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-Sout…

Question

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)

Accepted Answer

Correct answer: D. (d) 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. KE= work done \&W= 0.05em . dθ = MB( cos ⁡ _1- cos ⁡ _2) KE= MBcos ⁡0^ =2.5 .3 10^-5 =7.5 10^-5 J=75 μJ

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

Solution

More Magnetism & Matter PYQs