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

Concept:
For an electromagnetic wave propagating in free space:

Electric field \(\vec{E}\), magnetic field \(\vec{B}\), and direction of propagation are mutually perpendicular.
Magnitudes of fields are related by: \[ E = cB \] where \(c = 3 \times 10^{8}\,\text{m/s}\).

Step 1: Identify Direction of Propagation
Given: \[ \vec{E} = 54\sin(kz-\omega t)\,\hat{i} \] The phase \((kz-\omega t)\) indicates propagation along the \(\hat{k}\) (z-axis). Thus: \[ \vec{E} \perp \vec{B} \perp \text{direction of propagation} \] Hence, magnetic field must be along \(\hat{j}\)-direction.
Step 2: Calculate Magnetic Field Amplitude
\[ B = \frac{E}{c} = \frac{54}{3 \times 10^{8}} = 18 \times 10^{-8}\,\text{T} \]
Step 3: Write Magnetic Field Expression
\[ \vec{B} = 18 \times 10^{-8}\sin(kz-\omega t)\,\hat{j} \] \[ \boxed{\vec{B} = 18 \times 10^{-8}\sin(kz-\omega t)\,\hat{j}} \]