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

We know that the relationship between the magnetic field \( \vec{B} \) and electric field \( \vec{E} \) in an electromagnetic wave is given by: \[ \vec{E} = \vec{B} \times \hat{c} \] and \( \vec{E} = B_0 c \), where \( c \) is the speed of light. We are given: \[ \vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left( \omega \left( t - \frac{z}{c} \right) \right) \] We calculate \( \vec{E} \) using the cross product and the fact that \( \vec{E} = B_0 c \). \[ \vec{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30 c \sin \left( \omega \left( t - \frac{z}{c} \right) \right) \]