a square coil of 600 turn, each side 20cm, is placed with its plane inclined at 30degreeto a uniform magnetic field of 4.5 * 10^-2 . find the flux through the coil...

Dear Student,

N=600side of square=20 cm=0.2 mArea, A=0.22A=0.04 m2B=4.5×10-2 Tθ=30°ϕm=NBAcosθϕm=600×4.5×10-2×0.04×cos30°ϕm=1.08×32ϕm=0.9342 Wb

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