Question

A solenoid that is 93.9 cm long has a cross-sectional area of 17.3 cm2. There are 1270 turns of wire carrying a current of 7.80 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).

Answer 1

`65.6\ \text{J/m}^3`

`0.11\ \text{J}`

Step-by-step explanation:

B = Magnetic field =
`\mu_0 (N)/(l)I`

`\mu_0` = Vacuum permeability =
`4\pi10^(-7)\ \text{H/m}`

N = Number of turns = 1270

`l` = Length of solenoid = 93.9 cm = 0.939 m

`I` = Current = 7.8 A

A = Area of solenoid =
`17.3\ \text{cm}^2`

Energy density of a solenoid is given by

`u_m=(B^2)/(2\mu_0)\\\Rightarrow u_m=((\mu_0 (N)/(l)I)^2)/(2\mu_0)\\\Rightarrow u_m=(\mu_0N^2I^2)/(2l^2)\\\Rightarrow u_m=(4\pi* 10^(-7)* 1230^2* 7.8^2)/(2* 0.939^2)\\\Rightarrow u_m=65.6\ \text{J/m}^3`

The energy density of the magnetic field inside the solenoid is
`65.6\ \text{J/m}^3`

Energy is given by

`U_m=u_mAl\\\Rightarrow U_m=65.6* 17.3* 10^(-4)* 0.939\\\Rightarrow U_m=0.11\ \text{J}`

The total energy in joules stored in the magnetic field is
`0.11\ \text{J}`.