If the magnetic field of a plane electromagnetic wave is given by (The speed of light = $\left.3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) \mathrm{B}=100 \times 10^{-6} \sin \left[2 \pi \times 2 \times 10^{15}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)\right]$ then the maximum electric field associated with it is:

(B) $3 \times 10^{4} \mathrm{~N} / \mathrm{C}$

(A) $6 \times 10^{4} \mathrm{~N} / \mathrm{C}$

(C) $4 \times 10^{4} \mathrm{~N} / \mathrm{C}$

(D) $4.5 \times 10^{4} \mathrm{~N} / \mathrm{C}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · EM Waves

If the magnetic field of a plane electromagnetic wave is given by (The speed of light =

3 \times 1 0^{8} m / s) B = 100 \times 1 0^{- 6} sin [2 π \times 2 \times 1 0^{15} (t - \frac{x}{c})]

then the maximum electric field associated with it is:

(A) $6 \times 1 0^{4} N / C$
(B) $3 \times 1 0^{4} N / C$ ✓ Correct
(C) $4 \times 1 0^{4} N / C$
(D) $4.5 \times 1 0^{4} N / C$

Answer: (B)

Solution

In a plane electromagnetic wave the field amplitudes are locked together by

E_{0} = c B_{0}

. Reading the magnetic amplitude straight off the given expression,

B_{0} = 100 \times 1 0^{- 6}

T, so

E_{0} = (3 \times 1 0^{8}) (100 \times 1 0^{- 6}) = 3 \times 1 0^{4}

N/C. The huge factor

c

is why, in SI units, the electric amplitude dwarfs the magnetic one even though the two carry equal energy.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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