A plane electromagnetic wave of frequency 35MHz travels in free...

At every instant, the ratio of the magnitude of the electric field to the magnitude of the magnetic field in an electromagnetic wave equals the speed of light.

Therefore, $\dfrac{E}{B} = c$

$⇒ \dfrac{E}{B} = 3 \times {10}^{8}$

$⇒ B = \dfrac{E}{3 \times {10}^{8}} = \dfrac{9 . 6}{3 \times {10}^{8}}$

$⇒ B = 3 . 2 \times {10}^{- 8}T$

Now, direction of magnetic field can be calculated as

\overset{^}{B} = \overset{^}{v} \times \overset{^}{E}

= \overset{^}{i} \times \overset{^}{j} = \overset{^}{k}

So, \overset{\to}{B} = 3 . 2 \times {10}^{- 8}\overset{^}{k}T