Solve and don't give any links Q 8 Starting from the expression for the energy W = 12 - Physics - Electromagnetic Induction

Question

Solve and don't give any links
Q 8 Starting from the expression for the energy W = 1 2   L I
2 , stored in a solenoid of self-inductance L to build up the
current I, obtain the expression for the magnetic energy in terms
of the magnetic field B, area A and length l of the solenoid having
n number of turns per unit length
Hence, show that the energy density is given by B 2 2 μ 0

Shobhit Mittal · Accepted Answer

Dear Student, $begin{displaymath} W = frac{1}{2},L,I^2 end{displaymath}$ This energy is actually stored in the magnetic field generated by the current flowing through the inductor In a pure inductor, the energy is stored without loss, and is returned to the rest of the circuit when the current through the inductor is ramped down, and its associated magnetic field collapses Consider a simple solenoid $begin{displaymath} W = frac{1}{2} ,L,I^2 =frac{mu_0,N^2,A}{2,l} ,left(frac{B,l}{mu_0,N} right)^2, end{displaymath}$ which reduces to $begin{displaymath} W = frac{B^2}{2,mu_0} ,,l,A end{displaymath}$ This represents the energy stored in the magnetic field of the solenoid However, the volume of the field-filled core of the solenoid is

, so the magnetic energy density (i e , the energy per unit volume) inside the solenoid is

, or $begin{displaymath} w = frac{B^2}{2,mu_0} end{displaymath}$