Moving Charges & Magnetism — PYQ Formula Sheet

Worked past-year questions

25 problems · 2013–2016

01NEET Phase II 2016

Wire re-bent into n turns

Same wire ⇒ length fixed ⇒ each new turn has radius R/n.

n(2πr) = 2πR ⇒ r = Rn B′ = μ₀nI2r = n²·μ₀I2R

B B′ = n²B

02NEET Phase II 2016

Frequency of revolution of an electron

Speed and radius cancel out — the period depends only on B and m/q.

evB = mv²r ⇒ T = 2πmeB f = eB2πm = em·B2π

A f = 1 GHz

03NEET 2016

Square loop beside a long straight wire

Sides ⟂ to the wire cancel. Near side is pulled, far side pushed — subtract.

F₁ = μ₀Iiπ − F₂ = μ₀Ii3π

D F = 2μ₀Ii3π toward the wire

04NEET 2016

Thick wire — inside vs outside

Enclosed current scales with area, so r/2 catches only I/4.

B_in = μ₀Ir2πa² , B_out = μ₀I2πr

B B(a/2) : B(2a) = 1 : 1

05CBSE AIPMT 2015

Semicircle joined to two long wires

Semicircle gives the î term, the two half-wires give the k̂ term.

μ₀I4R (semicircle) + 2·μ₀I4πR (two half-wires)

C B = −μ₀I4πR(πî + 2k̂)

06CBSE AIPMT 2015

Electron making n rotations per second

A revolving charge is a current: it passes any point n times a second.

I = qT = ne ⇒ B = μ₀I2r

D B = μ₀ne2r

07AIIMS 2015

Loop of two arcs, radii R and 2R

Radial straight bits give zero. Each arc uses its own radius.

B_arc = θ2π·μ₀i2R 23·μ₀i2R + 13·μ₀i4R

B B = 5μ₀i12R into the page

08UK PMT 2015

Acceleration of a proton

Horizontal velocity ⟂ vertical field, so sin θ = 1. Then divide by mass.

F = qvB , a = Fm e = 1.6×10⁻¹⁹ C, m_p = 1.67×10⁻²⁷ kg

C a = 5.8×10¹¹ m/s²

09Kerala CEE 2015

Ratio B′ : B for a re-bent coil

Identical to problem 1 — the word “bent” is the whole question.

r = Rn ⇒ B′B = n²

C n² : 1

10Guj. CET 2015

Two parallel wires, same direction

Like currents attract — the opposite of like charges.

FL = μ₀I₁I₂2πd

A They attract each other

11CG PMT 2015

Which statement about the B–x graph is false?

Slope zero only at the centre; curvature zero at the inflection points.

B = μ₀IR²2(R²+x²)^3/2 d²B/dx² = 0 at x = ±R/2

C “dB/dx = 0 at A, A′” is false

12WB JEE 2015

Mass ratio from two path radii

Same V ⇒ heavier particle moves slower ⇒ r ∝ √m, not m.

r = 1B√ 2mV/q

C m₁/m₂ = (R₁/R₂)²

13UP CPMT 2015

Spinning ring of charge

Convert λ and ω into a current first; the radius then cancels out.

Q = λ·2πr , I = Qω2π = λrω

C B = μ₀λω2

14Manipal 2015

Mass ratio, numerical version

The charge value (+6 C) is a decoy — it cancels in the ratio.

m_Am_B = (23)²

C 4 : 9

15KCET 2015

Radius from specific charge

Wb/m² = tesla. Keep q/m intact as one block.

r = v(q/m)B

A r = 100 m

16KCET 2015

What can a cyclotron accelerate?

Both forces carry q. Sign is irrelevant; zero charge is fatal.

f = qB2πm , K_max = q²B²R²2m

C Both positive and negative charges

17Manipal 2015

Two parallel positron beams

A free beam isn’t neutral, so electric repulsion survives — and wins.

F_BF_E = μ₀ε₀v² = v²c² < 1

B They repel each other

18KCET 2015

Two concentric coils at right angles

Fields are ⟂ to their own planes, so ⟂ to each other — Pythagoras.

B = μ₀2R√ 3²+4² , R = 2π cm

A B = 5×10⁻⁵ Wb/m²

19WB JEE 2015

When is an emf induced?

Only if B, A or θ changes. Circle has max area for a fixed perimeter.

ε = −dΦdt , Φ = BA cos θ

B, D Rotation about a diameter; deformation

20KCET 2014

Field inside a solenoid

n is turns per metre. The radius plays no part at all.

B = μ₀nI , n = NL B_end = μ₀nI2

B B = 6.28×10⁻³ T

21Kerala CEE 2014

Field in the open space inside a toroid

An Amperian loop in the hole encloses no wire, so B = 0. The data is bait.

B_core = μ₀NI2πr ; hole & outside: I_enc = 0

B Zero

22NEET 2013

Finding both E and B in a room

“At rest” isolates E, since magnetism cannot touch a stationary charge.

F⃗ = q(E⃗ + v⃗ × B⃗) north × down = west

B E = ma₀/e west; B = 2ma₀/ev₀ down

23UK PMT 2014

Axial field → field at the centre

Divide the two formulas: μ₀ and I vanish, leaving pure geometry.

