A magnetic needle lying parallel to a magnetic field required w unit of work to turn it through 60° . the torque needed to maintain the needle in the position will be [AIEEE 2003]
Answers
Answer:
The torque needed to maintain the needle in the position will be W√3
Explanation:
W = MB (1-cosθ )
=> W = MB(1-cos60° ) [∵ θ= 60 ° ]
- or, W = MB/2 or MB = 2W ____(1)
- Torque τ = MBsin60°
- = MB √3 /2 = 2W √3/2 From (1)
- = W/√3 Answer
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Answer:
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