For a plane eletromagnetic wave, the electric field is given by {E^{\to}}=90 \sin (0.5 \\times10^{3} {x}+1.5 \\times10^{11} {t}) {k} \.{v} \/ {m}. The corresponding magnetic field \overrightarrow{{B}} will be

Correct Answer is : →B=3×10−7sin(0.5×103x+1.5 ×1011t)^jT

\overrightarrow{B}=3 \\times10^{-7} \sin (0.5 \\times10^{3} x+1.5 \times10^{11} t) \stackrel{\^}{i} {T}

\overrightarrow{B}=3 \\times10^{-7} \sin (0.5 \\times10^{3} x+1.5 \times10^{11} t) \stackrel{\^}{j} {T}

\overrightarrow{B}=27 \\times10^{9} \sin (0.5 \\times10^{3} x+1.5 \times10^{11} t) \stackrel{\^}{j} {T}

\overrightarrow{B}=3 \\times10^{-7} \sin (0.5 \\times10^{3} x+1.5 \times10^{11} t) \stackrel{\^}{k} {T}

WBJEE 2021 Physics and Chemistry Solved Paper

Category II (Q31 to Q 35)

Carry 2 marks each and only one option is correct. In case of incorrect answer or any combination of more than one answer, ½ mark will be deducted

Question : 35 of 80

Marks: +1, -0

For a plane eletromagnetic wave, the electric field is given by

→

=90‌sin(0.5×103x+1.5×1011t) k.v∕m. The corresponding magnetic field

→

will be

Go to Question:

Prev Question

Next Question