A solenoid of length $4.00 \mathrm{~cm}$ and cross-sectional area $2.00 \mathrm{~cm}^{2}$ has 80 turns. What is its self-inductance?
A) $120 \mu \mathrm{H}$
B) $40.2 \mu \mathrm{H}$
C) $80.0 \mu \mathrm{H}$
D) $160 \mu \mathrm{H}$
E) $0.500 \mu \mathrm{H}$

Question

Quizplus · Accepted Answer

The self-inductance $L$ of a solenoid can be calculated using the formula $L = \mu_0 N^2 A / l$, where $\mu_0 = 4\pi \times 10^{-7} \mathrm{~T\cdot m/A}$ (the permeability of free space), $N$ is the number of turns, $A$ is the cross-sectional area in square meters, and $l$ is the length of the solenoid in meters. Plugging in the given values ($N = 80$, $A = 2.00 \mathrm{~cm}^2 = 2.00 \times 10^{-4} \mathrm{~m}^2$, $l = 4.00 \mathrm{~cm} = 4.00 \times 10^{-2} \mathrm{~m}$), we get $L = 4\pi \times 10^{-7} \times 80^2 \times 2.00 \times 10^{-4} / 4.00 \times 10^{-2} = 40.2 \mu \mathrm{H}$.

A Solenoid of Length 4.00 cm4.00 \mathrm{~cm}4.00 cm And Cross-Sectional Area 2.00 cm22.00 \mathrm{~cm}^{2}2.00 cm2 Has 80 Turns

A Solenoid of Length $4.00 \mathrm{~cm}$ And Cross-Sectional Area $2.00 \mathrm{~cm}^{2}$ Has 80 Turns