Question

a rectangular surface of sides 40cm x 20 cm is placed in an uniform field of strength 2x10⁴ NC-1 such that it i normal to the field.Calculate the electric flux linked with the frame.Also calculate the flux linked with tha frame when it is converted into i) a circular frame ii) a square​

Answer 1

Answer:

1600, 2292, 1800 Nm²/C

Explanation:

Rectangle:

E=2×10⁴N/C

length=40cm=0.4m

Breadth=20cm=0.2m

Surface area, S = 0.4×0.2=0.08m²

Electric flux = E×S=0.08×2×10⁴=1600Nm²/C

Cicular frame:

Circumference = sum of sides of rectangle = 2×0.4+2×0.2 = 1.2m

2πr=1.2

r=0.6/π

Area, S = πr² = 11.46m²

Electric flux = E×S = 2×10⁴ × 11.46 = 2292 Nm²/C

Square:

Perimeter = sum of sides of rectangle = 2×0.4+2×0.2 = 1.2m

Side of square = 1.2÷4= 0.3m

Area, S = (side)² = 0.3² = 0.09m²

Electric flux = E×S = 2×10⁴ ×0.09 = 1800Nm²/C