Question:hard

The magnetic field of an electromagnetic wave is given by : $\vec{B} =1.6 \times10^{-6} \cos\left(2\times10^{7} z +6 \times10^{15}t\right)\left(2\hat{i} +\hat{j}\right) \frac{Wb}{m^{2}} $ The associated electric field will be :-

Updated On: Apr 3, 2026

$\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z + 6\times10^{15}t\right)\left(\hat{i} -2\hat{j}\right) \frac{V}{m} $
$\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z - 6\times10^{15}t\right)\left(2\hat{i} + \hat{j}\right) \frac{V}{m} $
$\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z - 6\times10^{15}t\right)\left( - 2\hat{i} + \hat{j}\right) \frac{V}{m} $
$\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z + 6\times10^{15}t\right)\left( -1 \hat{i} + 2\hat{j}\right) \frac{V}{m} $

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The Correct Option is A

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Questions Asked in JEE Main exam

Six point charges are kept $60^\circ$ apart from each other on the circumference of a circle of radius $ R $ as shown in figure. The net electric field at the center of the circle is ___________. ($ \varepsilon_0 $ is permittivity of free space)