The Earth's magnetic field passes through a square tabletop with a magnitude of 4.95 × 10^-⁵ T and is directed at an angle of 165° relative to the normal of the tabletop.If the tabletop has 1.50-m sides,what is the magnitude of the magnetic flux through it? A)1.08 × 10^-⁴ Wb B)7.11 × 10^-⁵ Wb C)2.88 × 10^-⁵ Wb D)1.92 × 10^-⁵ Wb E)3.30 × 10^-⁶ Wb

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Quizplus · Accepted Answer

The magnetic flux through the tabletop is given by the product of the magnetic field strength, the area of the tabletop, and the cosine of the angle between the magnetic field and the normal to the tabletop. In this case, the magnetic field strength is 4.95 × 10^-5 T, the area of the tabletop is 1.50 m^2, and the angle between the magnetic field and the normal to the tabletop is 165°. Therefore, the magnetic flux through the tabletop is:$$\Phi_B = B A \cos	heta = (4.95 	imes 10^{-5} 	ext{ T})(1.50 	ext{ m}^2)(\cos 165°) = 1.08 	imes 10^{-4} 	ext{ Wb}$$

The Earth's Magnetic Field Passes Through a Square Tabletop with a Magnitude