Step 1: Identify the given magnetic field wave equation.
The magnetic field is: \[ \vec{B} = (3 \times 10^{-7}\,\text{T})\sin(3 \times 10^4 x + 9 \times 10^{12} t)\,\hat{j} \] The form $\sin(kx + \omega t)$ represents a wave travelling in the negative $x$-direction.
Step 2: Find the amplitude of the electric field.
For an electromagnetic wave in free space: \[ E_0 = c B_0 \] where $c = 3 \times 10^8\,\text{m/s}$. Given $B_0 = 3 \times 10^{-7}\,\text{T}$: \[ E_0 = 3 \times 10^8 \times 3 \times 10^{-7} = 90\,\text{V m}^{-1} \]
Step 3: Determine the direction of wave propagation.
Since the argument is $(kx + \omega t)$ with positive signs for both $x$ and $t$, the wave propagates in the $-x$ direction. So the unit vector of propagation is $\hat{k}_{\text{prop}} = -\hat{i}$.
Step 4: Use the relationship between $\vec{E}$, $\vec{B}$, and the direction of propagation.
For an EM wave: $\hat{k}_{\text{prop}} = \dfrac{\vec{E} \times \vec{B}}{|\vec{E} \times \vec{B}|}$. Given $\vec{B}$ is along $\hat{j}$ and propagation is along $-\hat{i}$: \[ \vec{E} \times \hat{j} \propto -\hat{i} \] If $\vec{E}$ is along $\hat{k}$: $\hat{k} \times \hat{j} = -\hat{i}$ (since $\hat{k} \times \hat{j} = -\hat{i}$). This gives propagation in the $-\hat{i}$ direction.
Step 5: Write the complete electric field expression.
The electric field has the same phase as $\vec{B}$, amplitude $90\,\text{V m}^{-1}$, and direction $\hat{k}$: \[ \vec{E} = 90\sin(3 \times 10^4 x + 9 \times 10^{12} t)\,\hat{k}\,\text{V m}^{-1} \]
Step 6: State the final answer.
The electric field of the electromagnetic wave is: \[ \boxed{\vec{E} = 90\sin(3 \times 10^4 x + 9 \times 10^{12}\,t)\,\hat{k}\;\text{V m}^{-1}} \]