InterviewSolution
Saved Bookmarks
| 1. |
Demonstarate that the magnetic energy of interaction of two current-carrying loops located in vacumm can be represented as `W_(ia) = (1/mu_(0)) int B_(1) B_(2) dV`, where `B_(1)` and `B_(2)` are the magnetic inductions within a volume element `dV`, produced indiviually by the currents of the first and the secound loop respectively. |
|
Answer» The interaction energy is `(1)/(2 mu_(0)) int |vec(B_(1)) + vec(B_(1))|^(2) dV - (1)/(2 mu_(0)) int |vec(B_(1))|^(2) dV - (1)/(2 mu_(0)) int |vec(B_(2))|^(2) dV` `= (1)/(mu_(0)) int vec(B_(1)). vec(B_(2)) dV` Here, if `vec(B_(1))` is the magnetic field produced by the first of the current carrying loops, and `vec(B_(2))` that of the secound one, then the magnetic field due to both the loops will `vec(B_(1)) + vec(B_(2))`. |
|