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

Step-by-step 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=(\mu_oN^2A)/(l)`

`L=(4\pi* 10^(-7)* (645)^2* 4.75* 10^(-5))/(0.28)`

L = 0.0000886 H

`L=8.86* 10^(-5)\ H`

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

`U=(1)/(2)LI^2`

`U=(1)/(2)* 8.86* 10^(-5)\ H* (85\ A)^2`

U = 0.3200675 Joules

or

U = 0.321 Joules

Hence, this is the required solution.