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Magnetism and Matter question

2016 · Q125
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Magnetism and Matter question

2016 · Q125

NEETPhysicsMagnetism and MatterMCQ+4 / −1
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 60o is W. Now the torque required to keep the magnet in this new position is
  1. A
    W3{W \over {\sqrt 3 }}3​W​
  2. B
    3W\sqrt 3 W3​W
  3. C
    3W2{{\sqrt 3 W} \over 2}23​W​
  4. D
    2W3{{2W} \over {\sqrt 3 }}3​2W​
View written solutionFree

Correct answer: B

W = MB (cos 0° – cos 60°)

⇒\Rightarrow⇒ W = MB(1−12)=MB2\left( {1 - {1 \over 2}} \right) = {{MB} \over 2}(1−21​)=2MB​ ....(1)

Required torque for this position

τ\tau τ = MB sin θ\theta θ

= MB sin 60°

= 32MB{{\sqrt 3 } \over 2}MB23​​MB = 3W\sqrt 3 W3​W [From (1)]

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