A five-sided object, whose dimensions are shown in the drawing, is placed in a u

Question

A five-sided object, whose dimensions are shown in the drawing, is placed in a uniform magnetic field. The magnetic field has a magnitude of 0.31 T and points along the positive y direction. Determine the magnetic flux through each of the five sides.

(a) The flux through the front triangular surface is Number Units VV/mA?HTWbrad/sdeg (b) The flux through the back triangular surface is Number Units VV/mA?HTWbrad/sdeg (c) The flux through the bottom surface is Number Units VV/mA?HTWbrad/sdeg (d) The flux through the 1.2 m by 0.50 m surface is Number Units VV/mA?HTWbrad/sdeg (e) The flux through the 1.2 m by 0.30 m surface is Number Units VV/mA?HTWbrad/sdeg

Explanation / Answer

The general expression for the magnetic flux through an area A is given by the equation = BAcos , where B is the magnitude of the magnetic field and is the angle of inclination of the magnetic field with respect to the NORMAL to the surface.

a) to c) Since the magnetic field B is parallel to the surface for the triangular ends and the bottom surface, the flux through each of these three surfaces is 0 Wb.

d) For the 1.2 m by 0.50 m side, the area makes an angle with the magnetic field B, where

   = 90° - comp = 53°

  Therefore, = BAcos53 = 0.31*1.2*0.5*cos53 = 0.1119 Wb

e) The flux through the 1.2 m by 0.30 m face is

   = BAcos0 = 0.31*1.2*0.3 = 0.1116 Wb

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