InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Total energy of density of electromagnetic waves in vacuum is given by the relationA. `1/2.(E^(2))/(epsi_(0)) + (B^(2))/(2mu_(0))`B. `1/(2)epsi_(0)E^(2) + 1/2mu_(0)B^(2)`C. `(E^(2) + B^(2))/(c)`D. `(1)/(2) epsi_(0)E^(2) + (B^(2))/(2mu_(0))` |
|
Answer» Correct Answer - D The energy in EM waves is divided equally between the electric and magnetic fields. The total energy per unit volume is `u - u_(e) + u_(m)` `1/2 epsi_(0)E^(2) + 1/2 (B^(2))/(mu_(0))` |
|