Saved Bookmarks
| 1. |
The work done for rotating a magnet with magnetic dipole moment m, by 90° from its magnetic meridian is n times the work done to rotate it by 60°, find value of n. |
|
Answer» Solution :Suppose the WORK done for RELATING a magnet with magnetic dipole moment m, by 90° from its magnetic meridian is `W_1`, then `W_1 = mB (1- cos theta_(1) )= mB (1- cos 90^(@) )= mB` and the work done to rotate it by 60° is W `therefore W_(2) = mB ((1)/(2)) = (mB)/(2)` Now according to condition GIVEN `W_(1) = nW_2` `mB= n ((mB)/( 2))` `therefore n=2` |
|