Consider a tightly wound 100-turn coil of radius 10 cm, carrying a current of 1 A. What will be the magnitude of the magnetic field at the centre of the coil?
- (a)6.28 × 10⁻⁴ T
- (b)6.28 × 10⁻³ T
- (c)2.68 × 10⁻⁴ T
- (d)6.82 × 10⁻² T
Correct — A, 6.28 × 10⁻⁴ T.
A tightly wound coil of N turns behaves like N circular loops of practically the same radius sitting on top of one another, so the contribution of each turn at the centre adds. The magnitude is B = μ₀NI/(2R), directed along the coil's axis.
The constant collapses neatly. With μ₀ = 4π × 10⁻⁷ T·m·A⁻¹, the factor μ₀/2 becomes 2π × 10⁻⁷, so B = 2π × 10⁻⁷ × NI/R.
Now substitute, with the radius written in metres: N = 100, I = 1 A, R = 0.1 m. That gives B = 2π × 10⁻⁷ × 100 ÷ 0.1 = 2π × 10⁻⁴, which evaluates to 6.28 × 10⁻⁴ T.
Carry away the routine rather than the number: convert the radius to metres, finish the multiplication to fix the digits, then match the power of ten separately.
- (b)6.28 × 10⁻³ T — The digits are right and the exponent is one step too big, so this is exactly ten times the field the coil in the stem produces.
Run the formula backwards and it belongs to a 100-turn coil of about 1 cm radius carrying 1 A, since B varies as 1/R and shrinking the radius tenfold multiplies the field by ten.
The radius is therefore the one quantity here that a slipped decimal place punishes directly.
- (c)2.68 × 10⁻⁴ T — The power of ten matches the correct value, so this option survives a quick order-of-magnitude glance; the digits are where it parts company, being 6.28 with its first two figures interchanged.
Solved backwards, this field belongs to a 100-turn, 1 A coil of about 23 cm radius, not to the 10 cm coil in the stem.
The protection is to actually evaluate 2 × 3.1416 rather than to recall the string of digits attached to 2π.
- (d)6.82 × 10⁻² T — This one is wrong twice over: the digits 6.28 appear transposed as 6.82, and the exponent is two steps too large.
At the given radius and current it is the reading of a coil wound with roughly eleven thousand turns instead of a hundred.
A size check catches it. Against the 5 × 10⁻⁵ T of the Earth's field near the surface, 6.82 × 10⁻² T is on the order of a thousand times as strong, which is a lot to ask of a hundred turns carrying a single ampere.
A current-carrying wire sets up a magnetic field around it, and the Biot–Savart law adds up the contributions of every element of the wire.
For a single circular loop of radius R carrying current I, each element lies the same distance from the centre and each contribution points the same way along the axis. The sum is therefore clean: B = μ₀I/(2R).
Winding the wire into N tight turns of essentially the same radius simply repeats that loop N times, and the fields superpose. Hence B = μ₀NI/(2R) at the centre.
The direction follows the right hand: curl the fingers along the current in the loop and the extended thumb points along the field at the centre.
This sits in the magnetic effects of electric current, alongside the field of a long straight wire, of a solenoid and of a toroid. The four results share the constant μ₀ but differ in what sits under it, which is why they have to be kept apart rather than blended.
The circular-coil result is the working formula behind the tangent galvanometer and the Helmholtz pair used to make a small uniform field in a laboratory.
As an exam item it is a two-step calculation: recall the formula, then handle SI units. Both steps carry their own way of going wrong.
- The magnetic field at the centre of a single circular loop of radius R carrying current I is B = μ₀I/(2R).
- For N tightly wound turns of the same radius the contributions add, giving B = μ₀NI/(2R) at the centre.
- The permeability of free space μ₀ is 4π × 10⁻⁷ T·m·A⁻¹, so μ₀/2 equals 2π × 10⁻⁷ in SI units.
- Substituting N = 100, I = 1 A and R = 0.1 m gives B = 2π × 10⁻⁴ T, that is 6.28 × 10⁻⁴ T.
- The centre field varies directly with the number of turns and the current, and inversely with the coil's radius.
- The field at the centre lies along the coil's axis, its direction given by the right-hand curl rule.
- A single turn of this coil would give 6.28 × 10⁻⁶ T at the centre; the hundred turns supply the factor of 100.
- The Earth's magnetic field near the surface is about 5 × 10⁻⁵ T, so this coil's central field is roughly ten times larger.
- For a long solenoid the corresponding result is B = μ₀nI, where n is turns per unit length, not the total number of turns.
The radius is the step where centimetres must become metres; the last row fixes both the digits and the exponent.
- Substituting the radius as 10 rather than 0.1 — the formula needs metres, and a centimetre reading shifts the result by two powers of ten.
- All four options are built from the digits 6, 2 and 8, so the digits alone cannot separate them; the exponent has to be checked as a second step.
