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] \[I=\frac{E_{0}^{2}}{2c{{\mu }_{0}}}\] |
| \[\frac{P}{4\pi {{r}^{2}}}=\frac{E_{0}^{2}}{2c{{\mu }_{0}}}\] |
| \[E=\frac{P2c{{\mu }_{0}}}{4\pi {{r}^{2}}}\] |
| \[E=\frac{0.1\times 2\times 3\times {{10}^{-8}}\times 4\pi \times {{N}^{-1}}}{4\pi }\] |
| \[=\sqrt{6}=2.45\] |
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