1. Given magnetic field

$\vec B = 3\times 10^{-8}\sin\left[200\pi (y+ct)\right]\hat i\;\text{T}$

So,

• magnetic field amplitude is $B_0 = 3\times 10^{-8}\,\text{T}$
• direction of $\vec B$ is along $\hat i$.

2. Direction of propagation

The phase is

$200\pi (y+ct)$

A wave of the form $\sin(k y - \omega t)$ travels in $+y$ direction, while

$\sin(k y + \omega t)$

travels in the $-y$ direction.

Hence this electromagnetic wave propagates along

$-\hat j$

3. Relation between $\vec E$, $\vec B$ and direction of propagation

For an electromagnetic wave,

$\vec E \times \vec B$

gives the direction of propagation.

Here propagation is along $-\hat j$, and

$\vec B \parallel \hat i$

Let $\vec E$ be along $\pm \hat k$.

Now,

$\hat k \times \hat i = \hat j$

So to get $-\hat j$, we need

$(-\hat k)\times \hat i = -\hat j$

Therefore,

$\vec E \text{ is along } -\hat k$

4. Magnitude relation

For electromagnetic waves in free space,

$E_0 = c B_0$

So,

$E_0 = (3\times 10^8)(3\times 10^{-8}) = 9\,\text{V/m}$

5. Write the electric field

Since $\vec E$ has the same phase as $\vec B$ and points along $-\hat k$,

$\vec E = -9\sin\left[200\pi (y+ct)\right]\hat k\;\text{V/m}$

6. Check options

• A: magnitude wrong
• B: correct magnitude, correct direction, correct phase
• C: wrong direction
• D: wrong magnitude

Thus the correct option is

$\boxed{\text{B}}$