JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 6)
Intensity of sunlight is observed as 0.092 Wm$$-$$2 at a point in free space. What will be the peak value of magnetic field at the point?
($${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$$)
($${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$$)
2.77 $$\times$$ 10$$-$$8 T
1.96 $$\times$$ 10$$-$$8 T
8.31 T
5.88 T
Explanation
$${I \over C} = {1 \over 2}{\varepsilon _0}.E_0^2$$
$$ \Rightarrow {E_0} = \sqrt {{{2I} \over {C{\varepsilon _0}}}} $$
$${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C}$$
$$ \Rightarrow {B_0} = \sqrt {{{2I} \over {{\varepsilon _0}{C^3}}}} = \sqrt {{{2 \times 0.092} \over {8.85 \times {{10}^{ - 12}} \times 27 \times {{10}^{ + 24}}}}} $$
$$ = 2.77 \times {10^{ - 8}}$$ T
$$ \Rightarrow {E_0} = \sqrt {{{2I} \over {C{\varepsilon _0}}}} $$
$${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C}$$
$$ \Rightarrow {B_0} = \sqrt {{{2I} \over {{\varepsilon _0}{C^3}}}} = \sqrt {{{2 \times 0.092} \over {8.85 \times {{10}^{ - 12}} \times 27 \times {{10}^{ + 24}}}}} $$
$$ = 2.77 \times {10^{ - 8}}$$ T
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