A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60- The torque required to maintain the needle in this position will beA- WB- -3-2 WC- 2 WD- -3 W

The correct option is D √(3)W W=MB (cos θ_1 cos θ_2)=MB(cos 0^∘ cos 60^∘) =MB ( 1 1/2 )=MB/2 and τ =MB sin θ =MB sin 60^∘=MB √(3)/2 ∴τ = ( MB/2 )√(3)⇒τ =√(3 ...

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Byju's Answer

Standard XII

Physics

Solenoid Core in Magnetic Field

A magnetic ne...

Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $60^{\circ}$ . The torque required to maintain the needle in this position will be

\sqrt{3} W

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

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Solution

The correct option is A $\sqrt{3} W$

$W = M B (c o s θ_{1} - c o s θ_{2}) = M B (c o s 0^{\circ} - c o s 60^{\circ}) = M B (1 - \frac{1}{2}) = \frac{M B}{2}$
and $τ = M B s i n θ = M B s i n 60^{\circ} = M B \frac{\sqrt{3}}{2}$
$∴ τ = (\frac{M B}{2}) \sqrt{3} \Rightarrow τ = \sqrt{3} W$

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

The correct option is A √3W W=MB(cosθ1−cosθ2)=MB(cos0∘−cos60∘)=MB(1−12)=MB2 and τ=MBsinθ=MBsin60∘=MB√32 ∴τ=(MB2)√3⇒τ=√3W

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $60^{\circ}$ . The torque required to maintain the needle in this position will be

The correct option is A $\sqrt{3} W$

$W = M B (c o s θ_{1} - c o s θ_{2}) = M B (c o s 0^{\circ} - c o s 60^{\circ}) = M B (1 - \frac{1}{2}) = \frac{M B}{2}$
and $τ = M B s i n θ = M B s i n 60^{\circ} = M B \frac{\sqrt{3}}{2}$
$∴ τ = (\frac{M B}{2}) \sqrt{3} \Rightarrow τ = \sqrt{3} W$