A toroid of rectangular cross section with inner and outer radii 'a' and 'b' and height 'h' respectively consists of N no. of turns. find the total magnetic energy stored in the toroid.
Answers
Answer:
I'll mention permeability of free space as u here, current as C and symbol of integration as I to reduce typing error.
In order to maintain symmetry, let's assume that the path of magnetic conduction is circular. Hence, by Ampere's circuital law, we get I(closed loop) B.ds = B(2πr) = uNC. B and ds are vector quantities and will have arrows indicating directions.
Hence, B = uNC/2πr
Now, magnetic energy for a single field loop is given by m = B^2/2u
Total energy can found out by integrating it for the whole volume. Again, for ease of calculations, we assume the symmetry of the toroid as cylinder. So, V= πr^2h which gives dV = 2πrhdr as field is generated only within the radius.
Integrating magnetic energy w.r.t dV from one point x to other point y
I (m.dV) = I (B^2/2u).2πhdr = 2πh I [(uNC/2πr)^2/2u].r.dr
= [(uN^2C^2h)/4π] I (dr/r)
Integrating it from x to y we get the final expression for total energy as [(uN^2C^2h)/4π] ln(y/x)
Request to change the correct notations with the ones available in regular textbooks.