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A circularantenna of area 3 m^(2) is installed at a place in Madurai. The plane of the area of antennais inclined at47^(@) with the direction of Earth's Magnetic field.If the magnitude of Earth's field at that place is 40773.9 nT find the magnetic flux linked with the antenna . |
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Answer» SOLUTION :`B=40773.9 nT , theta = 90^(@) - 47^(@) = 43^(@) , A = 3m^(2)` We know that , `Phi_B= BA cos theta ` `B= 40773.9 xx 10^(-9) xx3 xx cos 43^(@) = 40.7739 xx 10^(-6) xx 3 xx 0.7314= 89. 47 xx 10^(-6) Wb` ` Phi_B= 89.47 mu ` Wb |
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