × (10⁻¹⁰)² = 10⁻²⁴, with leading digits ≈ 9, so the answer is ~10⁻²³. Worth knowing as a benchmark: this is essentially the Bohr magneton, μ_B = eh/4πm = 9.27 × 10⁻²⁴ A·m², which is exactly what ground-state hydrogen should give.

Answer(c) 1 × 10⁻²³

Q43

Total field at P from two staircase-shaped conductors

Given

Two infinitely long conductors bent into identical staircase shapes, all short sections of equal length. P lies symmetrically with respect to the two conductors, and each conductor alone produces field B at P.

Asked

The total field at P.

Concept

The whole question is which symmetry relates the two conductors.
180° rotation about P ⇒ contributions add. Mirror reflection ⇒ contributions cancel.
Every field at P from wires lying in the page must be perpendicular to the page, so the two contributions are always collinear.

Formula to be used

dB = μ₀I4πdl × r̂r²

under r̂ → −r̂ and dl → −dl: (−dl) × (−r̂) = +dl × r̂

Diagram

Steps involved

Read the figure: the second staircase is the first one rotated by 180° about P, not reflected in a line through P (both climb in the same sense and both currents point right).
Under a 180° rotation about P, an element at r̂ with direction dl maps to −r̂ with direction −dl.
Its contribution: (−dl) × (−r̂) = +dl × r̂ - both signs flip, so the cross product is unchanged.
Every element of conductor 2 therefore contributes in the same direction as its partner in conductor 1: B_total = B + B = 2B.

ShortcutFor two conductors placed symmetrically about P only two outcomes are possible: 180° rotation ⇒ 2B; mirror reflection ⇒ 0. The √2 B option is impossible on principle - both fields must be perpendicular to the page, hence collinear. The staircase shape and 'equal short sections' are deliberate distractions; you are handed the per-conductor result B.

Answer(d) 2B

Q44

Where the particle re-crosses the Y-axis

Given

Charge +q, mass m, launched from the origin with speed v₀ along +x̂; field B = −B₀k̂ (into the page, B₀ > 0).

Asked

The value of y at which the particle passes through (0, y, 0).

Concept

v ⊥ B ⇒ circular path. The centre lies one radius away along the initial force direction.
Starting on an axis moving perpendicular to it ⇒ the particle returns to that axis after half a circle, at a diameter.

Formula to be used

F = q(v × B), r = mvqB, y = 2r

Diagram

Steps involved

Initial force: F = q(v₀x̂) × (−B₀ẑ) = −qv₀B₀(x̂ × ẑ) = +qv₀B₀ ŷ - the particle curves towards +y.
Radius: r = mv₀/qB₀.
The centre of the circle sits at (0, r), directly along the centripetal direction.
The circle cuts the Y-axis at the start point (0, 0) and at the diametrically opposite point (0, 2r).
y = 2r = 2mv₀/qB₀.

ShortcutStart on an axis moving perpendicular to it ⇒ the particle returns after half a circle at twice the radius. Recognise 'diameter, not radius' and write the answer immediately. The sign is positive because the launch force pointed along +ŷ (remember x̂ × ẑ = −ŷ).

Answer(a) 2mv₀/qB₀

Q45

Torque on a 100-turn coil between magnet poles

Given

N = 100 turns, I = 2 A, B = 0.2 Wb/m², coil ABCD measuring 8 cm × 10 cm. The N and S poles sit left and right, so B lies horizontally, in the plane of the coil.

Asked

The magnitude and sense of the torque.

Concept

B in the plane of the coil ⇒ m ⊥ B ⇒ torque is maximum.
The factor of 100 (the turns) is the single biggest source of error here.

Formula to be used

τ = N m × B, τ = NIAB sin α = NIAB when α = 90°

Diagram

Steps involved

Area A = (0.08)(0.10) = 8 × 10⁻³ m².
τ = NIAB = (100)(2)(8 × 10⁻³)(0.2).
τ = 100 × 2 × 1.6 × 10⁻³ = 0.32 N·m.
Sense: take x̂ = right (along B, N→S) and ẑ = out of the page. For the current drawn, m = m ẑ, so τ = m ẑ × B x̂ = mB ŷ, i.e. upward.
A torque along +ŷ rotates points at −x out of the page; side AD is the left-hand side, so AD swings out of the page.

