A loop of wire has the shape shown in the drawing. The top part of the wire is b
ID: 2189584 • Letter: A
Question
A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius r = 0.22 m. The normal to the plane of the loop is parallel to a constant magnetic field (phi = 0?) of magnitude 0.89 T. What is the change ?? in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution
A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius r = 0.22 m. The normal to the plane of the loop is parallel to a constant magnetic field (phi = 0?) of magnitude 0.89 T. What is the change ?? in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolutionExplanation / Answer
The magnetic flux through a closed path is equal to the magnetic field strength multiplied by the area through which the flux is passing. In this case the area of the "enclosed path" would be the area of a semi-circle, (pi*r^2)/2. Multiply this by the magnetic field strength and you get the magnetic flux, phi=B*A=B*(pi*r^2)/2. When the loop has been turned through 1/2 rotation the flux through the semi-circle will be reversed and so the change in flux, Dphi, will be equal to 2x the original flux. Dphi=2*phi=2*B*(pi*r^2)/2=B*pi*r^2
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