Question

The electric field component of a monochromatic radiation is given by
\(\vec{E} = 2E_0\,\hat{i}\,\cos kz\,\cos\omega t\)
Its magnetic field \(\vec{B}\) is then given by :

• A. \(\frac{2E_0}{c}\,\hat{j}\,\sin kz\,\cos\omega t\)
• B. \(-\frac{2E_0}{c}\,\hat{j}\,\sin kz\,\sin\omega t\)
• C. \(\frac{2E_0}{c}\,\hat{j}\,\sin kz\,\sin\omega t\)
• D. \(\frac{2E_0}{c}\,\hat{j}\,\cos kz\,\cos\omega t\)

Hint:

Use Faraday's law \(\nabla\times\vec E = -\partial\vec B/\partial t\) with \(E_x(z,t)\) and \(\omega = ck\).

Answer 1

The Correct Option is C

Only \(E_x = 2E_0\cos kz\cos\omega t\) is non-zero and it depends only on z, so \((\nabla\times\vec E)_y = \frac{\partial E_x}{\partial z} = -2E_0 k\sin kz\cos\omega t\).
Faraday's law: \(\frac{\partial B_y}{\partial t} = -(\nabla\times\vec E)_y = 2E_0 k\sin kz\cos\omega t\).
Integrate over time: \(B_y = \frac{2E_0 k}{\omega}\sin kz\sin\omega t = \frac{2E_0}{c}\sin kz\sin\omega t\) (using \(\omega = ck\), integration constant zero for a pure oscillation).
\(\vec B = \frac{2E_0}{c}\,\hat j\,\sin kz\sin\omega t\).
Answer: (3) \(\frac{2E_0}{c}\,\hat{j}\,\sin kz\,\sin\omega t\)