Question

The magnetic field of an E.M. wave is given by: \[ \vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left( \omega \left( t - \frac{z}{c} \right) \right) \] The corresponding electric field in S.I. units is:

• A. \( \vec{E} = \left( \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t + \frac{z}{c} \right) \right) \)
• B. \( \vec{E} = \left( \frac{3}{4} \hat{i} + \frac{1}{4} \hat{j} \right) 30 c \cos \left( \omega \left( t - \frac{z}{c} \right) \right) \)
• C. \( \vec{E} = \left( \frac{\sqrt{3}}{2} \hat{i} - \frac{1}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t + \frac{z}{c} \right) \right) \)
• D. \( \vec{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t - \frac{z}{c} \right) \right) \)

Hint:

For an electromagnetic wave, the electric and magnetic fields are perpendicular and related through the speed of light.

Answer 1

The Correct Option is D

This problem involves electromagnetic fields, where we are given a magnetic field \( \mathbf{B} \) and need to calculate the electric field \( \mathbf{E} \) and other related quantities. Let's break it down step-by-step.

Step 1: Given Magnetic Field 

The magnetic field is given by: \[ \mathbf{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \frac{z}{c} \right) \right] \] where \( \hat{i} \) and \( \hat{j} \) are the unit vectors along the x-axis and y-axis, respectively, and \( \omega \) is the angular frequency, \( t \) is time, and \( z \) is the position.

Step 2: Electric Field Formula

The electric field \( \mathbf{E} \) is related to the magnetic field \( \mathbf{B} \) and the direction of wave propagation \( \mathbf{c} \) by the following equation: \[ \mathbf{E} = \mathbf{B} \times \mathbf{c}, \quad \mathbf{E} = B_0 c \] where \( B_0 \) is the magnitude of the magnetic field.

Step 3: Calculating \( \mathbf{E} \)

To find the electric field, we take the cross product of \( \mathbf{B} \) and \( \mathbf{c} \). We get: \[ \mathbf{E} = \left( \frac{\sqrt{3}}{2} \hat{i} - \hat{j} \right) + \frac{1}{2} \hat{i} \]

Step 4: Evaluating \( E_0 \)

Now, we can evaluate \( E_0 \), the electric field at \( t = 0 \). We have: \[ E_0 = 30c \] This gives the value of the electric field at \( t = 0 \).

Step 5: Final Expression for \( \mathbf{E} \)

The electric field \( \mathbf{E} \) can be written as: \[ \mathbf{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30c \sin \left[ \omega \left( t - \frac{z}{c} \right) \right] \]

Final Answer:

\[ \mathbf{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30c \sin \left[ \omega \left( t - \frac{z}{c} \right) \right] \]