JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 18)
A plane electromagnetic wave, has
frequency of 2.0 $$ \times $$ 1010 Hz and its energy density is 1.02 $$ \times $$ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to
( $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$$ and speed of light
= 3 $$ \times $$ 108 ms–1)
frequency of 2.0 $$ \times $$ 1010 Hz and its energy density is 1.02 $$ \times $$ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to
( $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$$ and speed of light
= 3 $$ \times $$ 108 ms–1)
190 nT
150 nT
160 nT
180 nT
Explanation
Energy density, $${{dU} \over {dV}} = {{B_0^2} \over {2{\mu _0}}}$$
$$ \Rightarrow $$ 1.02 $$ \times $$ 10–8 = $${{B_0^2} \over {2{\mu _0}}}$$
Also, c = $${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$
$$ \Rightarrow $$ $${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}}$$
$$ \therefore $$ $${B_0^2}$$ = 1.02 $$ \times $$ 10–8 $$ \times $$ 2 $$ \times $$ $${1 \over {{c^2}{\varepsilon _0}}}$$
= 1.02 $$ \times $$ 10–8 $$ \times $$ 2 $$ \times $$ $${{4\pi \times 9 \times {{10}^9}} \over {9 \times {{10}^{16}}}}$$
$$ \Rightarrow $$ B0 = 16 $$ \times $$ 10-8 T = 160 nT
$$ \Rightarrow $$ 1.02 $$ \times $$ 10–8 = $${{B_0^2} \over {2{\mu _0}}}$$
Also, c = $${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$
$$ \Rightarrow $$ $${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}}$$
$$ \therefore $$ $${B_0^2}$$ = 1.02 $$ \times $$ 10–8 $$ \times $$ 2 $$ \times $$ $${1 \over {{c^2}{\varepsilon _0}}}$$
= 1.02 $$ \times $$ 10–8 $$ \times $$ 2 $$ \times $$ $${{4\pi \times 9 \times {{10}^9}} \over {9 \times {{10}^{16}}}}$$
$$ \Rightarrow $$ B0 = 16 $$ \times $$ 10-8 T = 160 nT
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