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A 30.0-cm-diameter circular loop is rotated in a uniform electric field until the...

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pepsi99 | eNotes Newbie

Posted September 9, 2013 at 3:58 AM via web

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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?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 9, 2013 at 4:26 AM (Answer #1)

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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`

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