A magnetic needle lying parallel to the magnetic field required W units of work to turn it through 60 degree. The torque needed to maintain the needle in this position will be
Asked by hemanttkumarr
31st May 2018,
8:41 PM
Answered by Expert
Answer:
work done to turn the needle through angle θ= m•B = mBcosθ ........(1)
Torque required to turn angle θ = m×B = mBsinθ ...........(2)
from (1) and given workdone W units for 60º rotation, we have mB = W/cos60 = 2W
from (2), required Torque = mBsin60 = 2W×(√3/2) = √3×W
Answered by Expert
11th June 2018,
12:38 PM
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