A magnetic needle lying parallel to a magnetic field requires `W units` of work to turn it through `60^(@)`. The torque needed to maintain the needle in this position will beA. `sqrt(3)W`B. `W`C. `sqrt(3)/2W`D. `2W`

A magnetic needle lying parallel to a magnetic field requires `W units

1.	A magnetic needle lying parallel to a magnetic field requires `W units` of work to turn it through `60^(@)`. The torque needed to maintain the needle in this position will beA. `sqrt(3)W`B. `W`C. `sqrt(3)/2W`D. `2W`
Answer» Correct Answer - A `W=MB(costheta_(1)-costheta_(2))=MB(cos 0^(@)-cos 60^(@))` `=MB(1-1/2)=(MB)/2` and `tau=MB sin theta=MB sin 60^(@)=MBsqrt(3)/2` `:. Tau=((MB)/2)sqrt(3)impliestau=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 needed to maintain the needle in this position will beA. `sqrt(3)W`B. `W`C. `sqrt(3)/2W`D. `2W`

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