Answer to Question #170470 in Electric Circuits for Innocent

Question

A coil of radius 9cm and 45turns is 14cm long and has a current of 4.0A through it

1) calculate the self inductance of the coil

2) calculate the energy stored in the inductor

Answer 1

Explanations & Calculations

• The inductance & the stored energy of a coil are given by the equations $L= \large\mu\cdot\Large \frac{N^2A}{l}$ & $E=\Large\frac{LI}{2}$ where

$\small$ $\qquad \begin{aligned} \small \mu &= \small\text{permitivity of the core medium}\\ \small N&= \small \text{Number of turns in the coil}\\ \small A &=\small \text{Area of the coil}\\ \small l&= \small \text{Length of the coil} \end{aligned}$

• Since it is an air-cored ($\small \mu_a=1$) coil (as it says), the permittivity should be $\small \mu =\mu_0\mu_a=\mu_0$
• Then,

$\qquad\qquad \begin{aligned} \small L &= \small (4\pi\times10^{-7}Hm^{-1})\cdot\frac{45^2\times(\pi\times (0.09m)^2)}{0.14m}\\ &=\small 4.625\times10^{-4}H\\ &=\small \bold{0.4625\,mH} \end{aligned}$

2.

$\qquad\qquad \begin{aligned} \small E&=\small \frac{4.625\times 10^{-4}\times4A }{2}\\ &=\small 9.25\times10^{-4}J\\ &=\small \bold{0.925\,mJ} \end{aligned}$