How many turns are needed in a solenoid of radius 10 cm and length 20 cm for its self-inductance to be 6.0 H?
A)30
B)74
C)500
D)550
E)5500

Question

Quizplus · Accepted Answer

The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 N^2 A / l$, where $N$ is the number of turns, $A$ is the cross-sectional area, $l$ is the length of the solenoid, and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$). Rearranging the formula to solve for $N$, we get $N = \sqrt{Ll / (\mu_0 A)}$. Substituting $L = 6.0 \, \text{H}$, $l = 0.20 \, \text{m}$, $A = \pi \times (0.10)^2 \, \text{m}^2$, and $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$, we find $N$ to be approximately 5500 turns.

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