An ideal solenoid with 3000 turns is $70.0 \mathrm {~cm}$ long. If its self-inductance is $25.0 \mathrm { mH }$, what is its radius? $\left( \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } \right)$
A) $0.00199 \mathrm {~m}$
B) $52.0 \mathrm {~m}$
C) $327 \mathrm {~m}$
D) $0.0222 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The self-inductance $L$ of a solenoid is given by $L = \mu_0 \frac{N^2 A}{l}$, 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. Rearranging for $A$ and then solving for the radius $r$ (since $A = \pi r^2$), we can find the radius that corresponds to the given self-inductance. Plugging in the given values, we find that the radius that satisfies these conditions is $0.0222 \mathrm {~m}$.

An Ideal Solenoid with 3000 Turns Is 70.0 cm70.0 \mathrm {~cm}70.0 cm Long If Its Self-Inductance Is

An Ideal Solenoid with 3000 Turns Is $70.0 \mathrm {~cm}$ Long If Its Self-Inductance Is