A loop of a wire has the shape shown in the drawing. The top part of the wire is

Question

A loop of a wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius r = 0.20 m. The normal to the plane of the loop is parallel to a constant magnetic field (theta = 0

A loop of a wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius r = 0.20 m. The normal to the plane of the loop is parallel to a constant magnetic field (theta = 0 degree ) of magnitude 0.75 T. What is the change (delta phi) 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?

Explanation / Answer

Area of semi circle

A=(1/2)pir^2 =(1/2)pi*0.2^2 =0.063 m^2

X=phi

initial flux

Xi =BAcos0 =0.063*0.75*1 =0.047 wb

final flux

Xf =BAcos(180) =0.063*0.75*(-1) =-0.047 wb

so change in flux

dX =Xf-Xi =-0.047-(0.047)

dX =-0.094 wb

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