Question

Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in J) is stored in this inductor when 21.0 A of current flows through it? J (c) How fast (in s) can it be turned off if the induced emf cannot exceed 3.00 V? s

Answer 1

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$