A square loop of area 50 cm² is placed in a uniform electric field of intensity 50 N/C such that the angle between the surface vector and the electric field is 60°, then the electric flux associated with the loop will be:

CONCEPT:

Electric flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity.
• The SI unit of the electric flux is N-m2/C.
• If the electric field is E and the area is A then the electric flux associated with the area is given as,
  \(⇒ ϕ = \vec{E}.\vec{A}\)
  ⇒ ϕ = EA cos θ
  Where θ = angle between the surface and the electric field

CALCULATION:

Given A = 50 cm2 = 50 × 10^-4, E = 50 N/C, and θ = 60°

If the electric field is E and the area is A then the electric flux associated with the area is given as,

⇒ ϕ = EA cos θ     ---(1)

Where θ = angle between the surface and the electric field

So by equation 1, the electric flux associated with the square loop will be,

⇒ ϕ = EA × cos θ

⇒ ϕ = 50 × 50 × 10^-4 × cos 60° 

\(⇒ ϕ = 50×50×10^{-4}×\frac{1}{2}\)

⇒ ϕ = 1250 × 10^-4

⇒ ϕ = 0.125 V-m

• Hence, option 3 is correct.