The electric vector of an electromagnetic wave in vacuum is represented by E→=6.3 j^ V /m. The frequency of the wave is 20 MHz and it is propagating along the positive z-direction. At this point magnetic vector B→ is :

v^=E^×B^k^=j^×-i^ So direction of B^ is (−i^)EB=C⇒B=EC=6.33×108=2.1×10−8TB→=BB^=2.1×10−8(−i^)T

The electric vector of an electromagnetic wave in vacuum is represented by E→=6.3 j^ V /m. The frequency of the wave is 20 MHz and it is propagating along the positive z-direction. At this point magnetic vector B→ is :

v^=E^×B^k^=j^×-i^ So direction of B^ is (−i^)EB=C⇒B=EC=6.33×108=2.1×10−8TB→=BB^=2.1×10−8(−i^)T

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The electric vector of an electromagnetic wave in vacuum is represented by $\vec{E} = 6.3 \hat{j}$ V /m. The frequency of the wave is 20 MHz and it is propagating along the positive z-direction. At this point magnetic vector $\vec{B}$ is :

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- 2.1 x 10^-8 $\hat{i}$ T

+ 2.1 x 10^-8 $\hat{i}$ T

+ 4.0 x 10^-8 $\hat{i}$ T

- 18.9 x 10⁸ $\hat{i}$ T

answer is A.

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$\begin{aligned} \hat{v} = \hat{E} \times \hat{B} \\ \hat{k} = \hat{j} \times (- \hat{i}) \end{aligned}$

$So direction of \hat{B} is (- \hat{i})$

$\begin{aligned} \frac{E}{B} = C \Rightarrow B = \frac{E}{C} = \frac{6.3}{3 \times 10^{8}} \\ = 2.1 \times 10^{- 8} T \end{aligned}$

$\vec{B} = B \hat{B} = 2.1 \times 10^{- 8} (- \hat{i}) T$

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The electric vector of an electromagnetic wave in vacuum is represented by E→=6.3 j^ V /m. The frequency of the wave is 20 MHz and it is propagating along the positive z-direction. At this point magnetic vector B→ is :

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The electric vector of an electromagnetic wave in vacuum is represented by $\vec{E} = 6.3 \hat{j}$ V /m. The frequency of the wave is 20 MHz and it is propagating along the positive z-direction. At this point magnetic vector $\vec{B}$ is :