QJEEM14Phd758: JEE Main 2014 Physics - Class 12 JEEM14Phd758 Solution: During the propagation of electromagnetic waves in a medium :

Question

Verified Solution • 2025–26 Session moderate

During the propagation of electromagnetic waves in a medium :

Accepted Answer

E_{0} = C B_{0}

and

C = \frac{1}{\sqrt{μ_{0} ε_{0}}}

Electric energy density

= \frac{1}{2} ε_{0} {E_{0}}^{2} = μ_{E}

Magnetic energy density

= \frac{1}{2} \frac{{B_{0}}^{2}}{μ_{0}} = μ_{B}

Thus,

μ_{E} = μ_{B}

Energy is equally divided between electric and magnetic field

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