A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 600. The torque required to maintain the needle in this position will be xW the x is ___.

W=MB(cos⁡θ1−cos⁡θ2)=MB(cos⁡00−cos⁡60∘)=MB(1−12)=MB2and τ=MBsinθ=MBsin600=MB32∴τ=(MB2)3⇒τ=3W

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A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60⁰. The torque required to maintain the needle in this position will be $\sqrt{x} W$ the x is ___.

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answer is 3.

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Detailed Solution

$W = MB (\cos θ_{1} - \cos θ_{2}) = MB (\cos 0^{0} - \cos 60^{\circ})$
$= M B (1 - \frac{1}{2}) = \frac{M B}{2}$
and $τ = M B \sin θ = M B \sin 60^{0} = M B \frac{\sqrt{3}}{2}$
$∴ τ = (\frac{M B}{2}) \sqrt{3} \Rightarrow τ = \sqrt{3} W$

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