By measuring the heating of a material as it absorbs light from the sun, one finds that the intensity of sunlight at the surface of the earth is 1300 Wm -2. What will be magnetic field of sunlight?

Question

By measuring the heating of a material as it absorbs light from the sun, one finds that the intensity of sunlight at the surface of the earth is 1300 Wm -2. What will be magnetic field of sunlight? (A) 990 T (B) 3 × 10-6 T (C) 3.3 × 10-6 T (D) 9.9 × 10-6 T

Tardigrade Admin · Accepted Answer

Here, I=1300 Wm -2 As, I=(1/2) ε0 c E02 As, we know, electric field i.e., E0=√(2 I/ε0 c) =√(2(1300 Wm -2)/(8.85 × 10-12 C 2 Nm -2)(3 × 108 ms -1)) or E0=990 NC -1 So, magnetic field of sunlight i . e., B0=(E0/c)=(990 NC -1/3 × 108 ms -1)=3.3 × 10-6 T

Q. By measuring the heating of a material as it absorbs light from the sun, one finds that the intensity of sunlight at the surface of the earth is $1300 W m^{- 2}$ . What will be magnetic field of sunlight?

$990 T$

$3 \times 1 0^{- 6} T$

$3.3 \times 1 0^{- 6} T$

$9.9 \times 1 0^{- 6} T$

Q. By measuring the heating of a material as it absorbs light from the sun, one finds that the intensity of sunlight at the surface of the earth is 1300Wm−2. What will be magnetic field of sunlight?

990T

3×10−6T

3.3×10−6T

9.9×10−6T

Here, I=1300Wm−2 As, I=21​ε0​cE02​ As, we know, electric field i.e., E0​=ε0​c2I​​ =(8.85×10−12C2Nm−2)(3×108ms−1)2(1300Wm−2)​​ or E0​=990NC−1 So, magnetic field of sunlight i.e., B0​=cE0​​=3×108ms−1990NC−1​=3.3×10−6T

Q. By measuring the heating of a material as it absorbs light from the sun, one finds that the intensity of sunlight at the surface of the earth is $1300 W m^{- 2}$ . What will be magnetic field of sunlight?

$990 T$

$3 \times 1 0^{- 6} T$

$3.3 \times 1 0^{- 6} T$

$9.9 \times 1 0^{- 6} T$