A magnetic dipole of moment 2.5 Am ^2 is free to rotate about a vertical axis passing through its centre. It is released from East-West direction. Its kinetic energy at the moment, it takes North-South position is (B_H=3 10⁻⁵ ~T )

Step-by-step solution

When magnetic dipole is released from E - W a torque acts on it. So, in the displacement from E-W to N-S work is done by the torque. aligned & KE = work done & ~W = _ ₁ ^ ₂ . d & = MB ( ₁- ₂ ) & KE = MB 0^ & =2.5 3 10⁻⁵ & =7.5 10⁻⁵ ~J =75 10⁻⁶ J aligned