Answer:
The correct solution is "11.51 mA".
Explanation:
Given:
Time average power,
[tex]P_{avg}=0.05 \ W/m^2[/tex]
n = 377
As we now,
⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]
or,
⇒ [tex]E_0^2=0.05\times 377[/tex]
⇒ [tex]=4.341 \ V[/tex]
hence,
⇒ [tex]H_0=\frac{E_0}{n}[/tex]
By putting the values, we get
[tex]=\frac{4.341}{377}[/tex]
[tex]=11.51 \ mA[/tex]
