Question

The figure below shows three edge views of a square loop with sides of length 0.260 m in a magnetic field of magnitude 2.25 T. Calculate the magnetic flux (in wb) through the loop oriented perpendicular to the magnetic field, 600% from the magnetic field, and parallel to the magnetic field. 60.0 HINT (a) perpendicular to the magnetic field Wb (b) 60.0° from the magnetic field c) parallel to the magnetic field Wb

Answer 1

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straightphi Magnetic flux doublerightarrow B A cos theta where B = Magnetic field A = Area theta = angle between a perpendicular vector to the area and magnetic field. A = Area = (0.260)^2 = 0.0676 m^2 B = 2.25 T perpendicular to magnetic field doublerightarrow cos theta = 0 degree straightphi = 2.25 times 0.0676 times cos 0 degree = 2.25 times 0.0676 times 1 = 0.1521 Wb cos 60 degree = 1/2 straightphi = 2.25 times 0.0676 times 1/2 = 0.07605 Wb cos 90 degree = 0 straightphi = 2.25 times 0.0676 times 0 = 0 Wb