Question

A solenoid of 2100 turns, area 10 cm2, and length 30 cm carries a current of 4.0 A. (a) Calculate the magnetic energy stored in the solenoid from 1/2 LI 2. J [2 points] 0 attempt(s) made (maximum allowed for credit

Answer 1

E = 0.1472 J

Step-by-step explanation:

Given that,

The number of turns in the solenoid, N = 2100

Area of the solenoid, A = 10 cm² = 0.001 m²

The length of the solenoid, l = 30 cm = 0.3 m

Current in the solenoid, I = 4 A

We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :

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

Where

L is self inductance of the solenoid,

`L=(\mu_oN^2A)/(l)\\\\L=(4\pi * 10^(-7)* 2100^2* 0.001)/(0.3)\\\\L=0.0184\ H`

So,

`E=(1)/(2)* 0.0184* 4^2\\\\E=0.1472\ J`

So, 0.1472 J of energy is stored in the solenoid.