The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10m is (x/10)√(μ 0 C/π )(V/m) . The efficiency of the bulb is 10 % and it is a point source. The value of x is .

Question

The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10m is (x/10)√(μ 0 C/π )(V/m) . The efficiency of the bulb is 10 % and it is a point source. The value of x is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I=(1/2)C∈ 0E02 (8/4 π × 102)=(1/2)× c× (1/μ 0 C2)× E02 E0=(2/10)√(μ 0 C/π ) ⇒ x=2

Q. The peak electric field produced by the radiation coming from the $8 W$ bulb at a distance of $10 m$ is $\frac{x}{10} \frac{μ _{0} C}{π} \frac{V}{m}$ . The efficiency of the bulb is $10%$ and it is a point source. The value of $x$ is _____.

$I = \frac{1}{2} C \in_{0} E_{0}^{2}$
$\frac{8}{4 π \times 1 0 ^{2}} = \frac{1}{2} \times c \times \frac{1}{μ _{0} C ^{2}} \times E_{0}^{2}$
$E_{0} = \frac{2}{10} \frac{μ _{0} C}{π}$
$\Rightarrow x = 2$

Q. The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10m is 10x​πμ0​C​​mV​ . The efficiency of the bulb is 10% and it is a point source. The value of x is _____.

I=21​C∈0​E02​ 4π×1028​=21​×c×μ0​C21​×E02​ E0​=102​πμ0​C​​ ⇒x=2

Q. The peak electric field produced by the radiation coming from the $8 W$ bulb at a distance of $10 m$ is $\frac{x}{10} \frac{μ _{0} C}{π} \frac{V}{m}$ . The efficiency of the bulb is $10%$ and it is a point source. The value of $x$ is _____.

$I = \frac{1}{2} C \in_{0} E_{0}^{2}$
$\frac{8}{4 π \times 1 0 ^{2}} = \frac{1}{2} \times c \times \frac{1}{μ _{0} C ^{2}} \times E_{0}^{2}$
$E_{0} = \frac{2}{10} \frac{μ _{0} C}{π}$
$\Rightarrow x = 2$