ShortcutDo the magnitude in one line - NIAB with sin α = 1 - and the exponent alone eliminates half the options: 0.32 vs 0.0032 is a factor of 100, exactly the number of turns. Then only the rotation sense remains, decided by the arrowheads. Master rule: B in the plane ⇒ τ = NIAB (maximum); B perpendicular to the coil ⇒ τ = 0 (stable equilibrium).

Answer(a) 0.32 N·m, tending to rotate side AD out of the page

No.	Answer	Topic
Q1	(a) B is perpendicular to v	Biot–Savart law — direction of B due to a moving electron
Q2	(a) towards north	Field above a wire carrying current east → west
Q3	(a) undergoes acceleration all the time	Cyclotron — behaviour of the charged particle
Q4	(d) continues to move with uniform velocity along the axis	Electron projected along the axis of a solenoid
Q5	(c) Z-axis	Deflection of a proton in a perpendicular field
Q6	(b) and (c) — both are correct (multiple-correct item)	Dimensional formulae of ε₀ and μ₀
Q7	(a) outside the cable	Coaxial cable — where is B zero?
Q8	(a) the magnitude of the magnetic moment diminishes (m → m/√2)	Half of a circular loop bent into the perpendicular plane
Q9	(c) q and m	Ratio of magnetic moment to angular momentum
Q10	(d) 40 MeV	Maximum proton energy from the same cyclotron
Q11	(b) flat at zero up to R, then a 1/r decay	B-r graph for a long thin hollow cylinder
Q12	(c) 12.5 Wb/m²	Magnetic field at the centre of the hydrogen orbit
Q13	(b) and (d) - both correct (multiple-correct item)	Undeflected passage through crossed E and B (velocity selector)
Q14	(c) 2	Ratio of fields at the centres of two coils
Q15	(b) towards C	Resultant force on the middle wire B
Q16	(d) 1/π	Magnetic moment of a wire bent into a circle
Q17	(a) B′ = 4B	Same wire: one turn versus two turns
Q18	(d) (e/m)₁ + (e/m)₂ = 0	Identical helical paths traversed in opposite sense
Q19	(b) 9B₁	One loop versus three loops from the same wire
Q20	(c) equal to g	Charge released from the top of a vertical solenoid
Q21	(a) 4.35 m	Distance advanced along B after two complete circles
Q22	(d) zero	Charge on the axis of a circular wire fed at diametrically opposite points
Q23	(b) q(b − a)B/m	Minimum speed to cross a magnetic field slab
Q24	(c) helix of radius 0.07 m, period 0.5 × 10⁻⁶ s	Path of a proton projected at 60° to the X-axis
Q25	(d) zero	Field at the centre of an equilateral triangle of uniform wire
Q26	(a) (μ₀2I/4πR)(π + 1)	Infinitely long conductor bent into a circle (tangent loop)
Q27	(d) μ₀I/4R + μ₀I/4πR	Field at O due to an infinite wire with a semicircular bend
Q28	(d) √2 : 1	Two identical coils with planes at right angles
Q29	(b) is doubled	Tension in a current-carrying flexible loop
Q30	(c) √(v/B)	Radius of an electron orbit from the field it produces
Q31	(c) π : 4	Magnetic moment - square versus circle from the same wire
Q32	(c) (R₁/R₂)²	Ratio of masses from radii after acceleration through the same potential
Q33	(d) zero	Field at O due to two bent wires ABC and DEF
Q34	(d) zero	Circular conductor fed by a battery - field at the centre
Q35	—	not supplied
Q36	(c) F₁ = F₂	Forces on wire A from wires B and C
Q37	(a) 1.4 × 10⁻⁴ N towards the right	Force on 10 cm of the middle wire Q
Q38	(a) 30°	Angle between B and I from the force
Q39	(c) all ions deflect towards the −y-direction	Ionised gas in crossed E and B fields
Q40	—	not supplied
Q41	(a) IAB along the negative Z-axis	Torque about the Z-axis on a rectangular coil
Q42	(c) 1 × 10⁻²³	Magnetic moment of the orbiting electron in hydrogen
Q43	(d) 2B	Total field at P from two staircase-shaped conductors
Q44	(a) 2mv₀/qB₀	Where the particle re-crosses the Y-axis
Q45	(a) 0.32 N·m, tending to rotate side AD out of the page	Torque on a 100-turn coil between magnet poles