A) 5.48 V/m
B) 7.75 V/m
C) 1.73 V/m
D) 2.45 V/mions /
Correct Answer: D
Solution :
[d] \[I=\frac{E_{0}^{2}}{2c{{\mu }_{0}}}\] |
\[\frac{P}{4\pi {{r}^{2}}}=\frac{E_{0}^{2}}{2c{{\mu }_{0}}}\] |
\[E=\frac{P2c{{\mu }_{0}}}{4\pi {{r}^{2}}}\] |
\[E=\frac{0.1\times 2\times 3\times {{10}^{-8}}\times 4\pi \times {{N}^{-1}}}{4\pi }\] |
\[=\sqrt{6}=2.45\] |
You need to login to perform this action.
You will be redirected in
3 sec