Question

A solenoid that is 112 cm long has a cross-sectional area of 14.9 cm2. There are 1310 turns of wire carrying a current of 7.32 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

We know, magnetic field B for a solenoid having length L ,area of cross section A and no of turns N is given by

B = u_{​​​​​​o}​​​​​NI/L ....eqn 1

Where I is the current through the solenoid.

Given, N = 1310

I = 7.32 A

L = 112 cm = 1.12 m

Also, u_{​​​​​​o}​​​ = 4π*10^-7 T A^{​​​​​​-1}​ m i.e. the absolute permittivity of free space.

Using these values in eqn 1

B = 4π*10^-7* 1310*7.32/1.12

= 120440.352*10^-7 /1.12

= 107536.029*10^-7

B = 0.01075 T.

Energy density U corresponding to this magnetic field is given by

U = B^{​​​​​​2}/2u_o

=(0.01075)²/2*4π*10^-7

= 0.000115/8*3.14*10^-7

⁼ 0.000004578*10⁷

U = 45.78 J/m^{​​​​​​3}​​​​​m3

Now, in lrder to calculate the magnitude of total energy stored in this solenoid in form of magnetic field can wb calculated by multiplying the energy density U with volume V

Volume, = Area of cross section* length

Now, Area A = 14.9 cm^{​​​​​2}​​​​ = 14.9*10^-4 m^{​​​​​​2}

So Volume V = A*L

= (14.9*10^-4)*1.12

= 16.688*10^-4 m^{​​​​​​3}​​​

Hence , Energy E = U*V

= 45.78*16.688*10^-4

= 763.976*10^-4

= 76.39*10^-3 J

Energy E = 76.39 mJ