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(B) 3 (C) 2 (D) 1 51. A square frame of side 0.05 m is situated with its positive normal making an angle of 45 deg with a uniform electric field of 10√2 Vm¹. The flux of the electric field through the surface bounded by the square frame is sent lllar (D) 100 V m 2 3 4 (C) 10 V m (B) 25 mV m (A) 250 mV m A?
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(B) 3 (C) 2 (D) 1 51. A square frame of side 0.05 m is situated with i...
Calculation of Electric Flux through the Square Frame

Given data:
- Side of the square frame (a) = 0.05 m
- Angle between the normal to the frame and the electric field (θ) = 45 deg
- Electric field strength (E) = 10√2 V/m

Calculating the Area of the Square Frame:
- Area of a square = side x side
- Area = 0.05 x 0.05 = 0.0025 m²

Calculating the Component of Electric Field Perpendicular to the Frame:
- E_perpendicular = E x cos(θ)
- E_perpendicular = 10√2 x cos(45 deg)
- E_perpendicular = 10 V/m

Calculating Electric Flux:
- Electric flux (Φ) = E_perpendicular x Area
- Φ = 10 x 0.0025
- Φ = 0.025 V m²
Therefore, the electric flux through the surface bounded by the square frame is 0.025 V m².
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(B) 3 (C) 2 (D) 1 51. A square frame of side 0.05 m is situated with its positive normal making an angle of 45 deg with a uniform electric field of 10√2 Vm¹. The flux of the electric field through the surface bounded by the square frame is sent lllar (D) 100 V m 2 3 4 (C) 10 V m (B) 25 mV m (A) 250 mV m A?
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