The length of a wire required to make a solen | vedclass

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

Question

(A) The self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A = \pi r^2$ is the cross-sectional ar

Vedclass · Accepted Answer

$\sqrt{\frac{4 \pi L l}{\mu_0}}$

Vedclass · Answer

(A) The self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A = \pi r^2$ is the cross-sectional area,and $l$ is the length of the solenoid.From this,$N^2 = \frac{L l}{\mu_0 A} = \frac{L l}{\mu_0 \pi r^2}$,so $N = \frac{1}{r} \sqrt{\frac{L l}{\mu_0 \pi}}$.The total length of the wire used is $W = N 	imes (2 \pi r)$.Substituting the value of $N$,we get $W = \left( \frac{1}{r} \sqrt{\frac{L l}{\mu_0 \pi}} ight) 	imes (2 \pi r)$.$W = 2 \pi \sqrt{\frac{L l}{\mu_0 \pi}} = \sqrt{\frac{4 \pi^2 L l}{\mu_0 \pi}} = \sqrt{\frac{4 \pi L l}{\mu_0}}$.

The length of a wire required to make a solenoid of length $l$ and self-induction $L$ is

Similar Questions

The unit of self-inductance of a coil is

$A$ $10\, V$ battery connected to $5\,\Omega$ resistance coil having inductance $10\, H$ through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is $2\, ms$. The average $emf$ induced across the coil is

The current in a coil changes from $4 \, A$ to $0 \, A$ in $0.1 \, s$. If the average induced $e.m.f.$ is $100 \, V$,what is the self-inductance of the coil in $H$?

If in a coil,the rate of change of area is $5 \, m^2/ms$ and the current changes from $2 \, A$ to $1 \, A$ in $2 \times 10^{-3} \, s$. If the magnitude of the magnetic field is $1 \, T$,then the self-inductance of the coil is ... $H$.

An $e.m.f.$ of $12$ $V$ is induced in a coil when the current in it changes at the rate of $48$ $A/min$. The self-inductance of the coil is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module