1. Given electric field

   $\vec E = \hat i\,40\cos\omega\left(t-\frac{z}{c}\right)\,\text{N C}^{-1}$

   So:

   • the electric field is along the $x$-axis ($\hat i$ direction),
   • the phase is $\omega\left(t-\frac{z}{c}\right)$, which represents a wave traveling in the $+z$ direction.

2. Direction of magnetic field

   For an electromagnetic wave,

   $\vec E \perp \vec B \perp \text{direction of propagation}$

   and

   $\vec E \times \vec B$

   gives the direction of propagation.

   Here propagation is along $+\hat k$, and $\vec E$ is along $\hat i$.

   So we need:

   $\hat i \times \vec B = \hat k$

   Since

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

   therefore $\vec B$ must be along $\hat j$.

3. Magnitude relation between $E$ and $B$

   In free space,

   $E = cB$

   Hence,

   $B = \frac{E}{c} = \frac{40}{c}\cos\omega\left(t-\frac{z}{c}\right)$

4. Write the full magnetic field

   Therefore,

   $\vec B = \hat j\,\frac{40}{c}\cos\omega\left(t-\frac{z}{c}\right)$

5. Check options

   • A: $\vec B=\hat j\frac{40}{c}\cos\omega\left(t-\frac{z}{c}\right)$ ✅ Correct
   • B: Along $\hat i$, same as $\vec E$ ❌
   • C: Magnitude should be $E/c$, not $E$ ❌
   • D: Along $\hat k$, but $\vec B$ must be perpendicular to propagation direction ❌

Final Answer

The correct option is:

$\boxed{A}$