Question

Using the magnetic field and self-inductance of a long cylindrical solenoid derived in class, show that the magnetic field energy density inside the solenoid is given by

uB = B^2/2µ0 .

This result, even though you obtained it for a particular case, is general and it represents the energy density of the magnetic field in vacuum. In fact, it is completely analogous to uE = ε0E^2/2, the energy density of the electric field in vacuum.

Answer 1

solution Let Area of cross section of salenoid and it's length = e Let n= number of turns ferunit S length the self inductance = L and L = lon²al. Let I be the current in the solenoid the energy stored a L12 OR E = moshal I 2 so energy density = 2 = b Mo na 12 Al Volume = ě 40² 3272 Uo As B = lo n I. As the medium inside the solenoid in air which has also equal value of u as the Vaccum,