Multiple Choice
A ring of radius 0.5m lies in the xy-plane. If a magnetic field of magnitude 2 T points at an angle of 22° above the x-axis, what is the magnetic flux through the ring?
30. Induction and Inductance
Magnetic Flux
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A square conducting loop with sides equal to lies such that the plane of the loop is perpendicular to a magnetic field. What is the flux through the loop?
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Verified step by step guidance1
First, understand that magnetic flux (Φ) through a surface is given by the formula: Φ = B * A * cos(θ), where B is the magnetic field, A is the area of the surface, and θ is the angle between the magnetic field and the normal to the surface.
Since the plane of the loop is perpendicular to the magnetic field, the angle θ is 0 degrees. Therefore, cos(θ) = cos(0) = 1.
Calculate the area (A) of the square loop. The side of the square is given as 36 cm, which needs to be converted to meters for consistency in SI units. So, 36 cm = 0.36 m. The area A = side^2 = (0.36 m)^2.
Substitute the values into the magnetic flux formula: Φ = B * A * cos(θ). Here, B = 50 μT = 50 x 10^-6 T, A = (0.36 m)^2, and cos(θ) = 1.
Perform the multiplication to find the magnetic flux Φ. This will give you the flux through the loop in Weber (Wb).
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