Irodov 3.324: Stretch the Coil, Raise the Current – Amit & Deepali Sinha Foundation

Problem Irodov 3.324

A long solenoid whose winding is closed on itself is made of wire in the superconducting state, and a current $I_0 = 1.9$ A flows in it. The solenoid is then stretched so that its length increases by $\eta = 5\%$ . Find the current in it afterwards.

The superconducting solenoid of length l with the current I_0 (sense chosen). — The superconducting solenoid of length $l$ with the current $I_0$ (sense chosen).

The same winding stretched to the length (1+η) l. — The same winding stretched to the length $(1+\eta)\,l$ .

Given

$I_0 = 1.9$ A
length increased by $\eta = 5\% = 0.05$
superconducting, closed winding
cross-section and number of turns unchanged (assumed)

Find

the current

I

after the stretch

Solution

Step 1The flux linkage cannot change

The winding has no resistance, so the emf induced in it must vanish, and with it the rate of change of its flux linkage $LI$ :

\begin{align*} -\frac{d(LI)}{dt} &= RI = 0 \\ LI &= L_0 I_0 \tag{1}\end{align*}

Stretching changes the inductance, so the current must change to keep the product.

Step 2The inductance before and after

Let the solenoid have $N$ turns and cross-section $S$ , both unchanged by the stretch (assumed). A long solenoid has $L = \mu_0 N^2 S/l$ , inversely proportional to its length, so stretching to $(1+\eta)l$ gives

L = \frac{\mu_0 N^2 S}{(1+\eta)\,l} = \frac{L_0}{1+\eta} \tag{2}

Step 3The current

Substituting (2) into (1):

\frac{L_0}{1+\eta}\,I = L_0 I_0

Solving for $I$ , with the numbers:

I = (1.9\text{ A})(1 + 0.05)

I = I_0\,(1+\eta) \approx 2.0\text{ A}

(3)

The current keeps its sense and grows by $5\%$ .

Check

Energy. The magnetic energy of the coil, $(LI)^2/2L$ with $LI$ fixed, grows as $1/L$ , by the factor $1+\eta$ . That is right in sign: adjacent turns carry parallel currents and attract, so stretching the coil takes positive work, and with no resistance that work can only go into the field. The current also behaves sensibly in the limit $\eta \to 0$ , where (3) returns $I_0$ .

Remark

The field inside does not change at all. With $B = \mu_0 N I/l$ , the factor $1+\eta$ in the current cancels the factor $1+\eta$ in the length, and $B$ stays at its old value; this is no accident, since the flux through each turn, $BS$ , is what the superconductor holds fixed. Only the volume of the field grows, and with it the energy.