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  • Give answer! 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° is W. Now the torque required to keep the magnet in this new position is

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

  • Option 1)

    \frac{W} {{\sqrt \text{3} }}

  • Option 2)

    \sqrt \text{3} W

  • Option 3)

    \frac{{\sqrt \text{3} W}} {2}

  • Option 4)

    \frac{{2W}} {{\sqrt \text{3} }}

 

Answers (1)

best_answer

As we learnt in

Potential Energy -

U= -MBcos Theta =-vec{M}cdot vec{B}

- wherein

Theta - angle made by the dipole with the field

 

 and

 

Torque -

vec{ au }= vec{M} imes vec{B}

au = MBsin Theta

- wherein

Torque

 

 At Equilibrium

Ui  = -MBH

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

 


Option 1)

\frac{W} {{\sqrt \text{3} }}

Incorrect option

Option 2)

\sqrt \text{3} W

Correct option

Option 3)

\frac{{\sqrt \text{3} W}} {2}

Incorrect option

Option 4)

\frac{{2W}} {{\sqrt \text{3} }}

Incorrect option

Posted by

Aadil

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