Chapter 4: Q15E (page 1014)

(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this solenoid is given by the equation $L = \frac{μ_{0} A N^{2}}{l}$ . Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because B is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically $5 c m$ long and $0.15 c m$ in diameter and has $50$ coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

Expert verified

a) $L = \frac{μ_{0} {AN}^{2}}{l}$

b)role="math" localid="1664194101205" $L = 1.11 \times 10^{- 7} H$

Step by step solution

For a long solenoid of $N$ turns, length $i$ and current $i$ , the magnetic field is given by $B = μ_{0} n i = μ_{0} \frac{N}{I} i$ .

For the area of cross-section $A$ , the magnetic flux is given by role="math" localid="1664194375207" $ϕ = B A = \frac{μ_{0} NAi}{l}$ .

The self-inductance is defined as $L = \frac{N ϕ_{B}}{i}$ .

Clearly, after substituting the flux equation, we will get $L = \frac{μ_{0} N^{2} A}{l}$ .

Given that $I = 5 c m$ and $D = . 015 c m$ $a n d N = 50$ .

role="math" localid="1664194642248" $A = \frac{π D^{2}}{4} A = \frac{π \times 0 . 0015^{2}}{4} A = 1.767 \times 10^{- 6} m^{2}$

The self-inductance is given by

$L = \frac{4 π \times 10^{- 7} \times 50^{2} \times 1.767 \times 10^{- 6}}{0.05} L = 1.11 \times 10^{- 7} H$

So, the self-inductance is $1.11 \times 10^{- 7} H$ .

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