JEE Main 2019 — Physics solution

JEE Main2019Physics

If the magnetic field of a plane electromagnetic wave is given by (The speed of light = 3 × 10⁸/m/s) $B = 100 \times 1 0^{- 6} sin [2 π \times 2 \times 10^{15} (t - \frac{x}{c})]$ then the maximum electric field associated with it is :

A
4 × 10⁴ N/C
B
4.5 × 10⁴ N/C
C
6 × 10⁴ N/C
D
3 × 10⁴ N/C

Solution

$E_{m} = C B_{m} = 3 \times 1 0^{8} \times 100 \times 1 0^{- 6} = 3 \times 1 0^{4} N / C$

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