- Reaching for the solenoid result B = μ₀nI, in which n is turns per unit length, when the stem gives a total of 100 turns on a coil.
- Using the straight-wire form μ₀I/(2πr): the 2π sits in the denominator for the wire, while the centre of a loop carries a plain 2.
- Forgetting the factor N and computing for one turn, which gives 6.28 × 10⁻⁶ T here.
- Treating 2π × 10⁻⁴ as a finished answer and matching it to an option by memory instead of evaluating it as 6.28 × 10⁻⁴.
The same relation turns up in several disguises. One is a straight substitution like this stem, where the arithmetic and the unit conversion carry the whole mark.
Another is a proportionality: the current, the turns or the radius is changed by a stated factor and the new field is asked for, which needs only B ∝ NI/R and no constants at all.
A third is formula identification, where the loop, straight-wire, solenoid and toroid expressions are placed side by side. The item can also be inverted, supplying the field and asking for the radius or the number of turns.
GEO_GS_2020_Q1042020The same centre-of-loop relation, approached from the other end. That item doubles the current and the turns and asks how the field changes, so B ∝ NI settles it without any constant; this stem asks for a number, so μ₀/2 and the radius in metres both have to be handled. Its wording also describes the turns as turns per unit length, while the UKPSC stem fixes a total of 100 turns on a 10 cm coil.
NDA_GAT_2022_I_Q672022Both items turn on an inverse dependence on distance, but the geometry differs. That one is the field of a long straight wire at a point outside it, falling as 1/r; here the same inverse dependence appears on the coil's own radius. The expressions are not interchangeable — the straight wire carries 2π in its denominator, the centre of a loop a plain 2.
NDA_GAT_2025_II_Q622025Same chapter, different geometry and a different property asked. That item is about the uniformity of the field inside a long solenoid rather than its magnitude. It is worth reading beside this one because the solenoid result B = μ₀nI uses turns per unit length, whereas the coil formula used here uses the total number of turns.
- practice — not a real PYQ
A tightly wound circular coil of 200 turns and radius 20 cm carries a current of 2 A. The magnitude of the magnetic field at its centre is:
- (a)1.26 × 10⁻³ T
- (b)6.28 × 10⁻⁴ T
- (c)2.51 × 10⁻³ T
- (d)1.26 × 10⁻⁵ T
Answera — B = μ₀NI/(2R) = 2π × 10⁻⁷ × 200 × 2 ÷ 0.2 = 4π × 10⁻⁴, which is 1.26 × 10⁻³ T.Option b is what the same coil would give at 1 A, so it is the figure left behind by ignoring that the current is 2 A.
Option c uses 0.1 m for the radius instead of 0.2 m and so comes out twice too large. Option d leaves the radius as 20, that is in centimetres, and lands two powers of ten low.
- practice — not a real PYQ
The current through a tightly wound circular coil is doubled and its radius is also doubled, the number of turns remaining unchanged. The magnetic field at the centre of the coil will:
- (a)become four times the original
- (b)become twice the original
- (c)remain unchanged
- (d)become half the original
Answerc — the centre field goes as B = μ₀NI/(2R), so B ∝ I/R when N is fixed. Doubling I doubles the numerator, doubling R doubles the denominator, and the two changes cancel exactly.Option a would need both changes to raise the field, which ignores that R sits in the denominator. Option b counts the current and forgets the radius; option d counts the radius and forgets the current.
- practice — not a real PYQ
Which expression gives the magnitude of the magnetic field at the centre of a tightly wound circular coil of N turns and radius R carrying a current I?
- (a)μ₀NI/(2R)
- (b)μ₀I/(2πR)
- (c)μ₀NI/(2πR)
- (d)μ₀nI, with n the number of turns per unit length
Answera — the Biot–Savart contributions of a circular loop add at the centre to μ₀I/(2R), and N tight turns multiply that by N.Option b is the field of a long straight wire at a perpendicular distance R from it. Option c is the field inside a toroid of mean radius R wound with N turns.
Option d is the field inside a long solenoid, which is set by turns per unit length and is independent of the coil radius.
- practice — not a real PYQ
A tightly wound circular coil of 50 turns carrying a current of 1 A produces a magnetic field of 3.14 × 10⁻⁴ T at its centre. The radius of the coil is:
- (a)5 cm
- (b)10 cm
- (c)20 cm
- (d)50 cm
Answerb — rearranging B = μ₀NI/(2R) gives R = μ₀NI/(2B) = 2π × 10⁻⁷ × 50 × 1 ÷ (3.14 × 10⁻⁴) = 0.1 m, that is 10 cm.Because the field varies as 1/R, the other radii miss the stated value: 5 cm would double it to about 6.28 × 10⁻⁴ T, 20 cm would halve it to about 1.57 × 10⁻⁴ T, and 50 cm would leave about 6.28 × 10⁻⁵ T.