JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 22)
The electric field in a plane electromagnetic wave is given by
$$\overrightarrow E = 200\cos \left[ {\left( {{{0.5 \times {{10}^3}} \over m}} \right)x - \left( {1.5 \times {{10}^{11}}{{rad} \over s} \times t} \right)} \right]{V \over m}\widehat j$$. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is $${x \over {{{10}^9}}}{N \over {{m^2}}}$$. Find the value of x .
$$\overrightarrow E = 200\cos \left[ {\left( {{{0.5 \times {{10}^3}} \over m}} \right)x - \left( {1.5 \times {{10}^{11}}{{rad} \over s} \times t} \right)} \right]{V \over m}\widehat j$$. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is $${x \over {{{10}^9}}}{N \over {{m^2}}}$$. Find the value of x .
Answer
354
Explanation
E0 = 200
$$I = {1 \over 2}{\varepsilon _0}E_0^2.C$$
Radiation pressure
$$P = {{2I} \over C}$$
$$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$$
$$ = {\varepsilon _0}E_0^2$$
$$ = 8.85 \times {10^{ - 12}} \times {200^2}$$
$$ = 8.85 \times {10^{ - 8}} \times 4$$
$$ = {{354} \over {{{10}^9}}}$$
$$I = {1 \over 2}{\varepsilon _0}E_0^2.C$$
Radiation pressure
$$P = {{2I} \over C}$$
$$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$$
$$ = {\varepsilon _0}E_0^2$$
$$ = 8.85 \times {10^{ - 12}} \times {200^2}$$
$$ = 8.85 \times {10^{ - 8}} \times 4$$
$$ = {{354} \over {{{10}^9}}}$$
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