Saved Bookmarks
| 1. |
A vibration magnetometer consists of two identical bar magnets, placed one over the other, such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 4 second. If one of the magnets is taken away, find the period of oscillation of the other in the same field. |
|
Answer» Solution :For a vibration magnetometer, we KNOWN that `T = 2pi SQRT((I)/(MB))` LET M bet the MAGNETIC moment and I be the moment of INERTIA of each magnet then , `M^(1) = sqrt(M^(2) + M^(2)) = sqrt(2)M and I^(1) = I + I =2I` `:. T^(1) = 2pi sqrt((2I)/(sqrt(2)MB)) = 2pi xx sqrt((sqrt(2)I)/(MB))`.......... (1) when one of the magnet is taken away then , `M^(1) = M, I^(11) = I` ` :. T^(11) = 2pi sqrt((I)/(MB))`............ (2) `(eq(2))/(eq(1)) implies (T^(11))/(T^(1)) = (1)/((2)^(1//4)) of ` `T^(11) = (4)/((2)^(1//4)) = 3.36 ` sec. |
|