A freely suspended magnetic needle having moment of Inetia I and magnetic moment M in a magnetic field B is slightly deflected. Find the least time to align the magnetic needle in the direction of magnetic field.

When a bar magnet is suspended freely (attached at it center) in a uniform magnetic field B, the bar magnet aligns itself along the magnetic field, with the magnetic moment M in the direction of B.  The north pole is towards field B.  This is the stable equilibrium position for the bar magnet.  The bar magnet is oriented in a horizontal direction and free to rotate in horizontal direction.

 

   When the bar  magnet is deflected by an angle Ф, there is a restoring torque T developed  due to the magnetic force couple on the poles of the magnet.

     The torque vector = T = M X B

     Restoring Torque magnitude = M B Sin Ф

   T = I α = I * d² Ф/ dt² = - M B Sin Ф = - M B Ф  for small Ф

       ω² = - MB/I ,  I = moment of inertia of bar magnetic about its perpendicular bisector and perpendicular to the plane of rotation.

 

         Time period = 2 π √ (I/MB)

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     A second bar magnet is kept perpendicular to the first one and both are fixed to each other, there is no relative movement between them.  Both magnets become one system and rotate together as unit.

 

   First magnet is in stable equilibrium position and the second magnet is perpendicular to the external magnetic field B.  Hence the torque on it is maximum and hence it turns.  So the first bar magnet also turns.  They will both stabilize by making 45 degrees angle with the direction of the magnetic field B -  Like a scissors, their centers being the pivot.  In this stable equilibrium position, the restoration torque due to first magnet T = M B Sin 45⁰  vertically upwards cancels the opposite and equal torque on the second magnet.

 

  Now, if the system is twisted by a small angle Ф, there are two torques in opposite directions.   The angle between Moment of one bar magnet and B will be 45+Ф and the angle between the magnetic moment of the other bar magnet and B will be 45-Ф.

 

      T1 = - M B Cos(45-Ф)  this is higher and tries restore to the stable position.

     T2 =  M B Cos (45 + Ф)    this torque tries to move away from the stable position.

 

   Net restoration torque = T = T1 - T2 = - M B [ cos (45 - Ф) - Cos (45 +Ф) ]

             T = I d²Ф/dt² = -  √2 M B Sin Ф  = - √2 M B Ф, for small Ф.

 

           Hence ω²  =  √2 M B / I

 

       New Time period of oscillation =  2 π √[ I /√2 M B ]  = (1/2)^1/4 * 2π√(I/MB)

 

  The time period of oscillation of two magnets is slower, and is about 84.089 % of the time period of oscillation of the single bar magnet.