- Download PDF
A 30.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.58 105 N · m2/C. What is the magnitude of the electric field?
1 Answer | Add Yours
By DEFINITION the flux of a constant vector Athat passes through a surface S= S*n (n is the normal unit vector to the surface) is equal to the scalar product betweenA andS.
`Phi = A * S = AS*cos(theta)`
where `theta` is the angle between vectors A and n.
From the definition one can see that the maximum flux of a vector Ais obtained for angles `theta =0 degree` which means A is perpendicular to the surface S.
With the data in text one has
`S =pi*D^2/4 = pi*0.3^2/4 =0.0707 m^2`
The maximum flux of the electric field E is
`Phi = ES`
Which gives the electric field value
`E = Phi/S =(5.58*10^5)/0.0707 = 7.894*10^6 N/C`
We’ve answered 319,676 questions. We can answer yours, too.Ask a question