Answer:
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 = [tex]\pi r^{2}[/tex] = 3.142 x [tex]0.29^{2}[/tex] = 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 = [tex]cos^{-1}[/tex] 0.048
θ = 87.25°
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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