JEE Physics 2024 Question with Solution

Given: The electric field is $\vec{E} = E_0 \cos(\omega t - kx) \, \hat{i}$.

Find: The corresponding magnetic induction vector $\vec{B}$.

For an electromagnetic wave in free space:

• $\vec{E}$, $\vec{B}$, and the direction of propagation are mutually perpendicular.
• The direction of propagation is given by $\vec{E} \times \vec{B}$.
• Their magnitudes satisfy

$$E = C B$$

so

$$B = \frac{E}{C}$$

• Both fields are in the same phase.

From the phase term $\omega t - kx$, the wave propagates along the positive $x$-direction. Since $\vec{E}$ is along $\hat{i}$, the magnetic field must be perpendicular to both the propagation direction and $\vec{E}$, and its magnitude must be $\frac{E_0}{C}$ with the same phase.

Thus,

$$\vec{B} = \frac{E_0}{C} \cos(\omega t - kx) \, \hat{j}$$

Therefore, the correct option is B.

Given: $\vec{E} = E_0 \cos(\omega t - kx) \, \hat{i}$

Find: $\vec{B}$

Use the direct relation for an electromagnetic wave in free space:

$$B_0 = \frac{E_0}{C}$$

and $\vec{E}$ and $\vec{B}$ are in the same phase.

So the magnetic field must have the same cosine factor and amplitude $\frac{E_0}{C}$. Hence,

$$\vec{B} = \frac{E_0}{C} \cos(\omega t - kx) \, \hat{j}$$

Therefore, the correct option is B.