Physics · 2022
JEE Main · 30 June 2022, Shift 1 · Q44
An expression for oscillating electric field in a plane electromagnetic wave is given as E_z=300 sin (5 π × 10^3 x-3 π × 10^11 t ) Vm^-1 Then, the…
An expression for oscillating electric field in a plane electromagnetic wave is given as
$$\mathrm{E}_z=300 \sin \left(5 \pi \times 10^3 x-3 \pi \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}
$$
Then, the value of magnetic field amplitude will be :
(Given : speed of light in Vacuum $\displaystyle \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Official answer
(2)
$\displaystyle 5 \times 10^{-6} \mathrm{~T}$
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JEE Main 2022 Physics question, with the answer from NTA’s final answer key. Where our answers come from.