Saved Bookmarks
| 1. |
where will the new null points be located if the bar magnet in the previous example is rotated through180^(@) ? |
|
Answer» Solution :The new null point will be located on the perpendicular Magnetic field at a distance`r_(ax)`on the axs , `B_(H)=(mu_(0))/(4pi).(2p_(m))/(r_(ax)^(3))` Magnetic field at a distance`r_(EQ)`on the perpendicular bisector of the axis . `B_(H)=(mu_(0))/(4pi).(p_(m))/(r_(eq)^(3))` `therefore(mu_(0))/(4pi).(2p_(m))/(r_(ax)^(3))=(mu_(0))/(4pi).(p_(m))/(r_(eq)^(3))` `[therefore` at null point both the fields are EQUAL ] `therefore "" r_(eq)=(r_(ax))/(root(3)(2))=(14)/(root(3)(2))=11.1 "cm"` |
|