A long air-cored solenoid has N₁ = 2000 turns on a length of 0.50 m, and its cross-section has an…

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A long air-cored solenoid has $N_1 = 2000$ turns on a length of $0.50\ \text{m}$ , and its cross-section has an area of $4.0\ \text{cm}^2$ . A second coil of $N_2 = 100$ turns and radius $2.5\ \text{cm}$ is wound tightly round the middle of the solenoid, so that it surrounds the solenoid completely. Take $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$ , and treat the field of the solenoid as zero outside it. The current in the solenoid is raised steadily from zero to $5.0\ \text{A}$ in $0.025\ \text{s}$ . Mutual inductance is the flux linked with one coil for each ampere of current in the other coil.

What is the self-inductance of the solenoid?
What is the mutual inductance of the solenoid and the outer coil?
What emf is induced in the outer coil while the current rises?
What energy is stored in the solenoid when the current is $5.0\ \text{A}$ ?
The solenoid is now filled completely with a core of constant relative permeability $200$ . What emf is induced in the outer coil during the same rise of current?

Think about which area of the outer coil is crossed by the flux of the solenoid.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $N_1 = 2000$ , $N_2 = 100$ , $l = 0.50\ \text{m}$ , $A = 4.0\ \text{cm}^2 = 4.0 \times 10^{-4}\ \text{m}^2$ .

(a) Self-inductance. $L_1 = \dfrac{\mu_0 N_1^2 A}{l}$ . With $N_1^2 = 4.0 \times 10^{6}$ , the numerator is $4\pi \times 10^{-7} \times 4.0 \times 10^{6} \times 4.0 \times 10^{-4} = 2.011 \times 10^{-3}$ in SI units, and dividing by $0.50\ \text{m}$ gives $L_1 = 4.02\ \text{mH}$ .

(b) Mutual inductance. The field of the solenoid is $\mu_0 N_1 I / l$ and it crosses each turn of the outer coil over the area $A$ of the solenoid only, since there is no field outside it. So $M = \dfrac{\mu_0 N_1 N_2 A}{l}$ . Compared with $L_1$ , one factor $N_1$ is replaced by $N_2$ , so $M = L_1 \dfrac{N_2}{N_1} = 4.02\ \text{mH} \times \dfrac{100}{2000} = 201.1\ \mu\text{H}$ .

(c) Emf in the outer coil. The current rises at $\dfrac{5.0\ \text{A}}{0.025\ \text{s}} = 200\ \text{A/s}$ . The emf is $M \times 200\ \text{A/s} = 201.1\ \mu\text{H} \times 200\ \text{A/s} = 40.2\ \text{mV}$ .

(d) Energy. $U = \tfrac{1}{2} L_1 I^2 = \tfrac{1}{2} \times 4.02\ \text{mH} \times (5.0\ \text{A})^2 = 50.3\ \text{mJ}$ .

(e) With the core. The field is $\mu_r$ times as large, so $M$ becomes $200 \times 201.1\ \mu\text{H}$ and the emf is $200 \times 40.2\ \text{mV} = 8.04\ \text{V}$ .

Answer: (a) $4.02\ \text{mH}$ . (b) $201.1\ \mu\text{H}$ . (c) $40.2\ \text{mV}$ . (d) $50.3\ \text{mJ}$ . (e) $8.04\ \text{V}$ .

Check. The final field of the solenoid is $B = \dfrac{\mu_0 N_1 I}{l} = 25.13\ \text{mT}$ . One turn of the outer coil links $B A = 10.053\ \mu\text{Wb}$ , so $N_2$ turns link $1005.31\ \mu\text{Wb}$ , and $\dfrac{1005.31\ \mu\text{Wb}}{0.025\ \text{s}} = 40.2\ \text{mV}$ , as in (c). The energy density $\dfrac{B^2}{2\mu_0}$ times the volume $A l$ gives the same $50.3\ \text{mJ}$ as in (d).

Watch out. If you use the area of the outer coil, $19.6\ \text{cm}^2$ , you get an emf of $197\ \text{mV}$ in (c), about $4.9$ times too large. The field is confined to the solenoid, so only its area counts.

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