5.7 (a)

ii) A bar magnet of magnetic moment 1.5 TJT ^ {-1} lies aligned with the direction of a uniform magnetic field of 0.22 T.
 What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?

Answers (1)

The amount of work required for the given condition will be:- 

                                W\ =\ -MB\left [ \cos \Theta _2\ -\ \cos \Theta _1 \right ]                                

                                W\ =\ -MB\left [ \cos 180^{\circ}\ -\ \cos 0^{\circ} \right ]

or                                      =\ 2MB

or                                     =\ 2\times1.5\times0.22

or                                     =\ 0.66\ J

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