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A magnetic compass needle of magnetic moment 60 A-m^2 is placed at a place. The needle points towards the geographic north. Using the data given below, find the value of declination at that place. Horizontal component of earth's magnetic field = 40 xx 10^(-6) Wb m^(-2) and torque experienced by the needle = 1.2 xx 10^(-3) Nm. |
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Answer» Solution :Let angle of declination at the place be `D^@` Then toryue acting on magnetic compass NEEDLE when placed along GEOGRAPHIC north-south direction `tau = mB_H sin D`. Here `m = 60 A-m^2 , B_H = 40 xx 10^(-6) Wb m^(-2) and tau = 1.2 xx 10^(-3) Nm` `THEREFORE sin D =(tau)/(mB_H)=(1.2 xx 10^(-3))/(60xx40xx10^(-6))=0.5 impliestheta =30^@`. |
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