Irodov 3.239: The Winding Is One Turn Too – Amit & Deepali Sinha Foundation

Problem Irodov 3.239

A wooden ring-shaped core of very small cross-section carries $N = 2.5\cdot 10^{3}$ turns of wire wound evenly round it, and a current $I$ flows in the wire. Find the ratio $\eta$ of the magnetic induction inside the core to the magnetic induction at the centre of the ring.

The toroid seen along its axis: the thin core with its N turns, the current I in the leads and on two turns (sense chosen), and the centre O of the ring. — The toroid seen along its axis: the thin core with its $N$ turns, the current $I$ in the leads and on two turns (sense chosen), and the centre $O$ of the ring.

Given

$N = 2.5\cdot 10^{3}$
current $I$
wooden core, $\mu = 1$ (assumed)
very small cross-section

Find

\eta = B_{\mathrm{core}}/B_O

Solution

Step 1The field in the core

The field B in the core, along the ring, for the current drawn (sense chosen); R is the radius of the ring. — The field $\vect B$ in the core, along the ring, for the current drawn (sense chosen); $R$ is the radius of the ring.

The single turn of radius R that the winding makes by advancing once round the ring (sense chosen), and its B at O. — The single turn of radius $R$ that the winding makes by advancing once round the ring (sense chosen), and its $\vect B$ at $O$ .

Let $R$ be the radius of the ring. By symmetry the field in the core runs along the ring with the same magnitude all round. The circle of radius $R$ through the core links all $N$ turns, so Ampere’s law gives

B_{\mathrm{core}}\cdot 2\pi R = \mu_0 N I

B_{\mathrm{core}} = \frac{\mu_0 N I}{2\pi R}

(1)

Because the cross-section is very small, every point of the core is at the distance $R$ from the axis and this value holds across the whole section.

Step 2The field at the centre

The $N$ turns themselves give no field at $O$ . But the wire, going from turn to turn, also advances once round the whole ring, so the winding acts as the turns plus one loop of radius $R$ carrying $I$ . That loop’s field at its centre is the standard one, by symmetry along the axis:

B_O = \frac{\mu_0 I}{2R}

(2)

Step 3The ratio

Dividing (1) by (2):

\eta = \frac{\mu_0 N I/(2\pi R)}{\mu_0 I/(2R)}

Numerically, $N = 2.5\times 10^{3}$ , so

\eta = \frac{2.5\times 10^{3}}{\pi}

\eta = \frac{N}{\pi} = 8.0\times 10^{2}

(3)

Both $I$ and $R$ cancel.

Check

The $N$ turns spread over the length $2\pi R$ give $n = N/(2\pi R)$ turns per unit length, and (1) is then $\mu_0 n I$ , the field of a long straight solenoid, as a thin toroid should have. Both fields are $\mu_0 I$ over a length, so $\eta$ is a pure number.

Remark

The winding of a toroid is never a perfect toroidal current sheet: its pitch carries the current once round the ring, so the coil has a magnetic moment $\pi R^2 I$ along its axis and a dipole field outside, near the ring weaker than the core field by about the factor $1/\eta = \pi/N$ . Returning the wire once round the ring in the opposite sense, along the core, cancels that turn and leaves only the turns, whose field stays inside.