A spinning ring, a charged disc, an orbiting electron — all of them are current loops wearing a disguise.
Part 1 — The concept, explained simply
A charge going round in a circle is a current. Once you see that, a whole family of questions becomes ordinary loop problems.
1. The one idea behind the whole topic
Current means "how much charge passes a point each second". So take a charged ring spinning about its axis, and stand at one spot on the ring. Every time the ring completes a turn, the whole charge q sweeps past you.
Picture it: a fairground carousel with one rider. Stand beside the track. If the carousel goes round twice a second, the rider passes you twice a second. Speed the carousel up and "riders per second" rises — that is exactly what current measures.
Everything else follows from that single step. Once you have I, use the loop formulas you already know.
2. From spin to current
The ring turns once every T = 2π/ω seconds, so:
I = q / T = qω / 2π = qν
where ν is the number of turns per second. If a question gives the frequency directly, I = qν is even quicker.
3. Then use the loop results you already know
Quantity
Loop formula
Rotating ring result
Field at the centre
B = μ0I/2R
B = μ0qω / 4πR
Magnetic moment
m = IA
m = qωR² / 2
Do not memorise the right-hand column. Derive it each time: work out I, then substitute. Two lines, and no chance of mixing up a factor.
4. A charged disc is slightly different
A disc is not a single ring — it is many rings of different radii nested together. The outer rings carry more charge and enclose more area, so the sums come out different:
Disc: Bcentre = μ0qω / 2πR m = qωR² / 4
Compare with the ring: the disc gives twice the field and half the moment. Worth noticing, because questions like to pair them.
5. The orbiting electron
Same idea, applied to an atom. An electron going round a nucleus at radius r with speed v takes T = 2πr/v per orbit, so I = ev/2πr and:
m = I A = (ev/2πr)(πr²) = e v r / 2
This can also be written m = eL/2me using the angular momentum L = mevr — the form that appears in atomic physics.
Part 2 — Formula sheet
Situation
Formula
Note
Spin to current
I = qω/2π = qν = q/T
The step everything else depends on.
Ring: field at centre
B = μ0qω/4πR
From B = μ0I/2R.
Ring: magnetic moment
m = qωR²/2
From m = IA = IπR².
Disc: field at centre
B = μ0qω/2πR
Twice the ring value.
Disc: magnetic moment
m = qωR²/4
Half the ring value.
Orbiting charge: current
I = ev/2πr
T = 2πr/v.
Orbiting charge: moment
m = evr/2
The atomic-physics form.
In terms of angular momentum
m = eL/2me
L = mevr.
Orbiting charge: field at centre
B = μ0ev/4πr²
From B = μ0I/2r.
Frequency form
I = qν, ν = ω/2π
Use when frequency is given directly.
The two-step method — use it every time
Find the equivalent current. How much charge passes a point each second? I = q/T.
Treat it as an ordinary loop. Use B = μ0I/2R or m = IA as usual.
Two lines beats memorising four separate results.
Part 3 — 20 questions with step-by-step solutions
Attempt each on paper first. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Ring
A ring of radius r carries a uniformly distributed charge q and is rotated about its own axis with angular frequency ω. The magnetic field at the centre is:
(a) (μ0/4π)(qω/r)
(b) (μ0/4π)(r/qω)
(c) (μ0/4π)(q/rω)
(d) (μ0/4π)(ω/qr)
Show step-by-step solution
GivenRing of radius r, charge q, angular frequency ω
AskedField at the centre
ConceptConvert the spin into an equivalent current, then use the loop formula.
The magnetic moment of a ring of radius R carrying charge q and rotating with angular speed ω is:
(a) qωR²
(b) qωR²/2
(c) qωR²/4
(d) 2qωR²
Show step-by-step solution
GivenRing, charge q, radius R, angular speed ω
AskedMagnetic moment m
ConceptEquivalent current times area.
Formulam = IA, I = qω/2π
SolutionI = qω/2π. A = πR². m = (qω/2π)(πR²) = qωR²/2.
Answer: qωR²/2
Q3Current
A charge q revolves in a circle with frequency ν. The equivalent current is:
(a) qν
(b) q/ν
(c) 2πqν
(d) qν/2π
Show step-by-step solution
GivenCharge q, frequency ν revolutions per second
AskedEquivalent current
ConceptCurrent is charge per second, and the charge passes a point ν times each second.
FormulaI = q/T = qν
SolutionThe period is T = 1/ν. I = q/T = qν. (If ω is given instead, use ν = ω/2π.)
Answer: qν
Q4Numerical
A ring of radius 10 cm carrying a charge of 1 μC rotates at 100 rad/s about its axis. The field at its centre is:
(a) 1 × 10−10 T
(b) 2 × 10−10 T
(c) 1 × 10−9 T
(d) 5 × 10−11 T
Show step-by-step solution
GivenR = 0.1 m, q = 10−6 C, ω = 100 rad/s
AskedField at the centre
ConceptTwo steps: current first, then the loop formula.
