Question

Electric field of an EM wave is given as \(\vec{E} = 54\sin(kz-\omega t)\,\hat{i}\). Then what will be its corresponding magnetic field?

• A. \(18 \times 10^{-8}\sin(kz-\omega t)\,\hat{j}\)
• B. \(162 \times 10^{-8}\sin(kz-\omega t)\,\hat{j}\)
• C. \(18 \times 10^{-8}\sin(kz-\omega t)\,\hat{i}\)
• D. \(54 \times 10^{-8}\sin(kz-\omega t)\,\hat{i}\)

Hint:

In EM waves, always remember: \(\vec{E} \perp \vec{B}\), and \(E = cB\). Direction of propagation is given by \(\vec{E} \times \vec{B}\).

Answer 1

The Correct Option is A

The given problem involves determining the magnetic field of an electromagnetic wave when its electric field is known. The electric field of the EM wave is represented as:

\(\vec{E} = 54\sin(kz-\omega t)\,\hat{i}\)

To find the corresponding magnetic field, we use the relationship between the electric field \((\vec{E})\) and the magnetic field \((\vec{B})\) in an electromagnetic wave:

\(\frac{E}{B} = c\)

Here, \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^{8}\,\text{m/s}\).

Re-arranging the formula gives:

\(B = \frac{E}{c}\)

Given:

\(E = 54\,\sin(kz-\omega t)\,\hat{i}\)

Substitute the value of \(E\) and \(c\) into the equation:

\(B = \frac{54\,\sin(kz-\omega t)}{3 \times 10^{8}}\,\hat{j}\)

Simplifying gives:

\(B = 18 \times 10^{-8}\,\sin(kz-\omega t)\,\hat{j}\)

The direction of the magnetic field is perpendicular to both the electric field and the direction of wave propagation (usually in the \(\hat{z}\) direction). Given that the electric field is in the \(\hat{i}\) direction, the magnetic field will be in the \(\hat{j}\) direction based on the right-hand rule for cross-products.

Hence, the correct magnetic field is:

Correct Answer: \(18 \times 10^{-8}\sin(kz-\omega t)\,\hat{j}\)