A point source of electromagnetic radiation has an average power output of 1500 W. The maximum value of electric field at a distance of 3 m from this source in Vm-1 is

Question

A point source of electromagnetic radiation has an average power output of 1500 W. The maximum value of electric field at a distance of 3 m from this source in Vm-1 is (A) 500 (B) 100 (C)  (500/3)  (D)  (250/3)

Tardigrade Admin · Accepted Answer

E0=√(μ 0cPav/2π r2)=√((4π × 10-7)× (3× 108)× 1500/2π × (4)2) =100Vm-1

Q. A point source of electromagnetic radiation has an average power output of $1500 W$ . The maximum value of electric field at a distance of $3 m$ from this source in $V m^{- 1}$ is

500

100

$\frac{500}{3}$

$\frac{250}{3}$

$105$

$E_{0} = \frac{μ _{0} c P _{a v}}{2 π r ^{2}} = \frac{( 4 π \times 10 ^{- 7} ) \times ( 3 \times 10 ^{8} ) \times 1500}{2 π \times ( 4 ) ^{2}}$ $= 100 V m^{- 1}$

Q. A point source of electromagnetic radiation has an average power output of 1500W. The maximum value of electric field at a distance of 3m from this source in Vm−1 is

500

100

3500​

3250​

105​

E0​=2πr2μ0​cPav​​​=2π×(4)2(4π×10−7)×(3×108)×1500​​ =100Vm−1

Q. A point source of electromagnetic radiation has an average power output of $1500 W$ . The maximum value of electric field at a distance of $3 m$ from this source in $V m^{- 1}$ is

$\frac{500}{3}$

$\frac{250}{3}$

$105$

$E_{0} = \frac{μ _{0} c P _{a v}}{2 π r ^{2}} = \frac{( 4 π \times 10 ^{- 7} ) \times ( 3 \times 10 ^{8} ) \times 1500}{2 π \times ( 4 ) ^{2}}$ $= 100 V m^{- 1}$