A bar magnet of magnetic moment 6 \frac{J}{T} is aligned at 60^{\circ} with a uniform external magnetic field of 0\cdot 44\, T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field,(ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

 

 
 
 
 
 

Answers (1)
S safeer

Given,

Magnetic moment

m= 6\frac{J}{T}
Magnetic field      B= 0\cdot 44T
The angle between the Bar magnet and magnet field \theta = 60^{\circ}
a)
  Work done 

W= mB\left ( \cos \theta_1-\cos \theta _{2} \right )
i) normal to the magnetic field, then \theta_{2} = 90^{\circ}
Therefore,

W\doteq mB\left ( \cos 60-\cos 90 \right )
                     = 6\times 0\cdot 44\times \frac{1}{2}
                W= 1\cdot 32J
ii) Opposite to the magnetic field the \Theta _{2}= 180^{\circ}
Therefore,

W= mB\left ( \cos 60-\cos 180 \right ) 
                        = 6\times 0\cdot 44\left ( \frac{1}{2}-\left ( -1 \right ) \right )= 6\times 0\cdot 44\times \frac{3}{2}
                   W= 3\cdot 96 J
b) torque is given as,
                   \tau = m\times B= mB\sin \theta
in this orientation the angle between moment and magnetic field \theta \; is\; 180^{\circ}
\sin 180= 0
Therefore torque will be Zero

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