Question

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A bar magnet is hung by a thin cotton thread in uniform horizontal magnetic field and its equilibrium state. the energy a required to rotated by 60 degree is W. . now the torque required to keep the magnet in this new position is??

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Answer 1

At Equilibrium

Ui = -MB

H

Final Potential energy of dipole

Uf =-MBH Cos 60° =\frac{-MB_{H}}{2}

W= U_{f}-U_{i}=\frac{-MB_{H}}{2}-(MB_{H})=\frac{MB_{H}}{2}

Required torque T= MB_{H}\sin 60^{\circ} = 2W \times \frac{\sqrt{3}}{2}

T=\sqrt{3}W

Answer 2

Answer: 3−−√

Explanation:

Wext = Uf – Vi

= – MB cos 60° – (–MB)

= MB(1 – cos 60°) = MB/2 = W

r = MB sin 60° = MB 3−−√2 =3−−√W