A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P , placed at a distance of 2 m . The efficiency of the bulb is $1.25 \%$. The value of peak electric field at $P$ is $x \times 10^{-1} \mathrm{~V} / \mathrm{m}$. Value of $x$ is --------. (Rounded - off to the nearest integer) [ Take $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ ]

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · EM Waves

A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P , placed at a distance of 2 m . The efficiency of the bulb is

1.25%

. The value of peak electric field at

P

x \times 1 0^{- 1} V / m

. Value of

x

is --------. (Rounded - off to the nearest integer) [ Take

ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}, c = 3 \times 1 0^{8} ms^{- 1}

]

Answer: 137

Solution

Only

1.25%

of the bulb's power leaves as radiation, so the radiated power is

1000 \times 0.0125 = 12.5

W. Spread uniformly over a sphere of radius

2

m, the intensity at that distance is

I = \frac{12.5}{4 π ( 2 ) ^{2}} = \frac{12.5}{16 π} = 0.2487 W m^{- 2}

For an electromagnetic wave the intensity is

I = \frac{1}{2} ε_{0} c E_{0}^{2}

, so

E_{0} = \frac{2 I}{ε _{0} c} = \frac{2 \times 0.2487}{8.85 \times 1 0 ^{- 12} \times 3 \times 1 0 ^{8}} = 13.69 V/m

Written as

x \times 1 0^{- 1}

V/m this peak field is

136.9 \times 1 0^{- 1}

V/m, so to the nearest integer

x = 137

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · EM Waves

Solution

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