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At a certain location in the northern hemisphere, the earth.s magnetic field has a magnitude of 42 muTand points downwards at 53^@ to the vertical. Calculate the flux through a horizontal surface of area 2.5 m^2. [sin 53^@ = 0.8] |
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Answer» Solution :`phi_B = BA COS theta` `= 42 xx 10^(-6) xx 2.5 xx cos 53^@ = 63 mu Wb` |
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