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:
Show Hint
For an electromagnetic wave, the electric and magnetic fields are perpendicular and related through the speed of light.
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: \[ \mathbf{E} = \left( \frac{1}{2} \hat{i} - \frac{\sqrt{3}}{2} \hat{j} \right) 30c \sin \left[ \omega \left( t - \frac{z}{c} \right) \right] \]
