The electric field of an electromagnetic wave in free space is \overrightarrow{E}=57 \cos [7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)] (4 \stackrel{\^}{i}-3 \stackrel{\^}{j}) {\N} \/ {\C}. The associated magnetic field in Tesla is [23 Jan 2025 Shift 1]

Correct Answer is : →B=573×108cos[7.5×106t−5×10−3(3x+4y)](5^k)

\overrightarrow{B}=-\frac{57}{3 \times 10^8} \cos [7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)] (\stackrel{\^}{k})

\overrightarrow{B}=\frac{57}{3 \times 10^8} \cos [7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)] (5 \stackrel{\^}{k})

\overrightarrow{B}=-\frac{57}{3 \times 10^8} \cos [7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)] (5 \stackrel{\^}{k})

\overrightarrow{B}=\frac{57}{3 \times 10^8} \cos [7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)] (\stackrel{\^}{k})

JEE Main 23 Jan 2025 Shift 1 Paper

Section: Physics

Question : 45 of 75

Marks: +1, -0

The electric field of an electromagnetic wave in free space is

→

=57‌cos[7.5×106t−5×10−3(3x+4y)]
(4

−3

)N∕C. The associated magnetic field in Tesla is

[23 Jan 2025 Shift 1]

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