FormulaB = μ0qω/4πR
SolutionUse μ0/4π = 10−7. B = 10−7 × (10−6 × 100)/0.1 = 10−7 × 10−3 = 1 × 10−10 T
Answer: 1 × 10−10 T
Q5Disc
A uniformly charged disc of radius R and total charge q rotates with angular velocity ω. The field at its centre is:
(a) μ0qω/4πR
(b) μ0qω/2πR
(c) μ0qω/πR
(d) μ0qωR/2π
Show step-by-step solution
GivenDisc of radius R, charge q, angular velocity ω
AskedField at the centre
ConceptA disc is many nested rings; integrating gives twice the single-ring result.
FormulaB = μ0qω/2πR
SolutionSplit the disc into rings of radius x and thickness dx. Each contributes μ0dI/2x, and integrating from 0 to R gives μ0qω/2πR. This is exactly twice the value for a ring of the same radius and charge.
Answer: μ0qω/2πR
Q6Disc
The magnetic moment of a uniformly charged rotating disc, compared with a ring of the same charge, radius and angular speed, is:
(a) twice as large
(b) half as large
(c) equal
(d) four times as large
Show step-by-step solution
GivenDisc vs ring, same q, R, ω
AskedComparison of magnetic moments
ConceptThe inner parts of a disc enclose less area, pulling the average down.
Formularing qωR²/2; disc qωR²/4
SolutionRing: m = qωR²/2. Disc: m = qωR²/4. So the disc's moment is half. (Its central FIELD, by contrast, is twice as large.)
Answer: half as large
Q7Orbiting
An electron moves in a circular orbit of radius r with speed v. The equivalent current is:
(a) ev/2πr
(b) ev/r
(c) evr/2
(d) e/2πrv
Show step-by-step solution
GivenElectron, orbit radius r, speed v
AskedEquivalent current
ConceptFind the orbital period, then divide the charge by it.
FormulaI = e/T, T = 2πr/v
SolutionT = circumference/speed = 2πr/v. I = e/T = ev/2πr.
Answer: ev/2πr
Q8Orbiting
The magnetic moment of an electron in a circular orbit of radius r with speed v is:
(a) evr
(b) evr/2
(c) 2evr
(d) ev/2r
Show step-by-step solution
GivenElectron orbiting at radius r with speed v
AskedMagnetic moment
ConceptEquivalent current times orbital area.
Formulam = IA = (ev/2πr)(πr²)
SolutionI = ev/2πr. A = πr². m = (ev/2πr)(πr²) = evr/2.
Answer: evr/2
Q9Orbiting
In terms of the orbital angular momentum L of an electron of mass me, the magnetic moment is:
(a) eL/me
(b) eL/2me
(c) 2eL/me
(d) eL me/2
Show step-by-step solution
GivenElectron with orbital angular momentum L = mevr
AskedMagnetic moment in terms of L
ConceptSubstitute vr = L/me into m = evr/2.
Formulam = evr/2, L = mevr
SolutionFrom L = mevr, we get vr = L/me. m = e(vr)/2 = eL/2me. The ratio m/L = e/2me is called the gyromagnetic ratio.
Answer: eL/2me
Q10Numerical
An electron (e = 1.6 × 10−19 C) orbits at radius 0.53 × 10−10 m with speed 2.2 × 106 m/s. Its magnetic moment is about:
(a) 9.3 × 10−24 A m²
(b) 4.7 × 10−24 A m²
(c) 1.9 × 10−23 A m²
(d) 9.3 × 10−22 A m²
Show step-by-step solution
Givene = 1.6×10−19, r = 0.53×10−10, v = 2.2×106
AskedMagnetic moment
ConceptDirect substitution. This value is the Bohr magneton.
Formulam = evr/2
Solutionm = (1.6×10−19 × 2.2×106 × 0.53×10−10)/2 Numerator = 1.87 × 10−23 m = 9.3 × 10−24 A m².
Answer: 9.3 × 10−24 A m²
Q11Ring
If the angular speed of a rotating charged ring is doubled, the field at its centre:
(a) halves
(b) doubles
(c) becomes four times
(d) is unchanged
Show step-by-step solution
Givenω → 2ω, same q and R
AskedNew field
ConceptThe current is proportional to ω, and the field is proportional to the current.
FormulaB = μ0qω/4πR
SolutionI = qω/2π, so doubling ω doubles I. B ∝ I, so B doubles too.
Answer: doubles
Q12Ring
A charged ring rotating about its axis produces a magnetic moment m. If its radius is doubled while q and ω stay the same, the moment becomes:
(a) 2m
(b) 4m
(c) m/2
(d) m/4
Show step-by-step solution
GivenR → 2R, same q and ω
AskedNew magnetic moment
ConceptThe moment depends on R squared.
