If X amount of work is required to rotate a bar magnet in a magnetic field by $60^{0}$ , from a position parallel to the field, What is the torque required to maintain it in new position.

\sqrt{3} 5 X

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\sqrt{3} X

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\sqrt{3} 2 X

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\sqrt{3} 3 X

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Solution

The correct option is B $\sqrt{3} X$
work done in rotating a magnet from $0$ to $60$ will be,
$X =$ $M \times B \times (1 - c o s θ)$
= $M \times B \times (0.5)$
$2 X = M \times B$ taking $θ = 60$
torqur required to maintain in that position will be ,
$τ = M \times B \times S i n θ$
$τ = 2 X \times \frac{\sqrt{(} 3)}{2}$ taking $θ = 60$
= $\sqrt{3} X$

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If X amount of work is required to rotate a bar magnet in a magnetic field by 600, from a position parallel to the field, What is the torque required to maintain it in new position.

The correct option is B √3Xwork done in rotating a magnet from 0 to 60 will be, X=M×B×(1−cosθ) =M×B×(0.5) 2X=M×B taking θ=60 torqur required to maintain in that position will be , τ=M×B×Sinθ τ=2X×√(3)2 taking θ=60 =√3X

If X amount of work is required to rotate a bar magnet in a magnetic field by $60^{0}$ , from a position parallel to the field, What is the torque required to maintain it in new position.