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Find the torque on M_(1) due to M_(2) |
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Answer» Solution :DUE to `M_(2)`, magnetic fields at `S` and `N` of `M_(1)` are `B_(1)` and `B_(2)` RESPECTIVELY. The forces on `-m` and `+m` are `F_(1)` and `F_(2)` as shown in the figure.The torque (about the centre of the dipole `m_(1)`) will be `F_(1)a+F_(2)a=(F_(1)+F_(2))a` `[(mu_(0)/(4pi))M_(2)/((R-a))m+(mu_(0)/(4pi))M_(2)/((r+a))m]a` `(mu_(0)/(4pi))M_(2)m(1/r^(3)+1/r^(3))a :' alt LT r` `(mu_(0)M_(2)m)/(4pi)2/r^(3)a=(mu_(0)M_(1)M_(2))/(4pi^(3))` 
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