Question

(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.

Answer 1

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

$L=\dfrac{\mu_o N^2A}{l}$

A is area

$L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH$

(b) The energy stored in the inductor is given by :

$E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J$

Hence, this is the required solution.