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  1. Problems
  2. /A solenoid 0.30 m long and 1.2 cm in radius must have an inductance of 25 mH…
MediumPhysicsHigh School (11–12)Electricity & Magnetism › Electromagnetic InductionNumericalpythonBoard ExamAPA-LevelIBAEEE (Amrita)AMUEEE
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A solenoid 0.30 m long and 1.2 cm in radius must have an inductance of 25 mH…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A solenoid 0.300.300.30 m long and 1.21.21.2 cm in radius must have an inductance of 252525 mH. How many turns does it need if it has an air core? How long is the wire? How many turns are needed if an iron rod of relative permeability 400400400 fills it?

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Work it out on paper, then type your number. The answer itself stays hidden.

m
Need a reminder? The key ideaShowHide

The flux linkage of a coil is Nϕ=LIN\phi = LINϕ=LI. The self-induced emf is ε=−LdIdt\varepsilon = -L\dfrac{dI}{dt}ε=−LdtdI​, and it opposes the change of the current. For a long solenoid, L=μrμ0N2AlL = \dfrac{\mu_r\mu_0 N^2 A}{l}L=lμr​μ0​N2A​. The energy stored is U=12LI2U = \tfrac{1}{2}LI^2U=21​LI2, with μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}μ0​=4π×10−7 T m/A.

The worked solution

Have you tried it?

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. L=25×10−3L = 25 \times 10^{-3}L=25×10−3 H, l=0.30l = 0.30l=0.30 m, r=0.012r = 0.012r=0.012 m.

Plan. Find AAA. Then solve L=μrμ0N2AlL = \dfrac{\mu_r\mu_0 N^2 A}{l}L=lμr​μ0​N2A​ for NNN.

Step 1. The area. A=πr2=π×(0.012 m)2=4.52×10−4 m2A = \pi r^2 = \pi \times (0.012\ \text{m})^2 = 4.52 \times 10^{-4}\ \text{m}^2A=πr2=π×(0.012 m)2=4.52×10−4 m2.

Step 2. The air core. N=Llμ0A=25×10−3 H×0.30 m4π×10−7 T m/A×4.52×10−4 m2=3632N = \sqrt{\dfrac{Ll}{\mu_0 A}} = \sqrt{\dfrac{25 \times 10^{-3}\ \text{H} \times 0.30\ \text{m}}{4\pi \times 10^{-7}\ \text{T m/A} \times 4.52 \times 10^{-4}\ \text{m}^2}} = 3632N=μ0​ALl​​=4π×10−7 T m/A×4.52×10−4 m225×10−3 H×0.30 m​​=3632 turns.

Step 3. The wire. One turn is 2πr=0.07542\pi r = 0.07542πr=0.0754 m long. So the wire is 3632×0.0754 m=2743632 \times 0.0754\ \text{m} = 2743632×0.0754 m=274 m.

Step 4. The iron rod. LLL is 400400400 times larger for the same turns. So the turns fall by 400=20\sqrt{400} = 20400​=20: N=363220=182N = \dfrac{3632}{20} = 182N=203632​=182 turns.

Answer: 363236323632 turns and 274274274 m of wire; with iron, 182182182 turns.

Watch out. Iron lowers the number of turns by μr\sqrt{\mu_r}μr​​, not by μr\mu_rμr​. The inductance goes as N2N^2N2.

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