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A compass needle of magnetic moment60 A-m^(2) , pointing towards geographical north at a certain place where the horizontal component of earth's magnetic fieldis40 mu Wb//m^(2) experiences a torque of1.2 xx 10^(-3)Nm . Find the distance declination at that place. |
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Answer» Solution :If ` theta ` is the declination of the place , then the TORQUE acting on the needle is ` tau = M B_H sin theta ` ` implies sin theta = (tau)/(MB_H) = (1.2 xx 10^(-3))/(60 xx 40 xx 10^(-6)) =(1)/(2)` `implies :. theta = 30^(@)` |
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