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
Answers
Given info : a magnetic needle lying parallel to a magnetic field requires w units of work to turn it through 60°.
To find : The torque required to maintain the needle in this position will be ...
solution : work done on magnetic field is given by, W = MB(cosθ₁ - cosθ₂)
here, θ₁ = 0° and θ₂ = 60°
so, W = MB(1 - 1/2) = MB/2
⇒2W = MB ...(1)
now torque on magnetic field is given by, , τ = M × B
= MBsinθ
= 2W sin60°
= 2W × √3/2
= √3W
Therefore the torque required to maintain the needle in this position is √3W
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