A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60^ois W. Now the torque required to keep the magn…

Question

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60^ois W. Now the torque required to keep the magnet in this new position is

Accepted Answer

Correct answer: B. (b) At equilibrium, potential energy of dipole U_i=-MB_H Final potential energy of dipole, U_i=-MB_H cos60^o=- MB_H2 W=U_f-U_j=- MB_H2-(-MB_H)= MB_H2 ⋯ (i) =2W 32 [Using eqn. (i)] = 3W

$A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 6 0^{o} is W . Now the torque required to keep the magnet in this new position is$

Solution

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