Question

A 35.0-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 5.42 105 N · m2/C. What is the magnitude of the electric field? MN/C

Answer 1

To solve this problem we will apply the concept of Electric Flow, which is understood as the product between the Area and the electric field. For the data defined by the area, we will use the geometric measurement of the area in a circle (By the characteristics of the object) This area will be equivalent to,

`\phi = 35 cm`

`r = 17.5 cm = 0.175 m`

`A = \pi r^2 = \pi (0.175)^2 = 0.09621m^2`

Applying the concept of electric flow we have to

`\Phi = EA`

Replacing,

`5.42*10^5N \cdot m^2/C = E (0.09621m^2)`

`E = 5.6335*10^6N/C`

Therefore the magnitude of the electric field is
`5.6335*10^6N/C`