The magnetic field of an electromagnetic wave is given by: \overrightarrow{B}=1.6 \times 10^{-6} \cos (2 \times 10^{7} z+6\times10^{15} t )(2 {i^{\^}}+ {j^{\^}}) {W b}/{m^{2}} The associated electric field will be: [8 Apr 2019 Shift 2]

Correct Answer is : →E=4.8×102cos(2×107z+6×1015t)(−^i+2^j)Vm

\overrightarrow{E}=4.8\times 10^{2} \cos (2 \times 10^{7} z+6 \times 10^{15} t )( {i^{\^}}-2 {j^{\^}}) \frac{V}{m}

\overrightarrow{E}=4.8\times 10^{2} \cos (2 \times 10^{7} z-6 \times 10^{15} t )( -2 {j^{\^}}+{i^{\^}}) \frac{V}{m}

\overrightarrow{E}=4.8\times 10^{2} \cos (2 \times 10^{7} z-6 \times 10^{15} t )( 2 {i^{\^}}+{j^{\^}}) \frac{V}{m}

\overrightarrow{E}=4.8\times 10^{2} \cos (2 \times 10^{7} z+6 \times 10^{15} t )( - {i^{\^}}+{2j^{\^}}) \frac{V}{m}

JEE Main 8 Apr 2019 Paper 2 Solved Paper

Section: Physics

Question : 29 of 90

Marks: +1, -0

The magnetic field of an electromagnetic wave is given by:

→

=1.6×10−6‌cos(2×107z+6×1015t)(2

)‌

The associated electric field will be:

[8 Apr 2019 Shift 2]

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