Question

M A 40.0- 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 5.20 × 10⁵N . m²/C. What is the magnitude of the electric field?

Answer 1

By electric flux, the electric field is 41.36 x 10⁵ N/C

We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as

Φ = E . S = E . Scosθ

where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.

From the question above, we know that

Φ = 5.20 × 10⁵ N.m²/C

d = 0.4 m²

r = 0.2 m²

because of maximum flux, hence (cosθ = 1)

By substituting the following parameter, we get

Φ = E . Scosθ

5.20 × 10⁵ =E  . π.r²

5.20 × 10⁵ =E . π.0.2²

E = 41.36 x 10⁵ N/C

Hence, the electric field is 41.36 x 10⁵ N/C

For more on electric flux at: brainly.com/question/26289097

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