InterviewSolution
Saved Bookmarks
| 1. |
A magnet having magnetic moment of `2.0 xx 10^(4) (J)/(T)` is free to rotate in a horizontal plane where magnetic field is `6 xx 10^(-5)` T . Find the work done to rotate the magnet slowly from a direction parallel to field to a direction `60^(@)` from the field . |
|
Answer» `W = MB [ cos theta_(1) - cos theta_(2)]` `= 2.0 xx 10^(4) xx 6 xx 10^(-5) [ cos 0^(@) - cos 60^(@)]` `= 2 xx 10^(4) xx 6 xx 10^(-5) [ 1 - (1)/(2)]` `= (12)/(2) xx 10^(-1)` `= 0.6 J` |
|