Question

Obtain an expression for magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solonoid.
(b) How does this magnetic energy compare with the elecrostatic energy strored in a capacitor ?

Answer 1

(a) The magnetic energy is
`U_(B) = (1)/(2) LI^(2)`
As, `B = (mu_(0) NI)/(l)`, therefore,
`I = (B l)/(mu_(0) N) :. U_(B) = (1)/(2) L ((B l)/(mu_(0) N))^(2)`
Using `L = (mu_(0) N^(2) A)/(l)` , we get,
`U_(B) = (1)/(2) ((mu_(0) N^(2) A)/(l)) ((B l)/(mu_(0) N))^(2)`
`U_(B) = (B^(2))/(2 mu_(0)) A l`
(b) The volume that contains flux, `V = A xx l`
`:.` Magnetic energy per unit volume,
`u_(B) = (U_(B))/(V) = (B^(2))/(2 mu_(0))`
It is known that electrostatic energy stored per unit volume in a parallel plate capacitor is
`u_(E) = (1)/(2) in_(0) E^(2)`
In both the cases. energy is directly proportional to the square of the field strenth.
Note that Eqna. (ii) and (iii) have been derived for special cases -- a solenoid and a parallel plate capacitor respectively. But they are valid for any region of space in which a magnetic field `or//and` an electric field exists.