B_centreB_axis = (R²+x²R²)^3/2 = 12527

B 250 μT

24CBSE AIPMT 2014

Crossed wires, point off the plane

P is distance d from both wires, and the two fields are ⟂.

B = μ₀2πd√ I₁²+I₂²

D μ₀2πd(I₁²+I₂²)^1/2

25WB JEE 2014

Radius from kinetic energy

Momentum route is fastest: p = √(2mE), then r = p/qB.

r = pqB = √ 2mE qB

B r = √(2Em)/qB

Master formula bank

Learn these, derive the rest

Sources of magnetic field

Long straight wire (infinite): B = μ₀I2πd
Semi-infinite wire, point opposite its end: B = μ₀I4πd
Circular coil of n turns, at centre: B = μ₀nI2R
Arc of angle θ (radians): B = μ₀Iθ4πR → half: μ₀I4R, quarter: μ₀I8R
On the axis of a coil: B = μ₀IR²2(R²+x²)^3/2
Solenoid (inside / at the end): B = μ₀nI B_end = μ₀nI2 outside: 0
Toroid (within the core only): B = μ₀NI2πr hole & outside: 0
Thick wire, inside the metal: B = μ₀Ir2πa² (max at r = a)

Charge moving in a field

The one radius formula, four disguises: r = mvqB = pqB = √ 2mE qB = 1B√ 2mV/q
Period and frequency (both free of v and r): T = 2πmqB f = qB2πm
Cyclotron maximum energy: K_max = q²B²R²2m
Given q/m as one block: r = v(q/m)B f = (q/m)B2π

Forces

Lorentz force: F⃗ = q(E⃗ + v⃗ × B⃗)
On a current-carrying conductor: F = BIL sin θ
Between two parallel wires: FL = μ₀I₁I₂2πd — like currents attract
Free charged beams: F_B/F_E = v²/c² < 1 — repulsion always wins

Laws & conversions

Ampère’s circuital law: ∮B⃗·dl⃗ = μ₀I_enc ⇒ B·2πr = μ₀I_enc
Enclosed current in a uniform wire: I_enc = Ir²a²
Faraday’s law (Ch 6, but it appears here): ε = −dΦdt, Φ = BA cos θ
Rotating charge → current: I = QT = Qf = Qω2π
Ratio shortcut (axis vs centre): B_centreB_axis = (R²+x²R²)^3/2
Constants: μ₀ = 4π×10⁻⁷ · e = 1.6×10⁻¹⁹ C · m_p = 1.67×10⁻²⁷ kg · m_e = 9.1×10⁻³¹ kg

Combining two fields at the same point
Arrangement	Resultant
Same plane, same sense	B₁ + B₂
Same plane, opposite sense	B₁ − B₂
Planes at 90°	√ B₁²+B₂²
Planes at angle θ	√ B₁²+B₂²+2B₁B₂cos θ

Choosing the right form of r
The question gives you	Use	Dependence
speed v	r = mv/qB	r ∝ m
momentum p	r = p/qB	r ∝ p
kinetic energy E	r = √(2mE)/qB	r ∝ √m
accelerating voltage V	r = (1/B)√(2mV/q)	r ∝ √(m/q)

Five ideas that solved most of these

Pattern recognition beats calculation

Moving charge is a current. Whether it’s one electron orbiting (Q6) or a whole ring spinning (Q13), convert first with I = Qf = Qω/2π, then use the ordinary loop formula.
“Bent” means the radius shrinks. Q1 and Q9. Wire length is conserved, so r = R/n, and the answer picks up a square: n²B. If the question instead gives a fresh coil of the same radius, it’s only nB.
Match the radius formula to the data. Q12, Q14, Q15, Q25 are one formula in four costumes. The words “accelerated through a potential difference” always mean a square root is coming.
Perpendicular fields never add arithmetically. Q18 and Q24. If the two sources sit at 90°, it is Pythagoras — and examiners plant the arithmetic sum as a distractor every time.
Learn where the field is zero. Radial segments of a loop, points on a wire’s own line, the hole of a toroid, outside a solenoid, a stationary charge, and a rigid loop translated through a uniform field. Spotting a zero often is the question.

Traps that repeated

Where the marks leak

The mistake	Appeared in	Fix
Using μ₀I/2πd (straight wire) where μ₀I/2R (loop) belongs	Q6, Q13	Circle → 2R. Straight → 2πd.
Using one radius for both arcs	Q7	Each arc carries its own R.
Writing r ∝ m instead of r ∝ √m	Q12, Q14, Q25	v itself depends on m.
Adding perpendicular fields directly	Q18, Q24	3 + 4 = 7 is the trap; 5 is the answer.
Calculating when the answer is zero	Q21	Read the region before touching the numbers.
Confusing dB/dx = 0 with d²B/dx² = 0	Q11	Flat ≠ straight.
Forgetting to convert cm → m for turns per metre	Q20	n = N/L with L in metres.
Assuming charged beams behave like current-carrying wires	Q17	Wires are neutral; beams are not.