Question

A solenoid 28.0 cm long with a cross sectional area 0.475 cm^2 contains 645 turns and carries a current 85.0 A. a. Find the inductance of this solenoid. b. Find the total energy contained in the coil's magnetic field (assuming that the field is uniform throughout).

Answer 1

Explanation:

It is given that,

Length of solenoid, l = 28 cm = 0.28 m

Area of cross section, A = 0.475 cm² = 4.75 × 10⁻⁵ m²

Current, I = 85 A

(a) The inductance of the solenoid is given by :

$L=\dfrac{\mu_oN^2A}{l}$

$L=\dfrac{4\pi\times 10^{-7}\times (645)^2\times 4.75\times 10^{-5}}{0.28}$

L = 0.0000886 H

$L=8.86\times 10^{-5}\ H$

(b) Energy contained in the coil's magnetic field is given by :

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 8.86\times 10^{-5}\ H\times (85\ A)^2$

U = 0.3200675 Joules

or

U = 0.321 Joules

Hence, this is the required solution.