Question

A long solenoid has dimensions of length = 20cm, N = 1000 turns, diameter = 1cm, and a mu_r = 5 carries a current I = 200mA. What is the inductance of the solenoid? If the current increases to 1A, what happens to the inductance? Are there any issues that should be considered?

Answer 1

Length l = 20cm = 20* 10-2 m

No. of turns N= 1000

Area A = $\pi r^2= \pi (\frac{1 \times 10^{-2}}{2})^2= 0.78 \times 10^{-4} m^2$

$\mu = \mu_r \mu_0= 5 \times 1.26 \times 10^{-6}=6.3 \times 10^{-6}$

$L = \frac{N^2 \mu A}{l}= \frac{1000^2 \times 6.3\times 10^{-6} \times 0.78\times 10^{-4}}{20\times 10^{-2}}=0.2457 \times 10^{-2}$

the inductance is equal to 2.457 mF.

When the current is increased from 200mA to 1A, The magnetic flux $\phi_B$ increases indicating that the inductor is charging , because there is an increasing amount of energy being stored in its magnetic field. When the current is increased, it drops the voltage opposing the direction of electron flow, acting as a power load.

$L= \frac{N \phi_B}{I}$

$\therefore \phi_B= \frac{LI}{N}= \frac{200\times 10^{-3}A \times 2.457\times 10^{-3} F}{1000}=0.491\times 10^{-6}W$, when I=200mA

$\therefore \phi_B= \frac{LI}{N}= \frac{1A \times 2.457\times 10^{-3} F}{1000}=2.457\times 10^{-6}W$ when I= 1A, which is clearly greater.