JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 22)
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :
$${U_E} = {{{U_B}} \over 2}$$
$${U_E} > {U_B}$$
$${U_E} < {U_B}$$
$${U_E} = {U_B}$$
Explanation
Energy density of magnetic field (UB) = $${{{B^2}} \over {2{\mu _0}}}$$
Also,
Energy density of electric field
UE = $${1 \over 2}$$ $$\varepsilon $$0E2
= $${1 \over 2}$$ $$\varepsilon $$0 B2 C2 [as $${E \over B}$$ = C ]
= $${1 \over 2}$$ $$\varepsilon $$0 B2 $$ \times $$ $$\left( {{1 \over {{\varepsilon _0}{\mu _0}}}} \right)$$ [as C2 = $${{1 \over {{\varepsilon _0}{\mu _0}}}}$$]
= $${{{B^2}} \over {2{\mu _0}}}$$
$$ \therefore $$ UB = UE
Also,
Energy density of electric field
UE = $${1 \over 2}$$ $$\varepsilon $$0E2
= $${1 \over 2}$$ $$\varepsilon $$0 B2 C2 [as $${E \over B}$$ = C ]
= $${1 \over 2}$$ $$\varepsilon $$0 B2 $$ \times $$ $$\left( {{1 \over {{\varepsilon _0}{\mu _0}}}} \right)$$ [as C2 = $${{1 \over {{\varepsilon _0}{\mu _0}}}}$$]
= $${{{B^2}} \over {2{\mu _0}}}$$
$$ \therefore $$ UB = UE
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