JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 4)
The energy density associated with electric field $$\vec{E}$$ and magnetic field $$\vec{B}$$ of an electromagnetic wave in free space is given by $$\left(\epsilon_{0}-\right.$$ permittivity of free space, $$\mu_{0}-$$ permeability of free space)
$$U_{E}=\frac{\epsilon_{0} E^{2}}{2}, U_{B}=\frac{B^{2}}{2 \mu_{0}}$$
$$U_{E}=\frac{E^{2}}{2 \epsilon_{0}}, U_{B}=\frac{\mu_{0} B^{2}}{2}$$
$$U_{E}=\frac{\epsilon_{0} E^{2}}{2}, U_{B}=\frac{\mu_{0} B^{2}}{2}$$
$$U_{E}=\frac{E^{2}}{2 \epsilon_{0}}, U_{B}=\frac{B^{2}}{2 \mu_{0}}$$
Explanation
The energy density associated with the electric field $$\vec{E}$$ and magnetic field $$\vec{B}$$ of an electromagnetic wave in free space is given by:
For the electric field: $$U_{E} = \frac{1}{2} \epsilon_{0} E^2$$
For the magnetic field: $$U_{B} = \frac{1}{2} \frac{B^2}{\mu_{0}}$$
These expressions match Option A:
$$U_{E} = \frac{\epsilon{0} E^{2}}{2}, U_{B} = \frac{B^{2}}{2 \mu{0}}$$
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