An electromagnetic wave is traveling in a vacuum. The electric field E of the wa

Question

An electromagnetic wave is traveling in a vacuum. The electric field E of the wave is doubled. What happens to the magnetic field B, the total energy density u, and the intensity S of the wave?



B doubles, u doubles, S doubles

B doubles, u increases by a factor of four, S increases by a factor of four

B increases by a factor of four, u increases by a factor of four, S increases by a factor of four

B doubles, u increases by a factor of four, S doubles

B doubles, u decreases by a factor of two, S decreases by a factor of two

Explanation / Answer

we know manetic field B = E / c where c = speed of EM wave when E doubles then B doubles ---------------- total energy density U = eE ^ 2 where e = permitivity of free space So, U is increased to four times -------------------- intensity is proportional to E X H i.e., E X B when E doubles B also doubles.Therefore intensity becomes four times so, answer is B doubles, u increases by a factor of four, S increases by a factor of four

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