A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60. The torque needed to maintain the needle in this position will be

  • Option 1)

    \sqrt{3}W

  • Option 2)

    W

  • Option 3)

    \left ( \frac{\sqrt{3}}{2} \right )W

  • Option 4)

    2W

 

Answers (1)

As we learnt in

Torque -

\vec{\tau }= \vec{M}\times \vec{B}

\tau = MB\sin \Theta

- wherein

Torque

 

 and

Work done by bar magnet -

W= MB\left ( 1-\cos \Theta \right )

-

 

 

W=-MB(cos\, \theta _{2}-cos\, \theta _{1})

=-MB(cos\, 60^{0}-cos\, 0)=\frac{MB}{2}

\therefore \; MB=2W............(i)

Torque=\, MB\, sin\, 60^{0}=(2W)\, sin\, 60^{0}

=\frac{2W\times \sqrt{3}}{2}=\sqrt{3}W

 


Option 1)

\sqrt{3}W

Correct

Option 2)

W

Incorrect

Option 3)

\left ( \frac{\sqrt{3}}{2} \right )W

Incorrect

Option 4)

2W

Incorrect

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