Question

What is the average value of the magnitude of the Poynting vector S at 1 meterfrom a 100-watt lightbulb radiating in all directions?

Answer 1

Answer:

Answer

B=

c

E

⟹∣

p

∣=

μ

0

EB

=

cμ

0

E

2

=

3×10

8

×4π×10

−7

10

4

=26.5Wm

−2

Answer 2

Answer:-

The value is S = 7.96S= 7.96

The value is S = 7.96S= 7.96

Explanation:-

From the question we are told that

The power is P = 100 \ WP= 100 W

The radius is r = 1 \ mr= 1 m

Generally the average value of the magnitude of the Poynting vector is mathematically represented

S = \frac{P}{4 \pi r^2}S =

S = \frac{P}{4 \pi r^2}S = 4πr

S = \frac{P}{4 \pi r^2}S = 4πr 2

S = \frac{P}{4 \pi r^2}S = 4πr 2P

S = \frac{ 100 }{ 4 *3.142 *1^2 }S=

S = \frac{ 100 }{ 4 *3.142 *1^2 }S= 4∗3.142∗1

S = \frac{ 100 }{ 4 *3.142 *1^2 }S= 4∗3.142∗1 2100

S = \frac{ 100 }{ 4 *3.142 *1^2 }S= 4∗3.142∗1 2100S = 7.96S= 7.96