Irodov 3.338: The Gap Stores Three Quarters – Amit & Deepali Sinha Foundation

Problem Irodov 3.338

A thin ring of magnetic material has a mean diameter $d = 30$ cm and a cross-sectional area $S = 5.0$ cm $^2$ , and carries a winding of $N = 800$ turns. The ring is cut across, leaving a gap of width $b = 2.0$ mm. At the current flowing in the winding, the permeability of the material is $\mu = 1400$ . Neglecting the spreading of the flux at the edges of the gap, find:

the ratio of the magnetic energy in the gap to that in the material;
the inductance of the system, in two ways: from the flux, and from the energy of the field.

The cut ring (permeability μ, cross-section S) seen from above, with its winding of N turns carrying I, the gap b (widened for clarity) and the mean diameter d. The closed lines are B. — The cut ring (permeability $\mu$ , cross-section $S$ ) seen from above, with its winding of $N$ turns carrying $I$ , the gap $b$ (widened for clarity) and the mean diameter $d$ . The closed lines are $\vect B$ .

Given

$d = 30$ cm
$N = 800$
$S = 5.0$ cm $^2$
$b = 2.0$ mm
$\mu = 1400$
no spreading at the gap edges

Find

(a)

W_{\mathrm{gap}}/W_{\mathrm{m}}

· (b)

L

, from the flux and from the energy

Solution

Step 1The same B in the material and in the gap

With no spreading at the edges, the flux tube keeps the ring’s cross-section $S$ across the gap. The normal component of $\vect B$ is continuous at each face of the cut, and here it is the whole field, so the induction $B$ is the same in the material and in the gap. The thin ring lets us take it uniform over $S$ . The mean circle runs $\pi d - b$ through the material and $b$ across the gap, so the two volumes are

V_{\mathrm{m}} = S(\pi d - b), \qquad V_{\mathrm{gap}} = Sb \tag{1}

Step 2(a) The ratio of the energies

The energy density $BH/2$ is $B^2/2\mu\mu_0$ in the material and $B^2/2\mu_0$ in the gap. Multiplying each by its volume from (1) and dividing, $B^2$ , $\mu_0$ and $S$ cancel:

\frac{W_{\mathrm{gap}}}{W_{\mathrm{m}}} = \frac{(B^2/2\mu_0)\,Sb}{(B^2/2\mu\mu_0)\,S(\pi d - b)}

Numerically, with $\pi d = 0.942$ m,

\frac{\mu b}{\pi d - b} = \frac{(1400)(2.0\times10^{-3})}{0.942 - 0.002}

\frac{W_{\mathrm{gap}}}{W_{\mathrm{m}}} = \frac{\mu b}{\pi d - b} \approx 3.0

(2)

Since $b$ is $0.2$ percent of $\pi d$ , this is $\mu b/\pi d$ to that accuracy.

Step 3(b) The inductance from the flux

The circulation of $\vect H$ along the mean circle links the winding $N$ times. In the material $H = B/\mu\mu_0$ , in the gap $H_0 = B/\mu_0$ :

\frac{B}{\mu\mu_0}(\pi d - b) + \frac{B}{\mu_0}\,b = NI

Solving for $B$ :

B = \frac{\mu_0 NI}{b + (\pi d - b)/\mu} \tag{3}

Each of the $N$ turns is threaded by the flux $BS$ , so the inductance is the linkage $NBS$ per unit current. With (3) the current cancels:

L = \frac{NBS}{I} = \frac{N S}{I}\,\frac{\mu_0 NI}{b + (\pi d - b)/\mu}

Numerically, the numerator and the denominator are

\begin{aligned} \mu_0 N^2 S &= (4\pi\times10^{-7})(800)^2(5.0\times10^{-4}) = 4.02\times10^{-4}\text{ H m}\\ b + \frac{\pi d - b}{\mu} &= 2.0\times10^{-3} + \frac{0.940}{1400} = 2.67\times10^{-3}\text{ m} \end{aligned}

L = \frac{\mu_0 N^2 S}{b + (\pi d - b)/\mu} \approx 0.15\text{ H}

(4)

Dropping $b$ beside $\pi d$ gives the same $0.15$ H, with the denominator $b + \pi d/\mu$ .

Step 4(b) The inductance from the energy

Now add the energies of the two parts, using the volumes (1):

W = \frac{B^2}{2\mu\mu_0}\,S(\pi d - b) + \frac{B^2}{2\mu_0}\,Sb = \frac{B^2 S}{2\mu_0}\left[b + \frac{\pi d - b}{\mu}\right]

The same energy is $LI^2/2$ , so $L = 2W/I^2$ . Substituting $B$ from (3), one power of the bracket cancels:

\begin{aligned} \frac{2W}{I^2} &= \frac{S}{\mu_0 I^2}\,\frac{\mu_0^2N^2I^2}{\left[b + (\pi d - b)/\mu\right]^2}\left[b + \frac{\pi d - b}{\mu}\right]\\ &= \frac{\mu_0 N^2 S}{b + (\pi d - b)/\mu} \end{aligned}

which is the inductance (4) again.

Check

Limits of (4). Closing the gap, $b \to 0$ , gives $\mu\mu_0 N^2 S/\pi d$ , the inductance of a toroid filled with the material; taking $\mu \to \infty$ gives $\mu_0N^2S/b$ , the gap alone. Here the two terms of the denominator are $2.0$ and $0.67$ mm: their ratio is the energy ratio of (a), $3.0$ , as it must be, since the bracket in the energy of the last step shows each part’s energy in proportion to its term.

Remark

The energy and the flux routes agree because, with $\mu$ taken as fixed at this current, the system is linear and its energy is $LI^2/2$ whatever the shape of the field. Cutting a gap only $2$ mm wide lowers the inductance from $\mu\mu_0N^2S/\pi d = 0.60$ H to $0.15$ H, a factor of four, but it makes $L$ much less sensitive to $\mu$ : a ten percent change in $\mu$ now moves $L$ by only about two and a half percent, which is why chokes are built with a gap.