Irodov 3.336: Flux Times Current Gives the Energy – Amit & Deepali Sinha Foundation

Problem Irodov 3.336

A toroidal iron core carries a winding of $N = 500$ turns. When the winding carries a current $I = 2.0$ A, the magnetic flux through the cross-section of the core is $\Phi = 1.0$ mWb. Find the energy of the magnetic field.

The iron toroid seen from above, with its winding of N turns carrying I. The closed lines are B; their flux through the core's cross-section is Φ. — The iron toroid seen from above, with its winding of $N$ turns carrying $I$ . The closed lines are $\vect B$ ; their flux through the core’s cross-section is $\Phi$ .

Given

$N = 500$
$I = 2.0$ A
$\Phi = 1.0$ mWb
$B$ proportional to $H$ in the iron up to this current (assumed)

Find

the magnetic field energy

W

Solution

Step 1The inductance from the flux linkage

In a toroid the field lines stay inside the core, so every one of the $N$ turns is threaded by the whole flux $\Phi$ . The flux linked by the winding is $N\Phi$ , and the inductance is that linkage per unit current. We take the iron’s $B$ proportional to $H$ up to this current (assumed), so that $L$ does not depend on the current:

L = \frac{N\Phi}{I} \tag{1}

Step 2The energy of the field

The energy of the magnetic field of a winding with inductance $L$ carrying the current $I$ is $LI^2/2$ . Substituting (1), one factor of $I$ cancels:

W = \frac{LI^2}{2} = \frac{1}{2}\,\frac{N\Phi}{I}\,I^2

Numerically,

\frac{1}{2}N\Phi I = \frac{1}{2}\,(500)(1.0\times10^{-3})(2.0)

W = \frac{1}{2}N\Phi I \approx 0.50\text{ J}

(2)

This is the whole energy of the field, which lives inside the core.

Check

The same energy from the energy density $BH/2$ , for a core thin enough that $B$ and $H$ are uniform over its cross-section $S$ and along its mean length $l$ . The circulation of $\vect H$ round the mean circle gives $Hl = NI$ , the flux is $\Phi = BS$ , and the core’s volume is $Sl$ , so

W = \frac{BH}{2}\,Sl = \frac{1}{2}\,(BS)(Hl) = \frac{1}{2}\,\Phi NI,

which is (2). Units: $\text{Wb}\cdot\text{A} = \text{V s A} = \text{J}$ . The inductance itself is $L = N\Phi/I = 0.25$ H.

Remark

The energy is half the product of the flux linkage $N\Phi$ and the current, the magnetic twin of a capacitor’s $qV/2$ . Doubling the current in iron does not simply quadruple it: $\Phi$ grows more slowly than $I$ once the iron starts to saturate, so the energy then grows more slowly than $I^2$ , and the linear form holds only on the straight part of the magnetization curve.