The self-inductance of a solenoid is L, which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $r$ be the radius of the solenoid and $N$ be the number of turns.The length of the wire $l_w$ is given by $l_w = N(2\pi r)$, so $N = \frac{l_w

Vedclass · Accepted Answer

$\frac{\mu_0 l_w^2}{4\pi L}$

Vedclass · Answer

(D) Let $r$ be the radius of the solenoid and $N$ be the number of turns.The length of the wire $l_w$ is given by $l_w = N(2\pi r)$, so $N = \frac{l_w}{2\pi r}$.The self-inductance $L$ of a solenoid of length $\ell$ is $L = \frac{\mu_0 N^2 A}{\ell} = \frac{\mu_0 N^2 \pi r^2}{\ell}$.Substitute $N = \frac{l_w}{2\pi r}$ into the formula for $L$:$L = \frac{\mu_0 (l_w / 2\pi r)^2 \pi r^2}{\ell} = \frac{\mu_0 l_w^2 \pi r^2}{4\pi^2 r^2 \ell} = \frac{\mu_0 l_w^2}{4\pi \ell}$.Solving for the length of the solenoid $\ell$:$\ell = \frac{\mu_0 l_w^2}{4\pi L}$.

The self-inductance of a solenoid is $L$, which is made by a wire of length $l_w$. What is the length of the solenoid?

Similar Questions

Why is self-induced $emf$ known as back $emf$?

Calculate the self-inductance of a very long solenoid.

$A$ long solenoid has $500$ turns. When a current of $2 \,A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \,Wb$. The self-inductance of the solenoid is: (in $\,H$)

The self-inductance of a solenoid is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module