Question

A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 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

Answer:

(a) Energy density will be equal to $57.31J/m^3$

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area $A=15.9cm^2=15.9\times 10^{-4}m^2$

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H$

(b) Total energy stored in magnetic field $U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J$

(a) Energy density will be equal to

$U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3$