A magnetic needle suspended parallel to a magnetic field requires \sqrt{3}\text{J} of work to turn it through 60°. The torque needed to maintain the needle in this position will be:

  • Option 1)

    3 J

  • Option 2)

    \sqrt{3}\text{J}

  • Option 3)

    \frac{3}{2}\text{J}

  • Option 4)

    2\sqrt{3}\text{J}

 

Answers (1)

W= MB \left ( \cos \Theta _{1} \right-\cos \Theta _{2} )

\Theta _{1}= 0^{\circ}

\Theta _{2}= 60^{\circ}

W = MB \left [ 1-\frac{1}{2} \right ]=\frac{MB}{2} --------------------- (i)

The Torgue on the needle is 

\vec{T}= \vec{m\times \vec{b}}

in magnetic 

T= MB\sin \Theta = MB\sin 60^{\circ} = \frac{\sqrt{3}}{2}MB -----------------(ii)

 

\therefore from equ (i) and ii) 

\frac{T}{w} = \sqrt{3} = T=\sqrt{1w}

\sqrt{3}\times \sqrt{3j} = 3j


Option 1)

3 J

This option is correct.

Option 2)

\sqrt{3}\text{J}

This option is incorrect.

Option 3)

\frac{3}{2}\text{J}

This option is incorrect.

Option 4)

2\sqrt{3}\text{J}

This option is incorrect.

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