Formulam = qωR²/2
Solutionm ∝ R². Doubling R multiplies the moment by 4. m' = 4m.
Answer: 4m
Q13Numerical
A ring carries 2 μC and rotates at 50 revolutions per second. The equivalent current is:
(a) 1 × 10−4 A
(b) 2 × 10−4 A
(c) 1 × 10−5 A
(d) 5 × 10−5 A
Show step-by-step solution
Givenq = 2 × 10−6 C, ν = 50 rev/s
AskedEquivalent current
ConceptFrequency is given directly, so use I = qν.
FormulaI = qν
SolutionI = 2×10−6 × 50 = 1 × 10−4 A.
Answer: 1 × 10−4 A
Q14Concept
A charged ring rotating about its own axis produces a magnetic field because:
(a) static charge always makes a magnetic field
(b) the moving charge constitutes a current
(c) the ring is a magnet
(d) of the Earth's field
Show step-by-step solution
GivenRotating charged ring
AskedReason for the field
ConceptOnly moving charges make magnetic fields, and rotation is motion.
FormulaI = qω/2π
SolutionA stationary charge makes only an electric field. Rotation sets the charge moving, and moving charge is a current. That current then produces a magnetic field just like any loop.
Answer: the moving charge constitutes a current
Q15Comparison
A ring and a disc have the same charge, radius and angular speed. The ratio of the fields at their centres (ring : disc) is:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show step-by-step solution
GivenSame q, R, ω for both
AskedRatio of central fields
ConceptCompare the two standard results.
Formularing μ0qω/4πR; disc μ0qω/2πR
SolutionRing: μ0qω/4πR. Disc: μ0qω/2πR, which is twice as large. Ratio ring : disc = 1 : 2.
Answer: 1 : 2
Q16Numerical
A ring of radius 20 cm carries 4 μC and rotates at 200 rad/s. Its magnetic moment is:
(a) 1.6 × 10−5 A m²
(b) 3.2 × 10−5 A m²
(c) 8 × 10−6 A m²
(d) 1.6 × 10−4 A m²
Show step-by-step solution
GivenR = 0.2 m, q = 4×10−6 C, ω = 200 rad/s
AskedMagnetic moment
ConceptStraight substitution.
Formulam = qωR²/2
Solutionm = (4×10−6 × 200 × 0.04)/2 Numerator = 3.2 × 10−5 m = 1.6 × 10−5 A m².
Answer: 1.6 × 10−5 A m²
Q17Direction
The magnetic moment of a rotating positively charged ring points:
(a) along the axis, by the right-hand rule applied to the rotation
(b) radially outward
(c) opposite to the angular velocity
(d) in the plane of the ring
Show step-by-step solution
GivenPositively charged ring spinning about its axis
AskedDirection of m
ConceptThe rotation gives the current direction; then apply the loop rule.
FormulaRight-hand rule for a loop
SolutionA positive charge moving round the ring is a current in the same sense as the rotation. Curl the right fingers along that current; the thumb gives m. So m lies along the axis, parallel to the angular velocity. (For a negative charge it would be antiparallel.)
Answer: along the axis, by the right-hand rule applied to the rotation
Q18Method
The quickest reliable method for any rotating-charge question is:
(a) memorise all four results
(b) find the equivalent current first, then use the loop formulas
(c) use F = qvB
(d) use Ampere's law
Show step-by-step solution
GivenAny rotating charged body
AskedBest method
ConceptDeriving from I takes two lines and avoids mixing up factors of 2 and 4.
FormulaI = q/T, then B = μ0I/2R or m = IA
SolutionStep 1: how much charge passes a point per second? That is I. Step 2: it is now an ordinary current loop, so use the loop formulas. This works for rings, discs and orbiting electrons alike.
Answer: find the equivalent current first, then use the loop formulas
Q19Orbiting
An electron revolves in a circle of radius r making n revolutions per second. The magnetic field at the centre of the orbit is:
(a) μ0ne/2r
(b) μ0ne/r
(c) μ0ne/4πr
(d) μ0ner/2
Show step-by-step solution
GivenElectron, radius r, n revolutions per second
AskedField at the centre
ConceptThe frequency is given, so the current is ne.
FormulaI = ne; B = μ0I/2r
SolutionI = charge per second = ne. B = μ0I/2r = μ0ne/2r.
Answer: μ0ne/2r
Q20Trap
A student computes the field of a rotating charged disc using the ring formula. The answer will be:
(a) correct
(b) half the correct value
(c) twice the correct value
(d) zero
Show step-by-step solution
GivenRing formula applied to a disc
AskedNature of the error
ConceptA disc gives twice the ring's central field for the same q, R and ω.
Formularing μ0qω/4πR; disc μ0qω/2πR
SolutionCorrect disc value: μ0qω/2πR. Ring value used instead: μ0qω/4πR. That is half the correct answer.