Q&A - Ask Doubts and Get Answers
Q

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)
792 Views

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

U = -MBH

Final Potential energy of dipole

U=-MBCos 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

Exams
Articles
Questions