Question

A 295 turn solenoid has a radius of 5.25 cm and a length of 23.5 cm. (a) Find the inductance of the solenoid. mH (b) Find the energy stored in it when the current in its windings is 0.501 A.

Answer 1

a)
`L=40295.31* 10^(-7) H=4.0295\ mH`

b)
`E=5.0571\ J`

Step-by-step explanation:

Given:

• no. of turns in the solenoid,
  `N=295`

• radius of the solenoid,
  `r=0.0525\ m`

• length of the solenoid,
  `l=0.235\ m`

a)

The inductance of the solenoid is given as:

`L=(\mu.N^2.A)/(l)`

where:

`\mu=` permeability of the free space
`=4\pi* 10^(-7)\ T.m.A^(-1)`

`A=` cross sectional area of coil formed

`L=((4\pi* 10^(-7))*295^2 *(\pi* 0.0525^2))/(0.235)`

`L=40295.31* 10^(-7) H=4.0295\ mH`

b)

current in the coil,
`I=0.501\ A`

The energy stored in the solenoid can be given as:

`E=(1)/(2)* L.I^2`

`E=0.5 * 4.0295* 10^(-3)* 0.501^2`

`E=5.0571\ J`