A uniform electric field of magnitude E = 26 N/C points along the x-axis. A circular loop of radius R = 29 cm is centered at the origin with the normal to the loop pointing θ = 35 degrees above the x-axis. a. Calculate the electric flux in unit of Nm^2/C that passes through the loop.
b. To what angle in degrees relative to the x-axis should the normal of the loop be rotated so that the flux through the loop becomes ϕ=0.33Nm^2/C

A) The electric flux that passes through the loop is; ϕ = 6.8692 N.m²/C

B) The angle relative to the x-axis that the normal of the loop should be rotated so that ϕ = 0.33 N.m²/C is; θ = 87.25°

We are given;

Magnitude of electric field; E = 26 N/C

Radius; R = 29 cm = 0.29 m

Angle above the x-axis; θ = 35°

A) Formula for the Electric flux is; ϕ = EAcosθ

Where A is Area = πr²

A = π × 0.29²

A = 0.2642 m²

Thus;

ϕ = 26 × 0.2642 × cos 35

ϕ = 6.8692 N.m²/C

B) We are told that the electric flux is now ϕ = 0.33 N.m²/C

Thus;

0.33 = 26 × 0.2642 × cos θ

cos θ = 0.33/(26 × 0.2642)

cos θ = 0.048

θ = cos^(-1) 0.048

θ = 87.25°

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a) 5.62 Nm^2/C

b) 87.25°

Explanation:

Electric field strength E = 26 N/C

Radius r = 29 cm = 0.29 m

θ = 35°

Area of the loop = = 3.142 x = 0.2642 m^2

a) Electric flux  ϕ = EAcosθ

ϕ = 26 x 0.2642 x cos 35°

ϕ = 26 x 0.2642 x 0.81915 = 5.62 Nm^2/C

b) for the flux through the loop to be 0.33 Nm^2/C

from  ϕ = EAcosθ

0.33 = 26 x 0.2642 x cos θ

0.33 = 6.8692 x cos θ

cos θ = 0.33/6.8692 = 0.048

θ = 0.048

θ = 87.25°

i mean i gues

explanation: