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 o is W . Now, the torque required to keep the magnet in this new position is

Question

Aicharya · Accepted Answer

W = P E cos ⁡ θ 1 - cos ⁡ θ 2 W = P E cos ⁡ 0 - cos ⁡ 60 o = P E 2 . ⇒ P E = 2 W τ = P E sin ⁡ θ = 2 W sin ⁡ 60 o = 3 W .

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium…

Solution