JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 27)
A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x $$\times$$ 10$$-$$1 V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take $${\varepsilon _0} = 8.85 \times {10^{ - 12}}$$ C2N$$-$$1 m$$-$$2, c = $$3 \times {10^8}$$ ms$$-$$1]
Answer
137
Explanation
Intensity of electro magnetic wave is,
$$I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}$$
$${1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}$$
$${1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}$$
$$E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}$$
$$E_0^2 = {{375} \over 2} = 187.5$$
$${E_0} = 13.69$$
$${E_0} \approx 137 \times {10^{ - 1}}$$ v/m
$$I = {1 \over 2}c{\varepsilon _0}E_0^2 = {P \over {4\pi {r^2}}}$$
$${1 \over 2}4\pi {\varepsilon _0} \times c \times E_0^2 = {P \over {{r^2}}}$$
$${1 \over 2} \times {{3 \times {{10}^5} \times E_0^2} \over {9 \times {{10}^9}}} = {{1000 \times 1.25} \over {{{(2)}^2}}} \times {1 \over {100}}$$
$$E_0^2 = {{60 \times 1000 \times 1.25} \over {4 \times 100}} = {{125 \times 3} \over 2}$$
$$E_0^2 = {{375} \over 2} = 187.5$$
$${E_0} = 13.69$$
$${E_0} \approx 137 \times {10^{ - 1}}$$ v/m
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