Saved Bookmarks
| 1. |
Show the time period of oscillation when a bar magnet is kept in a uniform magnetic field is T = 2pi sqrt((l)/(p_(m)B)) . In second, where I represents moment of inertia of the bar magnet, p_(m) is the magnetic moment and is the magnetic field. |
|
Answer» Solution :The magnitude of deflecting torque ( the torque which makes the object rotate ) acting on the bar MAGNET which will tend to align the bar magnet parallel to the direction of the uniform magnetic field `vec(B)` is `|vec(tau)| = p_(m) B"sin" theta` The magnitude of restoring torque acting on the bar magnet can be WRITTEN as `|vec(tau)| = I (d^(2) theta)/(dt^(2))` Under equilibrium conditions, both magnitude of deflecting torque and restoring torque will be equal but act in the opposite directions, which means `I (d^(2) theta)/(dt^(2)) = - p_(m)" B sin " theta` the negative sign implies that both are in opposite directions. the above equation can be written as `(d^(2) theta)/(dt^(2)) = - (p_(m) B)/(I) sin theta` this is non-LINEAR second order homogeneous differential equation. In order to make it linear. we use small ANGLE approximation. i.e., sin `theta approx theta` , we get `(d^(2) theta)/(dt^(2)) = - (p_(m) B)/(I) theta` this inear second order homogeneous differential equation is a simple Harmonic differential equation. therefore, comparing with simple Harmonic Motion (SHM) differential equation `(d^(2)X)/(dt^(2)) = -omega^(2) x ` where `omega` is the angular frequency of the osxillation . `omega^(2) = (p_(m)B)/(I) rArr omega = sqrt((p_(m)B)/(I))` T = `2pi sqrt((I)/(p_(m)B))` T = `2pi sqrt((I)/(p_(m)B_(H)))` in second where, `B_(H)` is the horizontal component of Earth.s magnetic field. |
|